Group Theory – Fall 2014

GROUP THEORY – FALL 2014

THOMAS CREUTZIG

ABSTRACT. These are lecture notes in progressfor Math 328 – Group Theory. We will focus on finite groups. The highlights of the course are Sylow’s Theory and the classification of finitely generated abelian groups. The lecture concludes with applications of group theory.

December 1, 2014.

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CONTENTS 1. Introduction 3 2. Basic Deﬁnitions and Properties 4 3. Subgroups 7 4. Cosets 9 5. Normal subgroups 11 6. Homomorphisms 12 7. Exercises and Vocabulary 15 8. The Symmetric Group 17 9. Isomorphism Theorems 22 10. Group Actions 26 11. Sylow’s theory 30 12. Solvable Groups 37 13. Abelian Groups 42 14. Applications 46 15. Practice Questions 51 References 53 MATH 328 3

1. INTRODUCTION Groups are a set endowed with an operation satisfying certain properties. They are in some sense the most fundamental and most important objects in algebra. This course is an an introduction to this topic. Prerequisites are a good knowledge of linear algebra. There are many excellent books on algebra, and they usually start with a few chapters on group theory. I will mainly follow the book by Jantzen, Schwermer [JS]. Other good references are the book by Artin [A] and by Goodman [G]. In my opinion [JS] is an excellent resource, unfortunately it is not available in English. The book by Goodman is available for free at http://www.math.uiowa.edu/ goodman/algebrabook.dir/algebrabook.html. The author asks for a (voluntary) donation to a charity organization. I will also use some results of online notes of Keith Conrad [Co] and Wolf Holzmann [H]. If you are interested in algebra and you plan to continue in this direction, then surely [A] is a good (but expensive) investment. However, this course will be entirely based on these lecture notes and I donot require you to buy any book. All exercises and exams will be based on material covered in these notes. Groups are fundamental in both nature and mathematics. They appear as automorphism group or symmetry group of a complicated systems. One very well-known example is quantum ﬁeld theory or gauge theory. These theories often have a compact Lie group as symmetry, meaning that the ﬁelds describing fundamental particles like electrons, quarks or gluons carry an action of some Lie group. For the electron this group is called U(1) and for quarks and gluons it is called SU(3). MaPh 451 treats questions in that direction [Cr]. More examples are crystals, molecular structures and other symmetric materials. Since this is an algebra course, let me start with a relevant example. An algebra example are roots of unity. Take ξ = exp(2πi/p) a p-th root of unity for some prime number p. Every p-th root of unity is of the form ξ n for some (non-unique) positive integer n. Multiplying two such p-th roots of unity, we get another one ξ n · ξ m = ξ n+m. There is one relation, namely that thep-th power of a p-th root of unity is one, (ξ n)p = 1. So that the set of p-th roots of unity is 1,ξ ,ξ 2,...,ξ p−1 . This set is closed under multiplication, it has a multiplica tive identity namely the number 1, and every element ξ n has an inverse ξ p−n. This means that multiplication endows the set of p-th roots of unity with a group structure, called the (additive) group Z/pZ. Question for you: Why do you think it is called additive? Now, consider the map n mn σm : ξ 7→ ξ for positive integer m not divisible by p. Since ξ is a p-th root of unity, we have n (m+p)n mn mp mn n σm+p(ξ ) = ξ = ξ ξ = ξ = σm(ξ ) we see that the maps σm and σm+p coincide, so that the set of such maps is represented by

σ1,σ2,...,σp−1 . 4 T CREUTZIG

If we compose two such maps, we get n nm rmn n σr ◦ σm : ξ 7→ σr(ξ ) = ξ = σrm(ξ ). So that we get the composition law of maps

σr ◦ σm = σrm and we remember that we compute ”modulo p” meaning that we have the relations σm+p = σm. The set of these maps forms another group, called the group of multiplicative units of Z/pZ and denoted by (Z/pZ)∗. Groups of this type play a central role in cyclotomic ﬁelds and ﬁeld extensions.

2. BASIC DEFINITIONS AND PROPERTIES Let us start with basic definitions, some examples and proofs of properties. Definition 1. An assignment on a non empty set G is a map ◦ : G × G → G, (a,b) 7→ a ◦ b that assigns to each pair (a,b) of elements of the set G another element a ◦ b of G. A group G is a set G together with an assignment ◦ such that the following three properties hold: (1) The assignment ◦ is associative, that means that for every three elements a,b,c in G the identity (a ◦ b) ◦ c = a ◦ (b ◦ c) is true. (2) There is a unique element e in G such that e ◦ a = a ◦ e = a for all a in G. The element e is called the neutral element. (3) For every a in G there exists another element b in G with the property that a ◦ b = b ◦ a = e. The element b is called the inverse of a in G. A group G is called commutative or abelian if for all a,b in G a ◦ b = b ◦ a is true. The order of a group G is defined as the cardinality of the set G. Here are some examples that you probably know: Example 1. (1) The sets Z,Q,R,C of the integers, rational numbers, real numbers and complex numbers are commutative groups with assignment the usual addition of numbers. The neutral element is 0 and the inverse of a is −a. (2) The sets Q∗ = Q \{0},R∗ = R \{0},C∗ = C \{0} are commutative groups with as- 1 signment the usual multiplication. The neutral element is 1 and the inverse to a is a . (3) Let M be a non-empty set, then the set SM of bijective maps M → M is a group with assignment the concatanation of maps. This group is called the symmetric group of M. MATH 328 5

The neutral element is the idenitity map. If M = {1,2,...,n} then we write SM = Sn and calls Sn the symmetric group of degree n. Its order is n!. (4) Let k be a ﬁeld and V a k-vector space. Then the set GL(V ) of k-linear bijective maps from V to V is a group with assignment the concatanation of maps. The identity map is the neutral element. (5) Let k be a ﬁeld. Theset Mn(k) of n × n matrices with entries in k has an assignment given by matrix multiplication. The subset

GLn(k) := {A in Mn(k) |det(A) 6= 0 } of invertible matrices is a group. It is called the general linear group of degree n over k. (6) Let M be a non-empty set and G a group. Then the set of maps Map(M,G) from M to G is a group with assignment deﬁned by f ◦ g : M → G, m 7→ f (m)g(m). Exercise 1. Are the groups of the previous example point (3) commutative? Give a counter example or prove your answer. Note, that the answer might depend on M.

Solution. If M is a set consisting only of one or two elements, then SM is commutative. The reason is that a one element group consists of only the neutral element, while a two element group (2! = 2) consists of the neutral element e and one additional element a. Since a must have an inverse and e cannot be the inverse of a, the inverse of a must be a itself. Hence the group operations are e ◦ e = e, e ◦ a = a ◦ e = a, a ◦ a = e, so the group is commutative. If M has more than two elements, then choose three distinct elements m1,m2,m3 in M and consider a map σ that has the property m1 to itself and m2 to m3 and consider another map µ with the property that it maps m1 to m2. Then σ ◦ µ maps m1 to m3, while µ ◦ σ maps m1 to m2. Hence these maps donot commute and so SM cannot be commutative. We can relax the conditions in the deﬁnition of a group and still get a group. This goes as follows.

Theorem 1. Let G be a non-empty set with an assignment ◦ : G × G → G, such that (1) The assignment ◦ is associative. (2) There is a left neutral element e in G, satisfying e ◦ a = a for all a in G. (3) For every a in G there exists another element b in G with the property that b ◦ a = e. The element b is called the left inverse of a in G. Then G is a group with neutral element e and the left inverse of any element a is also its inverse. Proof. Consider an arbitrary a in G. We want to show that e is the neutral element, that means we have to show that a ◦ e = a. Let b be the left inverse of a and c the left inverse of b. Then by associativity a = e ◦ a =(c ◦ b) ◦ a = c ◦ (b ◦ a) = c ◦ e. 6 T CREUTZIG

Substituting e by e ◦ e and using above identity, we get a = c ◦ (e ◦ e)=(c ◦ e) ◦ e = a ◦ e. So that e is also right neutral in G. Since e is also right neutral, we have just shown that a = c, and hence a ◦ b = c ◦ b = e we observe that b is also right inverse to a. We thus have veriﬁed all group axioms. A very similar theorem is Theorem 2. Let G be a non-empty set with an assignment ◦ : G × G → G, such that (1) The assignment ◦ is associative. (2) There is a right neutral element e in G, satisfying a ◦ e = a for all a in G. (3) For every a in G there exists another element b in G with the property that a ◦ b = e. The element b is called the right inverse of a in G. Then G is a group with neutral element e and the right inverse of any element a is also its inverse. Exercise 2. Prove this theorem. Proceed in complete analogy to the proof of the previous theorem. Solution. Consider an arbitrary a in G. We want to show that e is the neutral element, that means we have to show that e ◦ a = a. Let b be the right inverse of a and c the right inverse of b. Then by associativity a = a ◦ e = a ◦ (b ◦ c)=(a ◦ b) ◦ c = e ◦ c. Substituting e by e ◦ e and using above identity, we get a =(e ◦ e) ◦ c = e ◦ (e ◦ c) = e ◦ a. So that e is also left neutral in G. Since e is also left neutral, we have just shown that a = c, and hence b ◦ a = b ◦ c = e we observe that b is also left inverse to a. We thus have veriﬁed all group axioms. Exercise 3. Let G be a group, prove that the inverse of any group element a is unique. Hint: Assume that a has two inverses, b and c, so that a ◦ b = b ◦ a = a ◦ c = c ◦ a = e, where e is the neutral element of G. Use this equation and associativity of G to prove the claim. Solution. We have to show that b = c, but b = e ◦ b =(c ◦ a) ◦ b = c ◦ (a ◦ b) = c ◦ e = c. We will now discuss how to perform manipulations in a group. There are two common ways of writing the group operation. Namely one likes to use addition or multiplication. If we use MATH 328 7 multiplication, then the identity might be denoted by 1 and the inverse of an element a by a−1. If we have such a multiplicative written group G, and we have an equation of the form xa = b, then x = ba−1. An equation of the form ay = b solves to y = a−1b. Observe, since a group is not necessarily commutative we have to be careful with the order of operations. Let us use from now on the multiplicative notation of group operation as it is probably most familiar.

3. SUBGROUPS Deﬁnition 2. A subgroup H of a group G is a subset H that is closed under multiplication (with a,b in H also ab is an element of H). that contains the identity e and for every a in H also a−1 is in H. Proposition 3. A subset H of a group G is a subgroup of G if and only if H 6= {} and ab−1 in H for all a,b in H. Proof. The conditions H 6= {} and ab−1 in H for all a,b in H are necessary conditions for H being a subgroup. They are also sufﬁcient conditions, since there exists at least one h in H, and hence also hh−1 = e, the neutral element, is in H. Further with a,b in H, also b−1 = eb−1 is in H and hence a(b−1)−1 = ab is in H. It is always good to be aware of a few examples. Example 2.

(1) Let M beasetand N ⊂ M a subset of M. Then the elements of SM that leave N pointwise fixed form a subgroup of SM. (2) Let A and B be elements of GLn(k) for some field k. Then we know from linear algebra, that det(A) det(B) = det(AB), hence the group of matrices of determinant one form a subgroup of GLn(k), called the special linear group SLn(k) of degree n over the field k. (3) Let G be a group, then G itself is a subgroup of G and also {e} is a subgroup of G. A subgroup H is called non-trivial subgroup if H 6= G. A subgroup H is called maximal, if it is a maximal non-trivial subgroup. In other words, there is no subgroup K of G that has H as subgroup. Exercise 4. Let M be a set and N ⊂ M a subset of M. Consider the subset

H = {σ ∈ SMk σ(n) in N, for all n ∈ N } ⊂ SM of SM. Prove that H is a subgroup of SM.

Proposition 4. Let J be a non-empty index set and let Hj j∈J be a collection of subgroups Hj of a group G. Then the intersection H := Hj \j∈J 8 T CREUTZIG is a subgroup of G Exercise 5. Prove this proposition.

Solution. There is at least one element in all Hj and hence also in H, the neutral element e. −1 −1 Further with a,b in Hj also ab in Hj for all j in J and hence ab is also in H. So that the Proposition follows from Proposition 3. Let M be a subset of a group G. We call the intersection of all subgroups of G (including G) that contain M the subgroup generated by M, hMi := H. M⊂\H≤G The set M is called generating system of a subgroup H if hMi = H. We alsosay that M generates H. If M is non-empty, then M generates the group consisting out of all products of elements in M and their inverses, that is

ε1 εk hMi = m1 ...mk | εi = ±1, mi ∈ M, k ≥ 0 . Here the case k = 0 is interpreted as the neutral element. G is called finitely generated if there is a finite subset M ⊂ G with hMi = G. The group generated by one element g in G is h{g}i := hgi = gi | i = 0,±1,±2,... with g0 = e. This group is called the cyclic subgroup generated by g. Definition 3. Let G be a group. A group element g of G is said to have the finite order n if the cyclic subgroup hgi is of order n. If hgi is of infinite order then we say that g is an element of infinite order. We denote the order of a group element g by ord(g). A group G is called cyclic group if there is an element g in G with hgi = G. Theorem 5. Let G be a group and g be an element of G. Then (1) If g is of finite order, then there exists a minimal positve integer n with gn = e. Then ord(g) = n and h{g}i := hgi = gi | i = 0,1,...,n − 1 . Further gm = e if and onlyif m is an integer multiple of n. (2) The element g has infinite order if and only if all its powers are distinct, that is gi 6= g j for i 6= j. (3) Denote by (n,s) the greatest common divisor of n and s. If g has finite order n, then gs for integer s has order n/(n,s). Proof. (1), (2): We proof the first two statements together. If gi = g j with i < j then g j−i = e. we can choose the smallest positive integer n with gn = e. The euclidean algorithm (division with remainder) implies that every integer m can be written as m = kn + r with integer k and r and 0 ≤ r ≤ n. Thus gm = gkn+r =(gn)kgr = gr and hence h{g}i := hgi = gi | i = 0,1,...,n − 1 . So that g has at most order n. Further, we have gm = e if and only if r = 0 meaning that n|m since n was chosen to be minimal with the property gn = e. It follows that gi = g j if and only if i = j mod n. Especially all e,g,g2,...,gn−1 are pairwise distinct, so that the order of g is exactly n. Hence hgi is infinite if all powers of g are pairwise distinct, while the converse is obvious (if all powers of g are distinct then g must have infinite order). MATH 328 9

(3): Let m = ord(gs) so that (gs)m = e and hence n divides ms. Sothat m is divided by n/(n,s). On the other hand (gs)n/(n,s) =(gn)s/(n,s) = e so that n/(n,s) is divided by ord(gs) = m. Theorem 6. Let G = hgi be a cyclic group. Every subgroup H ≤ G is cyclic. If G has the ﬁnite order n, then for every divisor d of n there is exactly one subgroup H of G of order d; namely H = hgn/di. In this way, one obtains all subgroups of G. Proof. Consider H ≤ G = hgi. If H = {e} then H is the cyclic group of order one. If H 6= {e}, then H contains at least one element gℓ 6= e. Since then also g−ℓ in H we can assume that ℓ > 0. So that we can set k := min ℓ > 0|gℓ ∈ H . We want to prove that H = hgki. Let g j be an arbitrary element of G. The euclidean algorithm allows us again to write j = qk + r for integers q,r with 0 ≤ r < k. If g j in H, then so is gr =(gk)−qg j. Minimality of k then implies that r = 0 so that we have proven that H = hgki. If the order of g is the ﬁnite number n, then we can apply above argument for j = n so that n = qk, that is k|n and |H| = n/k. Conversely, Theorem 5 tells us that for every divisor d of n, we have that hgn/di has order d.

4. COSETS Cosets are equivalence classes defined by a subgroup H of a group G. Recall the definition Definition 4. An equivalence relation on a set M is a relation on its elements, usually denoted by ∼. If two elements a,b are related via the relation, then we write a ∼ b. This relation has to satisfy • a ∼ a (Reflexivity) • if a ∼ b then b ∼ a (Symmetry) • if a ∼ b and b ∼ c then a ∼ c (Transitivity) The equivalence class of an element a is the set [a] = {b ∈ M|a ∼ b}. Proposition 7. Let G be a group and H a subgroup of G. Define a relation on G via a ∼ b ↔ there exists h in H with a = bh. This relation is an equivalence relation and the equivalence classes are [x] := xH = {xh|h ∈ H}. We call xH the left coset of H in G that contains the element x Distinct left cosets are disjoint and the cardinality of each left coset is H. We denote the set of all left cosets by G/H. Exercise 6. Prove this Proposition. Solution. We verify the three axioms of an equivalence relation. • The identity e of G is an element of every subgroup of G. Since a = ae, we have a ∼ a. • If a ∼ b, then there exists an h in H with a = bh and hence b = ah−1. Since H is a subgroup, with h in H also its inverse h−1 is in H, so that b ∼ a. • Let a ∼ b and b ∼ c, then there exist h and h′ in H with a = bh and b = ch′. Hence a = ch′h and transitivity follows since hh′ is also in H. Assume that two cosets xH and yH share a common element a. This means there are h and h′ in H with xh = a = yh′. Thus an arbitrary element xh′′ of xH satisfies xh′′ = xhh−1h′′ = yh′h−1h′′ 10 T CREUTZIG and hence is an element in the coset yH. Thus xH ⊂ yH, but a completely analogous argument also shows yh ⊂ xH. Finally, we need to show that the cardinality of any coset coincides with the cardianlity of the subgroup H. For this consider the map H −→ xH, h 7→ xh. The image of this map is xH, so it is surjective. Assume that there are h and h′ with xh = xh′ but then multiplying with the inverse of x from the left we get h = h′ hence the map is bijective and thus the cardinality of H and xH are the same.

For each left coset, choose a representative x of this equivalence class, and denote by T the set of these representatives (the left transversals). The group G itself is the disjoint union of its left cosets, so we can write it as G = xH. x[∈T In other words, every element g in G can be uniquely written as g = xh for a unique x in T and a unique h in H. We have choosed to consider left cosets. In complete analogy we can deﬁne right cosets as Hx = {hx|h ∈ H}. And we can deﬁne a set T ′ of right transversals representing the disjoint right cosets. Right and left transversals are related. Consider the bijective map G → G, g 7→ g−1. Short question: Why is it bijective? This map acts on our disjoint union of left cosets as G = xH 7→ G = Hx−1, x[∈T x[∈T i.e. it maps left cosets to right cosets, and representatives to representatives. Especially the set T −1 = x−1|x ∈ T is a set of right transversales. One conclusion that we ca draw from this is that the cardinality of the set of left and right cosets is the same. This cardinality is called the index of H in G. We denote it by [G : H]. There is a nice behaviour of indices of subgroups, namely

Theorem 8. Let M be a subgroup of N and N itself a subgroup of the group G. Let S be a left-transversal of N in G and T a left-transversal of M in N. Then the set ST = {st|s ∈ S, t ∈ T } is a left transversal of M in G, so that the indeices satisfy [G : M]=[G : N][N : M].

Proof. We have the disjoint unions G = sN and N = tM. s[∈S t[∈T MATH 328 11

Hence G = stM. s∈S t[∈T We have to show that this decomposition is also a disjoint one. Assume that this is not the case. Then there exist s,s′ in S and t,t′ in T with s′t′M = stM. It follows that s−1s′ is in N and hence sN = s′N. But S is a transversal of N in G and s and ′s represent the same coset, so they must coincide. It follows that tM = t′M so that also t = t′ must hold. Setting M = {e} in this theorem, we get the nice identity |G| =[G : H]|H|. (4.1) Especially, we see that the order of a subgroup H of a ﬁnite group G is a divisor of the order of G. This observation is called Lagrange’s Theorem. 0 1 1 0 Exercise 7. Let g = and h = . These matrices generate a soubgroup H of 1 0 1 −1 GL2(R). What is the order of g,h and gh? Show that |H| = 12. Use Lagrange’s Theorem. 1 0 Solution. Matrix multiplication tells us that g2 = h2 =(gh)6 = =: e. We also compute 0 1 that (gh)3 does not coincide with g or h. It follows that gh generates a cyclic subgroup of H of order 6, call it K. Note that the inverse of gh is hg, since ghhg = gg = e. Every element is a product whose factors are g′s and h′s. If the number of factors is even it is of the form (since g2 = h2 = e) ghgh...gh or hghg...hg and hence in the subgroup K generated by gh. Thus the only left cosets can be K,gK and hK. We have gK = hK, since ggh = h, so gK is the only non-trivial left coset and hence the index of K in H is two so that by Lagrange’s theorem |H| = 12.

5. NORMAL SUBGROUPS A special class of subgroups are normal ones. They obey one of the following equivalent properties. Lemma 9. Let G be a group and H a subgroup of G, then the following three statements are equivalent: (1) The equality of left and right cosets, gH = Hg for all g in G holds. (2) The identity g−1Hg = H holds for all g in G. (3) For all g in G and all h in H, g−1hg is an element of H. If these properties hold, then we call H a normal subgroup of G and we write H E G. Proof. Multiplying g−1 to the ﬁrst identity implies the second one. The third statement is an implication of the second one. We need to show that the third statement implies the ﬁrst one. −1 Consider g in G and h in H then gh = g−1 hg−1g ∈ Hg and hg = g g−1hg ∈ gH. Example 3. (1) Let G be a group and e the identity, Then G itself is a normal subgroup as well as the one element group {e}. If there are no other normal subgroups, then G is called simple. (2) If G is abelian, then every subgroup of G is normal. 12 T CREUTZIG

(3) Let k be a ﬁeld. The subgroup

SLn(k) := {A in Mn(k) |det(A) = 1 } of GLn(k) = {A in Mn(k) |det(A) 6= 0 } is normal. (4) A subgroup of index two of a group is a normal subgroup. (5) Let H be a normal subgroup of a group G. if H is cyclic of ﬁnite order then every subgroup of H is normal in G. Exercise 8. Prove the statements (2), (3), (4) and (5) of the previous example. Solution.

(2) Let G be abelian and H a subgroup of G. Then for every g in G and every h in H, we have g−1hg = hg−1g = h ∈ H and hence by the third property of normal subgroups H is normal in G. (3) Let A be in GLn(k) and B a matrix in SLn(k), then det A−1BA = det A−1 det(B)det(A) = det A−1 det(A) = det A−1(A = 1 −1 so that A BA satisﬁes the deﬁning property of SLn(k) and the claim (3) follows. (4) Let G be a group and H a subgroup of index two. Then G is the disjoint union of its two left cosets and also of its two right cosets and any g ∈/ H is a representative of the non-trivial coset, that is G = H ∪ gH = H ∪ Hg hence gH = Hg. (5) H is a cyclic subgroup, so there exists h in G with H =< h >. We have learnt (Theorem 5 and 6) that any subgroup of H is of the form < hs > for some divisor s of the order of H. Let n be the order of H and d the order of hs. Let g be an arbitrary element of G. Since H is normal, we know that ghsmg−1 in H for every m. We have to show that it is even in the subgroup generated by hs. This is equivalent (by Theorem 6) to showing that the order of this element is d. But this follows from d ghsmg−1 = ghsmdg−1 = e. Let G be a group and H be a normal subgroup of G. Then the left and right cosets coincide. In this case we will just call them cosets of H in G. The intersection of a family of normal subgroups is again a normal subgroup. For a subset M of G, one calls the smallest normal subgroup of G that contains M the normal closure of M, it is NC(M) = g−1mg|g ∈ G, m ∈ M .

6. HOMOMORPHISMS The notion of homomorphism is very important in algebra. Roughly speaking it means a map that preserves the algebraic properties. A group homomorphism is deﬁned as follows Deﬁnition 5. Let G and G′ be two groups. A map ϕ : G → G′ is called a homomorphism (of groups) if ϕ(gh) = ϕ(g)ϕ(h) MATH 328 13 for all g,h in G. The set of all homomorphisms from G to G′ is denoted by Hom(G,G′). Homomorphisms are useful in better understanding the subgroup and quotient structure of groups. Some properties are stated in the following Proposition. Proposition 10. Let e be the identity of the group G and e′ the one of the group G′. Let ϕ be a homomorphism from G to G′. Then (1) The image of the identity in G is the identity in G′, that is ϕ(e) = e′. (2) For every g in G, the image of the inverse is ϕ(g−1) = ϕ(g)−1. (3) The image of ϕ, imϕ := {ϕ(g)|g ∈ G} is a subgroup of G′. (4) The kernel of ϕ, kerϕ := {g ∈ G|ϕ(g) = e′} is a normal subgroup of G. Proof. (1) Let g be an arbitrary element of G, then ϕ(g) = ϕ(ge) = ϕ(g)ϕ(e), multiplying this equation from the left by ϕ(g)−1 we obtain ϕ(e) = e′. (2) Let g be an arbitrary element of G, then e′ = ϕ(e) = ϕ(gg−1) = ϕ(g)ϕ(g−1) and hence ϕ(g−1) is the inverse of ϕ(g). (3) We use Proposition 3 to prove this statement. We need to show that the image is non- empty and that with a = ϕ(g),b = ϕ(h) also ab−1 is in the image. But the image contains the identity e′, since ϕ(e) = e′, so it is non-empty. Further the computation ab−1 = ϕ(g)ϕ(h)−1 = ϕ(g)ϕ(h−1) = ϕ(gh−1) tells us that ab−1 is also in the image. (4) Again, we have to verify the conditions of Proposition 3. The kernel is non-empty, since e is in the kernel. Let g,h be in the kernel, then we know that ϕ(h−1) = ϕ(h)−1 = e′−1 = e′ so that also h−1 is in the kernel. Hence ϕ(gh−1) = ϕ(g)ϕ(h−1) = e′e′ = e′ is also in the kernel. We have shown that the kernel is a subgroup. It is even a normal subgroup, since for all h in G and all g in the kernel we have ϕ(h−1gh) = ϕ(h−1)ϕ(g)ϕ(h) = ϕ(h−1)e′ϕ(h) = ϕ(h−1)ϕ(h) = ϕ(h−1h) = ϕ(e) = e′.

A homomorphism from G to itself is called an endomorphism. A homomorphism ϕ is called injective if its kernel is the trivial one-element subgroup, that is kerϕ = {e} with e the identity of G. A homomorphism from G to G′ is called surjective, if its image is G′. A surjective and injective homomorphism is called bijective and also an Isomorphism and ﬁnally an isomorphism from G to itself is an automorphism. Proposition 11. (1) Let G,G′ and G′′ be groups and let ϕ : G → G′ and ψ : G′ → G′′ be homomorphisms, then the composition ψ ◦ ϕ : G → G′′ is also a homomorphism. (2) Let G and G′ be groups and let ϕ : G → G′ be an isomomorphism, then to each element x in G′ there is a unique element a in G with ϕ(a) = x. Deﬁne a map ϕ−1 : G′ → G by ϕ−1(x) = a. This map is an isomorphism of groups. Exercise 9. Prove this proposition. 14 T CREUTZIG

Solution.

(1) For any g,h in G, we have ψ(ϕ(gh)) = ψ(ϕ(g)ϕ(h)) = ψ(ϕ(g))ψ(ϕ(h)) which is the homomorphism property. (2) For x,y in G′ there are unique elements a,b in G with ϕ(a) = x and ϕ(b) = y. Hence xy = ϕ(ab) so that ϕ−1(xy) = ab = ϕ−1(x)ϕ−1(y). Hence ϕ−1 is a homomorphism. It is even an isomorphism, since the inverse of a bijective map must be bijective.

The composition of two automorphisms of a group G is again an automorphism, so that the set of automorphism together with assignment the composition is again a group, called the automorphism group aut(G) of G. The identity is the identity map and a special class of automorphisms are the maps −1 ig : G → G, h 7→ ghg for every g in G. The action of ig is called conjugation by g. The set of all conjugations is called the subgroup of inner automorphsisms, it is a subgroup, since one can verify that ig ◦ ih = igh. The kernel of the action of the group of inner automorphisms is the set of all g in G with h−1gh = g for any h in G or equivalently the set Z(G) = {g ∈ G|gh = hg for all h ∈ G}, called the center of G.

Example 4.

(1) Let G be a cyclic group generated by g in G. For all integers m the map m ϕm : G −→ G, ϕm(h) = h for all h in G is an endomorphism of G since G is abelian. Every endomorphism of G is of this form. To see this, let ϕ be an arbitrary endomorphism of G. It is uniquely specified by ϕ(g), since ϕ(gr) = ϕ(g)r for all r in Z. This m means if there ecists m in Z with ϕ(g) = g ; then it follows that ϕ = ϕm. The automorphisms of G are the endomorphisms ϕ of G for which ϕ(g) a generating m element of G is. These are the ϕm with G = hg i. The composition of automorphisms is ϕm ◦ ϕn = ϕnm so that we see that the automorphism group is commutative. If the order of G is infinite, then the automorphism group consists of the identity and the map h 7→ h−1 for all h in G. It is thus the cyclic group of order two. If the order of G is n for some positive integer n, then Theorem 5 tells us that the order of gm is n if and only if n and m are co-prime (two integers are called co-prime if their greatest common divisoris one). Hence the automorphism group of G consists of ϕm for 1 ≤ m < n and n and m co-prime. n (2) Let k be a field and let e1,...,en be the standard basis of the vector space k . For every permutation σ in Sn let π(σ) be the linear map n n π(σ) : k −→ k , π(σ)(ei) = eσ(i)

for all 1 ≤ i ≤ n. This deﬁnes an injective homomorphism π from Sn to GLn(k). MATH 328 15

(3) Let R be the abelian group of the real numbers with operation addition, and let S1 be the circle S1 := e2πix|0 ≤ x < 1 1 with operation multiplication. Both S and R are groups. The map ϕ : R −→ S1, x 7→ e2πix is a surjective group homomorphism with kernel the integers Z. The integers are a normal subgroup of the real numbers. Let a,b be arbitrary real numbers, then their cosets are a + Z = {a + n|n ∈ Z}, b + Z = {b + n|n ∈ Z}. Consider two elements a + n in a + Z and b + m in b + Z, that is n and m are integers. Then a + b + n + m =(a + b)+(n + m) ∈ (a + b) + Z. In other words, the sum of an element in the coset a + Z with an element in the coset b + Z is an element of the coset (a + b) + Z. This defines the operation • (a + Z) • (b + Z) :=(a + b) + Z on cosets which has identity Z and inverse −a + Z of a + Z. We thus get a group structure on the cosets. The set of cosets is {a + Z|0 ≤ a < 1} and the map {a + Z|0 ≤ a < 1}→ S1, a + Z 7→ e2πia is an isomorphism of groups. This example makes first contact to Lie groups. The group of real numbers with operation addition is the non-compact real form GL(1,R) of the general linear Lie group GL(1,C), the circle S1 is its compact real form U(1). Lie groups are differentiable manifolds with a group structure that is comaptible with the smooth structure of the manifold. They appear frequently in nature, for example U(1) is the gauge group of the quantum field theory of the electron.

Exercise 10. Let k be a field and SLn(k) the group of invertible n × n matrices over k with determinant one and let GLn(k) be the group of invertible n × n matrices over k. Then for every A in GLn(k) we can define an action on SLn(k) via conjugation with A, since SLn(k) is a normal subgroup of GLn(k) this defines an automorphism of SLn(k). Show that this automorphism is inner. 1/n Solution. Let A in GLn(k), then conjugation by A and A/(det(A)) is the same but the latter is in SLn(k) provided that the n-th root is in our field k. So that we observe, that this exercise only has a solution if the field k contains all its n-th roots. This is the case, if k is algebraically closed, e.g. k = C.

7. EXERCISESAND VOCABULARY In this first part of the course, we have introduced groups and some important concepts. We first defined the algebraic object called a group as a set together with an associative operation, an identity element and inverse elements. We then have talked about subgroups. We have 16 T CREUTZIG said that any subset of a group generates a subgroup and the subgroup generated by just one element got the special name cyclic subgroup. The next important concept were cosets. These are equivalence classes, where elements in the same equivalence class are related by the action of some subgroup. We have then shown that every group can be written as a disjoint union of its (either left or right) cosets of some subgroup. Especially this allowed us to express the cardianlity of a group as the product of the cardinality of the subgroup and the number of distinct cosets (again either left or right cosets). There are left and right cosets, and a special class of subgroups are those for which left and right cosets coincide. These subgroups are called normal subgroups. Finally, we have defined homomorphisms of groups as maps between two groups that respect the group structure. Given a homomorphism, we have seen that the image of it is also a group, and the kernel is even a normal subgroup. This is actually a very good way to realize normal subgroups, as kernels of group homomorphisms. If you want to review this first part of the course, it is a good idea to recall the precise statements of what I said in the previous paragraph. Here are also a few exercises, but without solutions. Let me know if solutions to some of those would help.

Exercise 11. (1) Let A and B be subgroups of a group G. Prove that the union A∪B is a group if and only if either A ⊂ B or B ⊂ A. Conclude that a group cannot be the union of two non-trivial subgroups. (2) Let G be a group geberated by two elements g and h. Let r and s be positive integers with gr = e and hg = ghs. Show that G = {gmhn|m,n ∈ Z,0 ≤ m < r}. Show also that if s 6= 1 then h has ﬁnite order. (3) Let G be a group. Show that for all g,h in G that ord(gh) = ord(hg). (4) Let ζ = e2πi/p be a p-th root of unity and let p be a prime number. Consider the map ϕ : Z −→ S1 = eix|0 ≤ x < 1 , n 7→ ζ n. The integers are a group under addition, the unit circle S1 is a group under multiplication. Answer the following questions: (a) Show that ϕ is a group homomorphism. (b) Determine the kernel of ϕ. (c) Determine the image of ϕ. (d) Show that the image is isomorphic to the integers modulo p, that is Z/pZ. (5) Answer the following questions without proof (a) Give an example of a proper subgroup H of the integers Z. (b) Are the integers a group with operation multiplication? (c) Give an example of a normal subgroup of a group G. (d) Let G be a group of cardinality N and let H be a cyclic subgroup of order d. How many right cosets of H in G are there? (e) Let σ =(m1,...,mk) be a cycle of length k in Sn. Let π be another element of Sn. What is πσπ−1? (6) Let X be a set with |X| > 1. Deﬁne a relation on the set of subsets of X via A ∼ B ↔ A ∩ B 6= { }. MATH 328 17

Is this an equivalence relation? If yes, then prove your answer otherwise give a coun- terexample. (7) Let X be a set and Y ⊂ X a non-empty proper subset of X. Deﬁne a subset of the permutation group of X, SX as

H := {π ∈ SX |π(y) = y for all y ∈ Y }.

Prove that H is a subgroup of SX . Is H normal in SX ?

Solution. (1) Assume that A ∪ B is a group and neither A ⊂ B nor B ⊂ A. Then there exist a in A but not in B and also b in B but not in A. It follows since A ∪ B is a group that ab is in A ∪ B. But if ab = a′ in A, then b = a−1a′ is in A and if ab = b′ in B then a = b′b−1 is in B. In both cases we get a contradiction to our assumptions. Hence our assumptions were impossible. 2 (2) One proves via induction that hng = ghsn and also hbga = gahbs . This shows that G is of the claimed type. The finite order follows from above identity with b = 1 and a = r. (3) If gh has the finite order n, then (hg)n+1 = h(gh)ng = hg and hence the order of hg is at most n. The same argument changing the roles of hg and gh shows that the order of gh is at most the orer of hg. Hence the two must coincide. This argument also shows that if gh has infinite order hg cannot have finite order (and vice versa). (4) (a) ϕ(n + m) = ζ n+m = ζ nζ m = ϕ(n)ϕ(m) for every n,m in Z hence it is a group homomorphism (b) ζ n = 1 if and only if n = mp for an integer m, hence the kernel is pZ (c) The image is {ζ ,ζ 2,...,ζ p} that is the set of all p-th roots of unity. (d) The image is isomorphic to the factor group of G = Z by the kernel of the homomorphism. (5) (a) pZ for any integer p and also {0}. (b) No, zero has no inverse (c) SLn(k) inside GLn(k) (d) N/d (e) (π(m1)...π(mk)) (6) No, transitivity is not satisfied. Take B such that it has non-trivial intersection with A and C but A and C do not intersect. (7) Let π and µ be in H, since µ(y) = y for every y in Y , the same must also be true for µ−1. Hence π · µ−1(y) = y for all y in Y and we have proven in the leture that if for all a,b in H also ab−1 is in H then H is a subgroup. This subgroup is not normal: take x in X \Y and y in Y . Take π with π(x) = x′ 6= x and σ with σ −1(y) = x then σπσ −1(y) = σ(x′) 6= y (since x 6= x′ and σ(x) = y and σ is a permutation).

8. THE SYMMETRIC GROUP We have already seen the symmetric group of a set as the group of all its permutation. We will now study this group in some detail. The reason is the following theorem, called Cayley’s Theorem.

Theorem 12. Every group G is isomorphic to a subgroup of the symmetric group SG of the set G. 18 T CREUTZIG

Proof. For every g in G, we deﬁne a permutation on G by

πg : G → G, h 7→ gh. This deﬁnes a homomorphism, since the identity ′ ′ πg πg′ (h) = πg g h = gg h = πgg′ (h) for all g,g′,h in G implies that πg ◦ πg′ = πgg′ . Let e be the identity in G. This homomorphism is injective, since if πg is the identity map then the computation g = ge = πg(e) = e tells us that g must be the identity e.

We can conclude that a finite group of order n is isomorphic to a subgroup of Sn. There is a very convenient notation for permutations. This notation allows to perform compu- tations in an efficient way and it is called the cycle notation. Let M be a set and π a permutation of M, that is an element of SM. The support of π is the subset of M defined as supp(π) := {m ∈ M | π(m) 6= m}.

If the support of two permutations σ,τ in SM is disjoint, then these two elements commute, στ = τσ. The deﬁnition of a cycle is

Deﬁnition 6. Let M be a set. An element π in SM is called cycle of length k if |supp(π)| = k and if in addition one can number the support as

supp(π) = {m1,...,mk} such that π(mi) = mi+1 for i = 1,...,k−1 and π(mk) = m1. Thiscylceis alsocalled (m1m2 ...mk). A cycle of length 2 is called a transposition. The following theorem tells us that this is a useful deﬁnition.

Theorem 13. Every permutation π in Sn. π 6= Id (Id is the identity map) is a product of cycles with disjoint support; up to the order of cycles this representation of π is unique. Proof. We first prove that π can be written as a product of cycles with disjoint support. Let < π > be the cyclic group generated by π. We call m and m′ in M equivalent if there exists an integer j such that π j(m) = m′, in formulae m ∼ m′ ↔ there exists j ∈ Z with π j(m) = m′. We prove that this defines an equivalence relation: • the relation is reflexive since π0(m) = m for every m in M and hence m ∼ m. • If π j(m) = m′ then π− j(m′) = m and hence the relation is symmetric. • Let m ∼ m′ and m′ ∼ m′′, this means there exist j and i with π j(m) = m′ and πi(m′) = m′′. Hence πi+ j(m) = πi(m′) = m′′ so that also m ∼ m′′ and the relation is also transitive. We observe that all three properties of an equivalence relation are satisfied. Equivalence classes are disjoint, so that M is the disjoint union of all equivalence classes K1,...,Ks for some s. We consider all i in {1,...,s} with |Ki| ≥ 2 and define σi in Sn by σi(x) = x if x not in Ki and σi(x) = π(x) otherwise. σi is a cycle and this cycle is constructed as follows. Let a be an r element in Ki and let r be the smallest positive integer such that π (a) = a, then the equivalence class Ki is the set of all elements euivalent to a and this is r−1 Ki = {a,π(a),...,π (a)}. MATH 328 19

j−1 Set a j := π (a), so that Ki = {a1,a2,...,ar} and π(a j) = a j+1 for j = 1,...,r−1 and π(ar) = a1. These are the deﬁning properties of a cycle of length r so that we have seen that on the equivalence class K1 the permutation π coincides with the cycle (a1,...,ar). Since cycles with disjoint support commute and since the equivalence classes are disjoint it follows that

π = ∏σi where the product is taken over all i with |Ki| ≥ 2. It remains to prove that this decomposition is unique up to order of factors. Assume that there is a second decomposition π = τ1 ...τt of π into cycles with disjoint support. Then π acts on supp(t j) as t j, hence supp(t j) is an equivalence class Ki for some i and so t j = σi on Ki. Since both act trivially on the complement of Ki they coincide on M. In other words, every τ j is equal to some cycle σi of our previous decomposition and hence the second decomposition coincndes with the ﬁrst one up to ordering.

Theorem 14. The following statements are true for the permutation group Sn.

(1) Let σ =(m1,...,mk) be a cycle of length k and let π be an arbitrary element in Sn, then −1 π(m1,...,mk)π =(π(m1),...,π(mk)).

(2) The symmetric group Sn is generated by all transpositions. (3) The subsets A = {(12),(13),...,(1n)} and B = {(12),(23),...,(n − 1n)}

each form a generating set of Sn. Proof. (1) For each i − 1,...,k − 1 we have that −1 πσπ π(mi) = πσ(mi) = π(mi+1) −1 and πσπ π(mk) = πσ(mk) = π(m1) so that the claim follows. (2) Every permutation is a product of cycles with disjoint support. So we have to show that every cycle is a product of transpositions, which is true since

(m1,...,mk)=(m1m2)(m2m3)...(mk1 mk). (3) The previous statement tells us that it is enough to show that every transposition can be written as a product of elements in A respectively B. Let i, j 6= 1 then the ﬁrst statement implies that (ij)=(1i)(1 j)(1i), since also (1 j)=( j1) the statement for the set A follows. Every transposition ( jk) with k > j + 1 can be written as ( jk)=( jj + 1)( j + 1k)( jj + 1) by statement (1) (or by a dircet computation). Induction for k − j then shows that ( jk) is a product of elements in B.

We see that every permutation is a product of transpositions. It turns out that permutations that are products of an even number of permutations form a subgroup of the symmetric group. 20 T CREUTZIG

This can be seen as follows. Define the parity or signum of a permutation as π(i) − π( j) sgn(π) := ∏ . 1≤i< j≤n i − j The relation < defines an order on the set {1,...,n}. Rearranging the order of our elements then defines another order. The definition of the parity is independent of such a change of order since π(i) − π( j) π( j) − π(i) = . i − j j − i Especially for any permutation σ we have π(i) − π( j) sgn(π) = ∏ . i j 1≤σ(i)<σ( j)≤n − We thus get for the product of two permutations π(σ(i)) − π(σ( j)) sgn(πσ) = ∏ 1≤i< j≤n i − j π(σ(i)) − π(σ( j)) σ(i) − σ( j) = ∏ ∏ 1≤i< j≤n σ(i) − σ( j) 1≤i< j≤n i − j π(σ(i))− π(σ( j)) σ(i) − σ( j) = ∏ ∏ σ i σ j i j 1≤σ(i)<σ( j)≤n ( ) − ( ) 1≤i< j≤n − π(i) − π( j) σ(i) − σ( j) = ∏ ∏ 1≤i< j≤n i − j 1≤i< j≤n i − j = sgn(π)sgn(σ). We summarize Theorem 15. The map sgn : Sn −→{±1}, π 7→ sgn(π) is a group homomorphism. Its kernel is a normal subgroup, called the alternating group An.

Theorem 16. The alternating group An for n > 1 has the following properties: (1) An consists of the elements that can be written as a product of an even number of transpositions. Let τ be an arbitrary transpositon, then

Sn = An ∪ τAn. 1 Especially the order of An is 2 n!. (2) An is generated by all three cycles. (3) An is simple if and only if n 6= 4. Proof.

(1) By Theorem 14 we know that Sn is generated by all transpositions but only those elements that are a product of an even number of transpositions are in the kernel of the parity map. The only non-trivial coset of An inside Sn is the set of all elements that are a product of an odd number of transpositons. (2) It is enough to show the statement for a product of two transpositons (ab)(cd). If the support of these two transpositions is disjoint, then (ab)(cd)=(acb)(acd). Otherwise, MATH 328 21

there are two types of cases: (ab)=(cd) and hence (ab)(cd) = 1, or with out loss of generality a = d and b 6= c, but then (ab)(ac)=(cba). (3) One ﬁrst computes explicitely that A4 has a normal subgroup (see the Exercises). A2,A3 are the cyclic abelian one and three element groups. They have no non-trivial subgroups, so they must be simple. Consider now n ≥ 5. Conjugation acts transitively on three cycles, so with one three cycle in a normal subgroup N all three–cycles must be in this subgroup. Since An is generated by all three cycles this subgroup must already be all An. We thus have to show that every normal subgroup of An contains at least one three–cycle. Let σ in N be of the form

σ = ∏τi

with cycles τi with disjoint support. We have a few cases and in our computation we will frequently use Thoerem 14 (1). • Assume that at least one of the τi has length at least four. Say

τs =(m1,...,mr) with r ≥ 4. Then −1 −1 −1 σ ◦ (m1m2m3) ◦ σ ◦ (m1m2m3) = σ ◦ (m1m3m2) ◦ σ ◦ (m1m2m3)

=(mrm2m1) ◦ (m1m2m3)

=(m2m3mr).

Hence N = An • Assume that two of the cycles say τs and τt are of length three. Set τs =(m1m2m3) and τt =(m4m5m6), then −1 −1 −1 σ ◦ (m1m2m4) ◦ σ ◦ (m1m2m4) = σ ◦ (m1m4m2) ◦ σ ◦ (m1m2m4)

=(m3m6m1) ◦ (m1m2m4)

=(m1m2m4m3m6)

and hence by the previous case N = An. 2 • Assume that τs is a three cycle and all other are transpositions, but then σ is just a three cycle since every transpositions squares to the identity. Hence again N = An. • if all τi are transposition then let τ1 =(m1m2) and τ2 =(m3m4) and then −1 −1 −1 σ ◦ (m1m2m3) ◦ σ ◦ (m1m2m3) = σ ◦ (m1m3m2) ◦ σ ◦ (m1m2m3)

=(m2m4m1) ◦ (m1m2m3)

=(m1m4)(m2m3).

Let m5 another element distinct from mi,m2,m3,m4 (this exists since n ≥ 5). Then −1 (m1m5m2) ◦ (m1m4) ◦ (m2m3) ◦ (m1m5m2) =(m5m4) ◦ (m1m3), and (m1m4) ◦ (m2m3) ◦ (m5m4) ◦ (m1m3)=(m1m2m3m4m5)

so that the ﬁrst case again implies that N = An. This exhausts all possibilities and we can conclude that N = An. In other words An has no normal subgroup and hence is a simple group. 22 T CREUTZIG

2 Exercise 12. Let G = {e,a1,a2,a3} be a non-cyclic four element group. Show that ai = e and aia j = ak for every permutation i, j,k of 1,2,3. G is called a Klein four-group.

Solution. The order of a group element must divide the order of the group. Since only e has order one and since there is no element of order four, all ai can only have order two. Consider −1 a1a2. This cannot equal to e as that would imply that a1 = a2 and it would contradict the uniqueness of the inverse. It neither can be equal to a1 or a2 as a1a2 = a1 implies a2 = e and a1a2 = a2 implies that a1 = e. Hence the only possibility is a1a2 = a3. The same argument applies to any permutation of the indices.

Exercise 13. Show that S4 contains a normal subgroup isomorphic to the Klein four group of the previous example. Conclude that A4 is not simple.

Solution. Every group G is a subgroup of its permutation group as a set SG. The Klein four group has four elements, so it must be a subgroup of S4. Let us order the set, such that e is the ﬁrst element, and ai the i+1-th element. Then a1 =(12)(34), a2 =(13)(24) and a3 =(14)(23). 2 Then by construction ai = e and aia j = ak for every permutation i, j,k of 1,2,3 and this group is isomorphic to the Klein four group. It is normal since by Theorem 14 statement (1) −1 π(m1m2)(m3m4)π =(π(m1)π(m2))(π(m3)π(m4)) and hence conjugation leaves this subgroup invariant.

Exercise 14. Let ϕ : Q −→ Z be a group homomorphism (the group operation is addition of rational numbers respectively integers). Show that ker(ϕ) = Q.

Solution. Assume that m 6= 0 in Z is in the image of ϕ. This means there exists a rational number q with ϕ(q) = m. The homomorphism property means that for any positive integer n that q m ϕ = n n since q n q n q q m = ϕ(q) = ϕ n = ϕ ∑ = ∑ ϕ = nϕ . n i=1 n! i=1 n n Choosing an integer n that doesnot divide m gives a contradicition. Hence every element has to map to 0.

9. ISOMORPHISM THEOREMS We have seen in the example of the real numbers and the integers, that the integers are a normal subgroup and that one can deﬁne on the cosets a group structure which is isomorphic to the unit circle S1. One also calls this group the factor group of the integers in R. In general, the deﬁnition is

Deﬁnition 7. Let G be a group and N a normal subgroup of G. For two subsets X and Y of G, we deﬁne the subset XY := {xy|x ∈ X,y ∈ Y }. So that the product of two cosets of N is again a coset (aN)(bN) = a(Nb)N = a(bN)N = abN. MATH 328 23

The set of all cosets together with this assignment forms a group, called the factor group or quotient group G/N of N in G. The neutral element is the coset N and the inverse of aN is a−1N. We haven’t proven that the factor group is actually a group. Verifying the group axioms is a short exercise. The map π : G −→ G/N, a 7→ aN is called the natural projection from G to G/N. This map is a surjective group homomorphism, since (aN)(bN) = abN with kernel kerπ = N. There is the following universal property Theorem 17. Let ϕ : G −→ G′ be a group homomorphism, N a normal subgroup of G with N ⊂ ker(ϕ). Then there exists exactly one homomorphism ϕ¯ : G/N −→ G′ with ϕ¯ ◦ π = ϕ. In other words, we have the commutative diagram ϕ G G′ π ϕ¯ G/N

Proof. For every a in N, we have ϕ(a) = e and hence ϕ(ga) = ϕ(g)ϕ(a) = ϕ(g) for all g in G. Thus the map ϕ¯ (gN) = ϕ(g) is a well-defined map ϕ¯ : G/N −→ G′. This map satisfies ϕ¯ ◦ π = ϕ. It is a homomorphism since ϕ¯ (gNhN) = ϕ¯ (ghN) = ϕ(g)ϕ(h) = ϕ¯ (gN)ϕ¯ (hN). Since π is surjective, this is the only map with these properties. Corollary 18. Let ϕ : G −→ G′ be a group homomorphism. Then there exists an isomorphism of groups ϕ¯ : G/kerϕ −→ imϕ with ϕ¯ (gkerϕ) = ϕ(g) for all g in G. Proof. Since the kernel is a normal subgroup, we can apply the previous theorem with N = kerϕ. The image of ϕ¯ is by definition imϕ, while the kernel is the coset kerϕ which is the neutral element of G/ker. Hence we also have injectivity. Example 5. Let G =< g > be a cyclic group. The map r 7→ gr is a surjective group homomorphism ρ from Z to G. If ker ρ = {0} then G is isomorphic to Z =< 1 >. Otherwise there exists a smallest positive integer n with gn = e so that n in ker ρ. Let m be another element of ker ρ then by the Eucludean algorithm m = nq + r for 0 ≤ r < n. Since m and n are in the kernel of ρ, so must be r and minimality of n implies that r = 0. It follows that ker ρ = nZ and G is isomorphic to Z/nZ the group of integers modulo n for a unique integer n. The first isomorphism theorem is: Theorem 19. Is H a subgroup of a group G and N a normal subgroup of G, then NH is a subgroup of G and N ∩ H is a normal subgroup of H. The assignment h(N ∩ H) 7→ hN is an isomorphism ∼ H/(N ∩ H) −→ NH/N.

Proof. We ﬁrst prove that NH is a subgroup of G. For this let n1,n2 be in N and h1,h2 be in H, then −1 −1 −1 −1 −1 −1 −1 −1 n1h1(n2h2) = n1h1h2 n2 = n1 h1h2 n2 h1h2 h1h2 ∈ NH. 24 T CREUTZIG

So that the statement follows with Proposition 3. Note, that since N is normal in G it is also normal in NH. The map h 7→ hN is a surjective group homomorphism from H to NH/N. The kernel consists of all h in H that are mapped to the coset N, these are exactly the elements of N ∩ H. The theorem follows now from the previous corollary.

One often calls NH the product of the subgroups N and H (correctly one has to call it the semi-direct product). If also H is normal in G, then the product NH is normal in G too. The second isomorphism theorem is

Theorem 20. Let M and N be normal subgroups of a group G, and let N be a subgroup of M, then M/N is a normal subgroup of G/N and ∼ (G/N)/(M/N) −→ G/M.

Proof. The assignment gN 7→ gM is a surjective group homomorphism from G/N to G/M. The kernel of this map are all cosets that are mapped to M, these are exactly all cosets of M/N. The previous corollary again then implies the theorem.

We have seen that the product of a normal subgroup and another subgroup is again a subgroup. This is an example of a semi-direct product, that is NH is the semi-direct product of N and H. The deﬁnition is

Deﬁnition 8. Let G be a group and N a normal subgroup of G and H a subgroup of G. The group G is called the semi-direct product of N and H if NH = G. Let I be an index set and (Gi)i∈I a family of groups. The direct product

∏Gi i∈I is the group, that as a set is the set theoretic product of the sets Gi; and whose group operation is the componentwise multiplication, that is

(gi)i∈I · (hi)i∈I :=(gihi)i∈I.

Exercise 15. Consider the cycles σ =(135), τ =(2345) and µ =(534) of S5. (1) compute στµ and τ2σ? (2) which of σ,τ and µ are elements of A5?

Solution. στµ =(135)(2345)(534)=(13)(254) τ2σ =(2345)(2345)(135)=(15)(24).

Three-cycles generate A5 so σ and µ are in A5. στµ has odd signum and hence τ cannot be in A5.

Exercise 16. Let G be a ﬁnite group that contains an index two subgroup. Prove that elements of odd order generate a proper subgroup H of G. (A subgroup H of a group G is called a proper subgroup of G if H 6= G) Hint: You have proven earlier that an index two subgroup must be a normal subgroup, call it N. One possibility to prove this exercise is to use the group homomorphism G −→ G/N, g 7→ gN. MATH 328 25

Solution. Let N be an index two subgroup of G. We will prove that H ⊂ N. N is normal by a previous exercise. Let G = N ∪ gN be the coset decomposition of G. The map 1 if a ∈ N ϕ : G −→{±1}, a 7→ (−1 if a ∈ gN is a group homomorphism with kernel N. An element h of odd order n must be in the kernel of this map since 1 = ϕ(e) = ϕ(hn) = ϕ(h)n = ϕ(h). Hence H is a subgroup of N. Exercise 17. Let σ : G −→ G′ be a group homomorphism and let π : G −→ H be a surjective group homomorphism. Let ker π ⊂ ker σ. Show that there exists a homomorphism τ : H −→ G′ such that σ = τ ◦ π holds. In other words, the diagram σ G G′ π τ H commutes. Solution. Both ker π and ker σ are normal subgroups of G. By Corollary 18, H is isomorphic to G/ker π, since π is surjective, with isomorphism π¯ −1, where π¯ : G/ker π −→ H, g ker π 7→ π(g). As in the proof of Theorem 20 we can construct a homomorphism µ : G/ker π −→ G/ker σ, g ker π 7→ g ker σ, so that by Theorem 20 the kernel is ker σ/ker π. Finally, deﬁne as in Corollary 18 σ¯ : G/ker σ −→ G′, g ker σ 7→ σ(g). Then we set τ : H −→ G′, h 7→ σ¯ µ π¯ −1(h) and by construction τ has the desired property. (You can also construct the map σ¯ ◦ µ in one step if you prefer)

Exercise 18. Let n1,...,nk be integers that are divisible by three. Let Gi =< gi > be a cyclic group of order ni (you can think of Gi as Z/niZ). Let G := G1 × G2 ×···× Gk be the direct product of the k abelian groups Gi. (1) How many elements of order three has G? (2) How many subgroups of order three has G? (3) What happens if you replace 3 by an arbitrary prime p? (You don’t have to prove your answer for this last question as the proof would be very similar to the case of p = 3).

Solution. Let g =(a1,...,ak) be an arbitrary element of G, and let mi be the order of the element m m m m ai of Gi. Then its m-th power is g =(a1 ,a2 ,...,ak ) and this equals to the identity (e1,...,ek) of G if and only if m is an integer multiple of each mi. In other words it must be an integer multiple of the least common multiple of all mi. Hence the order of g is the least common multiple of all mi. ni/3 We know from Theorem 5 and 6 that the only elements of order three of Gi are hi := gi and 2 d1 d2 dk hi . By the above argument all elements of order three are those of the form (h1 ,h2 ,...,hk ) k for each di in {0,1,2} and at least one of the di has to be non-zero. These are 3 − 1 elements. 26 T CREUTZIG

Each of these 3k − 1 elements generates an order three subgroup. Each subgroup has two elements distinct from the identity so that there are (3k − 1)/2 such subgroups. Remark: You can replace three in this exercise by another prime number p and your answers will be pk − 1 and (pk − 1)/(p − 1) for number of order p elements and number of order p subgroups.

10. GROUP ACTIONS The importance of groups in nature is that they act as symmetries on interesting objects. We have already seen that the permutation group is the group of bijective maps from a given set to itself. Another example are matrix groups that act on vector spaces. Here, we will consider group actions on sets.

Deﬁnition 9. A group action of a group G on a set X is a map G × X −→ X, (g,x) 7→ g · x satisfying for all g,h in G and all x in X (1) (gh) · x = g · (h · x) (2) e · x = x.

A set together with a group action of a group G is also called a G-set. We will often write gx instead of g · x.

Proposition 21. Let G be a group and let X be a G-set. Then the following statements are true: (1) For every g in G, the map

τg : X −→ X, x 7→ gx is bijective. (2) The map τ : G −→ SX , g 7→ τg

is a group homomorphism to the symmetric group SX of X. (3) Every group homomorphism from G to SX deﬁnes a group operation of G on X.

Proof.

(1) We will show that τg−1 is the inverse of the map τg and hence these maps must be bijective. We have

−1 −1 τg−1 ◦ τg (x) = g (g · x) = g g · x = e · x = x, so indeed τg−1 ◦ τg is the identity map. (2) Each τg acts bijectively on X, so it is a permutation of X. We have

(τh ◦ τg)(x) = h · (g · x)=(hg) · x = τhg(x) for all h,g in G and all x in X, so that τ satisﬁes the homomorphism property. (3) Let ϕ : G −→ SX be a group homomorphism, then we deﬁne a map G × X −→ X, (g,x) 7→ ϕ(g)(x) MATH 328 27

the homomorphism properties of ϕ translate into the group action properties under this map, that is ϕ(gh)(x) = ϕ(g) · ϕ(h)(x), ϕ(e)(x) = Id(x) = x

where Id is the neutral element of SX .

Deﬁnition 10. Let X be a G-set for a group G, then the set Gx := { gx | g ∈ G } is called the orbit of the elment x of X under G.

Proposition 22. The relation x ∼ y ↔ Gx = Gy deﬁnes an equivalence relation on X. Especially, the set X is the disjoint union of the orbits under G.

Exercise 19. Prove this Proposition.

Deﬁnition 11. Let G be a group and X be a G-set, then (1) The action of G on X is called transitive if X is an orbit under G. (2) The operationof G on X is called faithful if the homomorphism τ : G −→ SX is injective. (3) A map α : X −→ Y of G-sets X and Y is called a G-morphism if α(g · x) = g · α(x) for all x in X and all g in G. Further a G-morphism is called an G-isomorphism if it is a bijective G-morphism.

Example 6.

(1) A group G operates via left-multiplication on itself, this is called left-translation. The associated group homomorphism G −→ SG is called the left-regular permutation representation of G. This group action is faithful and hence G is isomorphic to a subgroup of SG. We have already proven this statement earlier. The right-regular permutation representation is deﬁned via (g,x) 7→ xg−1, i.e. right- multiplication with the inverse (this is necessary to get a group action). (2) Another (more important) action of a group G on itself is conjugation. Deﬁne the map −1 ig : G −→ G, h 7→ ghg .

So ig is the inner automorphism that acts via conjugation by g. So that the map

G × G −→ G, (g,h) 7→ ig(h) deﬁnes a group action of G on itself. The orbits under conjugation in G are called conjugacy classes so that the conjugacy class of an elment h is the set ghg−1|g ∈ G .

Conjugation also deﬁnes an operation on the set of subgroups UG of G. Namely −1 G × UG −→ UG, (g,H) 7→ gHg . The orbits under this action are called conjugacy classes of subgroups of G. 28 T CREUTZIG

We need a number of deﬁnitions/statements that tell us that we can rephraze well-known deﬁnitions in terms of group actions.

Definition 12. Let X be a G-set, then (1) An element x in X is called a fixed-point of the operation of G on X if gx = x for all g in G. The set of all fixed-points is denoted by X G = {x ∈ X|gx = x for all g ∈ G}. (2) For every x in X, the group action axioms allow one to prove that

Gx := {g ∈ G|gx = x} is a subgroup of G. This subgroup is called the isotropy subgroup of x in X. Note, that the ﬁxed-points are exactly those x with Gx = G. −1 Also note, that for every x in X and every g in G that Ggx = gGxg . Hence all elments of an orbit Gx have isotropic subgroups that are conjugate to each other. (3) If G as in the previous example point (2) acts on itself via conjugation, then the isotropy subgroups are called centralizers. So the centralizer of h in G is

CG(h) := {g ∈ G|gh = hg}, i.e. it is the set of all elments in G that commute with h. The set of all ﬁxed-point is the center of G, that we have deﬁned earlier. It is Z(G) = {h ∈ G|gh = hg for all g ∈ G}.

(4) If we consider the operation of G via conjugation on the set of its subgroups UG, then the isotropy subgroup of a subgroup H is called the normalizer −1 NG(H) := g ∈ G|gHg = H of H in G. The ﬁxed-points under this operation are exactly the normal subgroups of G.

Example 7. Let G = SL2(R) and H := {z ∈ C|im(z) > 0} the upper haf of the complex plane. One can prove the following (1) The M¨obius transformation a b az + b a b · z = for all ∈ G, z ∈ H c d cz + d c d deﬁnes a G-action on H. (2) The isotropy subgroup of i is the special orthogonal group SO(2), that is all matrices in SL2(R) with the property that the row-vectors are orthonormal. (3) The operation is transitive. (4) A meromorphic function on H that is invariant under the subgroup SL2(Z) is called a modular function. Such functions are essential in number theory, string theory and monstrous moonshine.

Theorem 23. Let X be a G-set , then for every x in X there is a G-isomorphism

p : G/Gx −→ Gx, gGx 7→ gx.

The cardinality of the orbit Gx is thus equal to the index [G : Gx]. MATH 328 29

−1 Proof. the map is surjective by construction. Let g1x = g2x then g2 g1 in Gx hence the map is also injective. It is a G-morphism, since

p(g1(g2Gx)) = p(g1g2Gx)=(g1g2)Gx = g1(g2Gx) = g1 p(g2Gx).

We can draw a few conclusions from this theorem. X is the disjoint union of its orbits, so if X is a ﬁnite set, then the cardinality satisﬁes

|X| = ∑|Gxi| = ∑[G : Gxi ] i∈I i∈I where I is an index set labelling the orbits and xi is a representative of the i-th orbit. Every ﬁxed- point belongs to an orbit of length one and this orbit is thus represented by the ﬁxed-point. If we split all one element orbits, we get G |X| = |X | + ∑ [G : Gxi ]. i∈I G xi∈/X This observation applied to the set G with G-action conjugation gives the following important theorem, called the class equation.

Theorem 24. Let (xi)i∈I be a system of representatives for the conjugacy classes of G, then

|G| = |Z(G)| + ∑ [G : CG(xi)]. i∈I xi∈/Z(G) This theorem will be very useful for Sylow’s theory. Example 8. A group of order 55 operates on a set of 39 elements. Claim: the group operation has at least one ﬁxed-point. The orbit length is the index of the isotropy group in G. Hence it must divide 55. It cannot be 55 itself as the set has less elements. So an orbit has either length 1,5 or 11. Let n1 be the number of ﬁxed-points, n5 the number of orbits of length 5 and n11 the number of orbits of length 11. Then

39 = n1 + 5n5 + 11n11

A case-by-case analysis reveals that n1 = 0 is impossible. Exercise 20. Let G be a group and Z(G) the center of G. Prove: if the factor group G/Z(G) is cyclic then G is abelian. Solution. Let G/Z(G) be cyclic, then it is generated as a group by an element gZ(G). An arbitrary element of G can thus be written as grh for integer r and h in Z(G). Let gsh′ be another such element. Since h and h′ are in the center, we have grhgsh′ = grgshh′ = gsgrhh′ = gsgrh′h = gsh′grh and hence these elements commmute so that G must be abelian. Exercise 21. Prove that the index of the center of a group is never a prime number. Solution. The index of a normal subgroup is the number of cosets and hence the order of the factor group. A group of prime order p must be cyclic as the order of every element must divide the order and hence every element except for the neutral one must have order p. By the previous exercise G must be abelian so that Z(G) = G and the index is one. 30 T CREUTZIG

Exercise 22. A group of order 91 is acting on a set of 38 elements. Show that the group action has at least 3 ﬁxed-points. Solution. 91 is divided by 1,7,13,91, so the index of any isotropic subgroup can either be 1,7 or 13. Let n1 be the number of ﬁxed-points, n7 be the number of orbits of length 7 and n13 the number of orbits of length 13. Then

38 = n1 + 7n7 + 13n13 We have a few cases

(1) n13 = 2, then 12 = n1 + 7n7 and hence n1 in {5,12}; (2) n13 = 1, then 25 = n1 + 7n7 and hence n1 in {4,11,18,25}; (3) n13 = 0, then 38 = n1 + 7n7 and hence n1 in {3,10,17,24,31,38}; so that 3 is the minimal possible solution for n1.

11. SYLOW’S THEORY We will study so-called p subgroups of a group G. Definition 13. Let p be a prime number. A finite group H is called p-group if the order of H is a power of p, that is |H| = pk for a positive integer k. Theorem 25. Let G be a finite abelian group G and let the prime number p be a divisor of the order of G. Then there exists a g in G with ord(g) = p. Proof. We prove this theorem by induction for the order of G. If |G| = 1 the statement holds. Let |G| = N > 1 and let the theorem be true for all abelian groups of smaller order. Take h 6= e in G and set m = ord(h). If p|m, then hn with n = m/p is an element of order p in G. If m is not dived by p then we consider the cyclic group < h > generated by h and the factor group G/ < h >. Since |G| = | < h > ||G/ < h > | the order of the factor group must be an integer multiple of p. By our induction hypothesis, there exists g in G with ord(g < h >) = p. Let n = ord(g), then (g < h >)n = gn < h >=< h > and hence p|n. As in the beginning of this proof gr with r = n/p is an element of order p in G. The theorem of Cauchy generalizes this situation to the non-ableian case Theorem 26. Let G be a finite group G and let the prime number p be a divisor of the order of G. Then there exists a g in G with ord(g) = p. Proof. We again prove this statement by induction for the group order |G| = N. The case N = 1 is clear, so let N > 1 and let p be a divisor of N. If G contains a proper subgroup H whose order is divisible by p, then by our induction hypothesis H and thus also G has an element of order p. We can thus assume that G has no proper subgroup whose order is divisible by p. The centralizer CG(g) of an element g in G is a proper subgroup of G if g is not in the center Z(G) of G. Hence p doesnot divide the order of the centralizer of g. Since |G| is divisble by p also the index [G : CG(g)] : must be divisible by p. The class equation of G is

|G| = |Z(G)| + ∑ [G : CG(xi)]. i∈I xi∈/Z(G)

Here the xi represent the conjugacy classes in G. We know that all summands (except Z(G)) in this equation are divisible by p. This can only be true if also |Z(G)| is divisble by p. The center MATH 328 31 is a subgroup, but the only subgroup whose order is divisble by p is G itself. Hence Z(G) = G and G is abelian and the theorem follows by the previous one. This theorem tells us that a finite group G is a p-group if and only if the order of every element is a power of p. We already knew that the order of a group element must divide the group order. Now, we have also seen that for every prime factor appearing in the group order an element of that order exists in the group. Lemma 27. Let p be prime and G a finite p-group. If G acts on a finite set X, then the fixed-point set X G satisfies |X G| = |X| mod p. Exercise 23. Prove this Lemma. (Use results of the group action section) Solution. The order of X satisfies G |X| = |X | + ∑ [G : Gxi ]. i∈I G xi∈/X

Every index [G : Gxi ] is divisible by p since Gxi 6= G and G is a p-group. Hence modulo p this equation becomes |X G| = |X| mod p. Theorem 28. Let G 6= {e} be a finite p-group, then the center Z(G) is not the trivial one element subgroup. Exercise 24. Prove this theorem. (Use the previous Lemma) Solution. We can apply the previous lemma with X = G and the operation conjugation. Then Z(G) = G mod p. Since G 6= {e} the center must be divisible by p and hence is non-trivial. Definition 14. Let G be a finite group of order pmq with q and p co-prime and p prime. A subgroup S of G is called p-Sylowgroup of G if |S| = pm. Theorem 29. Let G be a finite group of order pmq, with p and q co-prime to each other and p prime. Then (1) For every k, with 1 ≤ k ≤ m there exists at least one subgroup in G of order pk. (2) If H is a p-subgroup of G and S a p-Sylowgroup of G, then there exists a g in G with H ⊂ gSg−1. (3) Let s be the number of distinct p-Sylowgroups of G, then s|q and s = 1 mod p. Proof. The statements are clear for m = 0. Let m > 0. (1) We again prove this statement via induction for the group order. Consider the action of G on itself via conjugation, then the class equation is

|G| = |Z(G)| + ∑ [G : CG(xi)]. i∈I xi∈/Z(G)

if p does not divide the order of Z(G), then at least one of the indices [G : CG(xi)] is m ′ ′ neither divisble by p. Hence |CG(xi)| = p q for some q coprime to p. By induction k hypothesis CG(xi) has an element of order p for each 1 ≤ k ≤ m. If p divides the order 32 T CREUTZIG

of Z(G) then we have proven that Z(G) contains an element g of order p. < g > is a normal subgroup of G and |G/ < g > | = qpm−1. By induction hypothesis there is a subgroup U of G/ < g > of order pk−1. U is of the form V / < g > for some subgroup V of G and hence |V| = |U|| < g > | = pk−1 p = pk. (2) Let H be a p-subgroup and S a p-Sylowsubgroup of G. H acts on the set X = G/S of left cosets of S in G via H × X −→ X, (h,gS) 7→ hgS. The cardinality of X is |X| = |G|/|S| = q. We are in the situation of the previous Lemma, hence |X H| = |X| = q mod p. Since q and p are co-prime to each other the order X H cannot be divisible by p. Espe- cially X H is non-empty and there exists g in G with gS in X H, hence hgS = gS for all h in H and thus g−1hg in S so that H ⊂ gHg−1. (3) Let S be a p-Sylowgroup of G. Let X be the set of all p-Sylowgroups of G. The second statement of this theorem implies that all p-Sylowgroups are conjugate to S. G acts on X via conjuagtion. We are in the situation of Theorem 23 with just one orbit. The fixed- point subgroup is the normalizer of S in G, hence s := |X| =[G : NG(S)]. The index satisfies q =[G : S]=[G : NG(S)][NG(S) : S] and hence s|q. Consider the operation of the group S on X via conjugation. We claim that S is the only fixed-point. If our claim is true, then the previous Lemma implies that |X| = s = 1 mod p. Let S′ be a fixed-point, then S is contained in the normalizer of S′, ′ ′ −1 ′ S ⊂ NG(S ) = {g ∈ G|gS g = S }. ′ We claim that if a p-group H of G is contained in the normalizer NG(S ) of a p- Sylowgroup of G, then H ⊂ S′. The reason is as follows: S′ is a normal subgroup of its normalizer, and hence HS′ is a subgroup of the normalizer. For this situation we have earlier proven the isomorphism HS′/S′ =∼ H/(H ∩ S′). But the latter is a factor group of a p-group and hence a p-group. The index [(HS)′ : S′] divides the index [G : S′] = q, so it is not divisible by p. This can only be possible if HS′ = S′ and hence H ⊂ S′. For our situation this means S ⊂ S′, but since both groups have the same cardinality they must coincide. Hence, we have proven the claim that S is the only fixed-point.

Corollary 30. the p-Sylowgroups of a group G form a class of conjugate subgroups. Corollary 31. If G has only one p-Sylowsubgroup, then this must be a normal subgroup. Remark. If G is not a finite group, then a subgroup S is called a p-Sylowsubgroup of G if S is maximal in the set of all p-subgroups of G. Such a group exists by the Lemma of Zorn: The set X of all p-subgroups is non-empty since {e} is in X. A chain of p-subgroups (Hi)i∈I has in X the p-group Hi [i∈I as upper bound. So that X has at least one maximal element. The number s of p-Sylowgroups is then either infinite or it satisfies s = 1 mod p. Theorem 32. Let p,q be prime with p < q and p 6|(q − 1). Then every group G of order pq is cyclic, that is G is isomorphic to Z/pqZ. MATH 328 33

Proof. Let S be a p-Sylowsubgroup and U a q-Sylowsubgroup of G. Then S ∩U = {e}. Let s be the number of p-Sylowsubgroups and r the number of q-Sylowsubgroups. Then r = 1 mod q, r|p and s = 1 mod p, s|q. Since p < q it follows that r = 1, so that U is normal in G. For s we either have s = q or s = 1. If s = q, then q = 1 mod p and hence p|(q − 1), a contradiction. Hence s = 1 and S is also normal. It follows that SU is a subgroup of G and hence G = SU = S ×U, since the product of two normal cyclic subgroups is abelian. Let g be a generator of S and h a generator of U, then < gh >= G since ord(gh) = pq. Exercise 25. Let p be a prime number. Show that every group of order p2 is abelian.

Solution. Let n1 be the order of Z(G), n2 be the number of G-orbits (on itself via conjugation) 2 of order p, and n3 the number of orbits of order p . Then the class equation is 2 2 p = n1 + pn2 + p n3.

The identity e is in the center, hence n1 ≥ 1, so that n3 must be zero. It follows that n1 must 2 be divisible by p, and since Z(G) is a subgroup of G its order must divide p . If n1 = p, then 2 G/Z(G) is abelian by Exercise 20. If n1 = p , then Z(G) = G and hence G is abelian. Exercise 26. Let G be a group with |G| = 20 or |G| = 100. Show in both cases that G has a unique and hence normal 5-Sylowgroup. Solution. Let s be the number of 5-Sylowsubgroups of G. In both cases s|4 and s = 1 mod 5, hence s = 1. Exercise 27. Let p be prime and G a ﬁnite p-group, that is |G| = pk for some k. Let H be a normal subgroup of G of order p. Show that H ⊂ Z(G). (Hint: This statement can be proven by induction for k. ) Solution. We prove this statement by induction for k. The case k = 1 is clear as G is then cyclic and hence abelian. Let k > 1 and the statement be true for all subgroups of order p of all p- groups of order at most pk−1. Since Z(G) contains the identity, the class equation tells us that the order of Z(G) must be pr for some 1 ≤ r ≤ k. The center is a normal subgroup, so we can consider the factor group G/Z(G). This group has order pk−r. Let h be a generator of H and g an arbitrary element of Z(G). Then (hg)p = hpgp = gp in Z(G), but also hZ(G) 6= Z(G). Hence hZ(G) has order p in the factor group. By induction hypothesis hZ(G) in Z(G/Z(G)). This means for every g in G, we have ghZ(G) = hgZ(G). In other words, there exists a in Z(G) with gha = hg. We have to show that a = e. But since H is normal in G, there exists s with hg = ghs, and hence gha = ghs implyimg that a = hs−1. If s 6= 1, then hs−1 and hence H is in the center, but if s = 1 then a = e and again H must be in the center. Exercise 28. Let G be a group of order 200. Show that G has an abelian normal subgroup. Solution. Let s be the number of 5-Sylowgroups, then s|8 and s = 1 mod 5. Hence s = 1. So the only 5-Sylowgroup is normal and its order is 52, hence by previous exercise it is abelian. Theorem 33. Let p and q be primes with p < q and q = 1 mod p. Let G be a group of order pq, then either G is isomorphic to Z/(pq)Z or to x y H = | x,y ∈ Z/qZ, xp = 1 mod q . 0 1 34 T CREUTZIG

This theorem can be proven using Chauchy’s theorem and that is work. I will not do that.

Theorem 34. Let p and q be two different prime factors appearing in |G|, where G is a ﬁnite group. Let there be only one p-Sylowsubgroup S in G and also only one q-Sylowsubgroup T in G. Then the elements of S commute with the elements of T .

Proof. Since p and q are co-prime, the intesection S ∩ T must be {e}. Both S and T are normal subgroups by Sylow’s theorem. Let g in S and h in T , then ghg−1h−1 = ghg−1 h−1 = g hg−1h−1 ∈ T ∩ S = {e}. So they commute.

Theorem 35. Let G be a group with |G| = 45, then G is abelian.

Proof. Let s be the number of 3-Sylow groups, and t the number of 5-Sylowgroups, then s|5, s = 1mod3, t|9, t = 1 mod5 hence s = t = 1. Let S be the unique 3-Sylowgroup and T be the unique 5-Sylowgroup. Then the elements of S and T commute with each other, but also with themselves, hence ST is abelian. Especially ST = S × T . The order of S × T is |S × T | = |S||T| = 45 and hence ST = G.

Theorem 36. Let G be a group with |G| = p2q, p and p prime and p < q and q 6= 1 mod p. Then G is abelian.

I don’t prove this theorem. The argument is the same as in the previous proof.

Theorem 37. Let p and q be prime and let G be a group with |G| = pq2, then G has a normal Sylowsubgroup.

Proof. Let s be the number of p-Sylowsubgroups and t the number of q-Slowsubgroups. We have to show that either s = 1 or t = 1 or both hold. Assume that this is not the case, since s|q2, s = 1 mod p, t|p, t = 1 mod q we have t = p and s = q or s = q2. We will show that these situations are impossible. So let s = q, then we have q = 1 mod p and p = 1 mod q. If q < p then the ﬁrst condition is impossible, if p < q then the second condition is impossible. Now, let s = q2. Then there are q2 p-Sylowgroups each of size p. Two such groups can only have trivial intersection. Hence, these q2 p-Sylowgroups consist out of q2(p − 1) + 1 distinct group elements. p and q Sylowgroups have also trivial intersection. While the intersection of two distinct q-Sylowgroups must be a proper subgroup and hence has at most q elements. There are p and hence at least two q- Sylowgroups, adding all these distinct group elements up, we get q2(p − 1) + 1 + q2 − 1 + q2 − q = q2 p + q2 − q > q2 p = |G| that is more than the group order pq2. A contradiction.

Classifying all groups of order 105 We will classify all groups of order 3 · 5 · 7 = 105. For this, we need a few Lemmata.

Lemma 38. Let N be a normal subgroup of a group G and S a normal p-Sylowgroup of N, then S is also normal in G. MATH 328 35

Proof. Let g in G be an arbitrary element. Then gSg−1 ⊂ gNg−1 = N. So gSg−1 is a subgroup of N with the same order as S. But there is only one such subgroup, namely S itself, since S is p-Sylowgroup. Lemma 39. Let G be a finite group and p the smallest prime factor of G. A subgroup of G of index p is normal in G. Proof. Let H be a subgroup of G of index p. We will construct a group homomorphism with kernel H, so that H must be normal. G acts on the set of left cosets M via left multiplication. This defines a permutation of the p-elements of M, and hence this map defines a group homomorphism from G to Sp. Let g be in the kernel of this map, then especially gH = H and hence g in H. Denote the kernel by K. Then G/K is isomorphic to a subgroup of Sp and hence its order has to divide p! by Lagrange’s theorem. On the other hand |G/K| =[G : K]=[G : H][H : K] = p[H : K] so that [H : K]|(p − 1)! But p is the samllest prime factor in the order of G and [H : K] is a factor of the order of G, so the only possibility is [H : K] = 1 and hence H = K. Lemma 40. Every group of order 105 has normal and hence unique 5-Sylow and 7-Sylow subgroups. Proof. We first prove that either the number s of 5-Sylowsubgroups is one or the number r of 7-Sylowsubgroups is one. We will then use that result to conclude that actually both must be one. We have s|21, s = 1mod5, r|15, r = 1 mod7. If s 6= 1, then s = 21 and if r 6= 1 then r = 15. Hence there are 21 · 4 = 84 elements of order 5 and 15·6 = 90 elements of order 7. But 90+84 > 105, so this situation is impossible. So either s or r (or both) must be one. We will now prove that G has a subgroup H of order 35. If s = 1, then there is a normal 5-Sylowsubgroup S. Let T be a 7-Sylowsubgroup, then ST is a subgroup of G of order at most 35. But ST has elements of order 7 and 5, so ST must have exactly order 35. If r = 1, the analogous argument again gives a subgroup of order 35. So, we have proven that there exists a subgroup N of order 35. By the previous lemma N is normal in G. It is also cyclic by Theorem 32 and hence every subgroup is normal in N. But by the first lemma of this subsection these subgroups must also be normal in G. We thus have proven that s = r = 1. Lemma 41. Let G be a group of order 105, then G is isomorphic to Z/5Z × H, where H is a group of order 21. Proof. Let R be a 3-Sylowsubgroup of G and let T be the normal 7-Sylowsubgroup of G. Then RT is a subgroup H of order 21. Let S be the normal 5-Sylowsubgroup of G, then SH has order 105 and hence SH = G. we have to show that SH = S × H. This means that elements of H commute with elements of S. Since S is normal H acts on S via conjugation. This defines a group homomorphism from H to Aut(S) =∼ (Z/5Z)∗ (the group of multiplicative units in Z/5Z). Since the order of H is 21 this homomorphism must be trivial, that is elements of H fix S pointwise, so they commute. Theorem 42. Up to isomorphism there are two groups of size 105. 36 T CREUTZIG

This follows with Theorem 33, and Theorem 33 even gives us the explicit form of H.

Exercise 29. Prove the following statements

(1) Let G be a group and let H1 and H2 be two subgroups of G, both of ﬁnite order. Let |H1| be the order of H1 and |H2| the order of H2 and |H1 ∩H2| the order of the intersection of these two subgroups. What is |H1 ∪ H2| in terms of the orders of these three subgroups? (2) Let G be a group and let H1 and H2 be two subgroups of G, bothoforder 8. Let H1 6= H2. What is the lower bound for |H1 ∪ H2|? (3) Let G be a group and H1,...,Hn a family of n pairwise distinct groups, each of order p. Let p ∈{7,11}. How many elements has n Hi? i[=1 3 (4) Let G be a group of order 616 = 2 · 7 · 11. Find all possible solutions for the number s2 of 2-Sylowsubgroups, s7 of 7-Sylowsubgroups and s11 of 11-Sylowsubgroups. (5) Prove that G is not simple. Hint: Provethat there is at least onenormal p-Sylowsubgroup. Assume that s2 6= 1,s7 6= 1,s11 6= 1 and ﬁnd a contradiction.

Solution.

(1) The order of H1 ∪H2 is the number of all elements of H1 plus the number of all elements of H2 minus the number of common elements, which is the number of elements in the intersection. Thus

|H1 ∪ H2| = |H1| + |H2|−|H1 ∩ H2|.

(2) The intersection of H1 ∩ H2 must be a proper subgroup of both. The maximal size of a proper subgroup is 4. Hence by the previous exercise

|H1 ∪ H2| = |H1| + |H2|−|H1 ∩ H2| ≥ 8 + 8 − 4 = 12. (3) Since all groups are distinct and theorder is primethe intersection of any of these groups is the trivial one element group. Hence n n Hi = ∑ |H − i| − 1 + 1 =(p − 1)n + 1. i=1 ! i[=1

(4) Let sp be the number of p-Sylowsubgroups of G. Then

s2|77, s2 = 1 mod2 ⇒ s2 ∈{1,7,11,77},

s7|88, s7 = 1 mod7 ⇒ s7 ∈{1,8,22},

s11|56, s11 = 1 mod 11 ⇒ s11 ∈{1,56}.

(5) By the previous problem if s2 6= 1,s7 6= 1,s11 6= 1, then s2 ≥ 7,s7 ≥ 8,s11 ≥ 56. There are thus at least 56 11-Sylowgroups having at least distinct (11 − 1)56 = 560 elements of order 11 (and one element of order one). There are at least 8 7-Sylowgroups having at least 8(7−1) = 48 distict elements of order 7. There are at least 2 (we don’t need more) 2-Sylowgroups having at least 11 elements whose order is 2r for some r in {1,2,3}. Hence, we have at least 560 + 48 + 11 + 1 = 620 MATH 328 37

elements. This is impossible as the group order is 616. So at least one of the sp has to be one. There is thus a normal subgroup and G cannot be simple.

12. SOLVABLE GROUPS We want to study the subgroup structure of groups, especially of its normal subgroups. For this we use normal series. Deﬁnition 15. Let G be a group. A normal series in G is a ﬁnite series of subgroups

{e} = G0 E G1 E G2 E ··· E Gn = G, such that each group Gi is normal in Gi+1. The quotient (or factor) groups Gi+1/Gi are called the factors of the series. The subgroups Gi are called terms of the normal series.

Note, that the Gi donot need to be normal in G, only in Gi+1. Every group has the trivial normal series {e} = G0 E G1 = G. Let H and G be two normal series of G. Then H is called a refinement of G if every term of G is also a term of H. We call H equivalent to G if there is a bijection of the set of factors of H and the set of factors of G, such that corresponding factors are isomorphic as groups. We want to compare different normal series of a given group. Lemma 43. Let U,V be subgroups of a group G, and let U ′ EU, V ′ EV be normal subgroups of U respectively V. Then U ′(U ∩V ′) EU ′(U ∩V ) and V ′(V ∩U ′) EV ′(U ∩V) and the factor groups have isomorphisms U ′(U ∩V )/U ′(U ∩V ′) =∼ (U ∩V )/(U ∩V ′)(U ′ ∩V) =∼ V ′(V ∩U)/V ′(V ∩U ′). Proof. The problem is symmetric in U and V, so we just have to prove the first isomorphism. Since U ∩V us a subgroup of U and since U ′ is a normal subgroup U ′(U ∩V ) is also a subgroup of U. The group U ′ is normal in U and hence also in the subgroups of U containing U ′, especially in U ′(U ∩V ). We thus have the factor group U ′(U ∩ V )/U ′. Further, since U ′ is a normal subgroup in U, U ′ ∩V must be a normal subgroup in U ∩V (and also U ∩V ′). Note that U ′ ∩U ∩V = U ′ ∩V . We are thus in the situation of the first isomorphismtheorem (Theorem 19), which tells us that H/(N ∩ H) =∼ NH/H for N a normal subgroup and H a subgroup of a group G. Setting U ′ = N and H = U ∩V , we get U ′(U ∩V )/U ′ =∼ (U ∩V )/(U ∩V ∩U ′)=(U ∩V )/(V ∩U ′). (12.1) Now, we apply the second isomorphism theorem, which was: let N,M normal subgroups of G and N also a normal subgroup of M, then (G/N)/(M/N) =∼ G/M. Let N =(U ′ ∩V ), M =(U ∩V ′)(U ′ ∩V) and G = U ∩V , then the assumptions of the second isomorphism theorem are satisfied and hence U ∩V/(U ∩V ′)(U ′ ∩V ) =∼ U ∩V/(U ′ ∩V) / (U ∩V ′)(U ′ ∩V )/(U ′ ∩V ) . The left-hand side is a factor group of (U ∩ V)/(U′ ∩V ) so that with (12.1) there exists a surjective group homomorphism α : U ′(U ∩V )/U ′ −→ U ∩V /(U ∩V ′)(U ′ ∩V ). 38 T CREUTZIG

The kernel of this map is U ′(U ∩V ′)/U ′, so that we get the isomorphism U ′(U ∩V )/U ′ / U ′(U ∩V ′)/U ′ =∼ U ′(U ∩V ) / U ′(U ∩V ′) =∼ U ∩V /(U ∩V ′)(U ′ ∩V). Theorem 44. Let G and H be two normal series of a group G. They have equivalent reﬁnements. Proof. Given two normal series

G : {e} = G0 E G1 E G2 E ··· E Gn = G and H : {e} = H0 E H1 E H2 E ··· E Hm = G.

Define Gij := Gi(Gi+1 ∩ Hj) and Hij := Hj(Gi ∩ Hj+1) for all i and j. We apply the previous ′ ′ lemma with U = Gi+1,U = Gi,V = Hj+1,V = Hj, so that Gij E Gi, j+1 and Hij E Hi+1, j and there is an isomorphism of factor groups ∼ Gi, j+1/Gij = Hi+1, j/Hij for all i and j. Since Hm = Gn = G it holds that Gim = Gi+1,0 and also Hnj = H0, j+1 as well as G = Gn−1,m = Hn,m−1. We have thus constructed a refinement of G given by the Gij and a refinement of H given by the Hij that give equivalent normal series of G. A composition series of a group G is a normal series of G that has no proper refinement. A finite group has always a composition series as refining a given normal series must terminate after a finite number of steps (an obvious upper bound of number of steps is |G|). There are two consequences of the previous theorem: Theorem 45. A normal series G of a group G is a composition series if and only if all factors are simple groups. Exercise 30. Prove this theorem. (You can use the fact that if M is a normal subgroup of G and N/M is a normal subgroup of G/M, then N has to be a normal subgroup of G)

Solution. If one of the factors, say Gi+1/Gi is not simple, then it possesses a non-trivial normal subgroup N/Gi with Gi ( N ( Gi+1. N is normal in Gi+1. Adding N in the normal series G gives a non-trivial reﬁnement. If the normal series has a non-trivial reﬁnement, then there exists an index i with a subgroup K satisfying Gi E K E Gi+1, so that the factor group K/Gi+1 is a non-trivial This gives us another characterization of a composition series: A normal series

G : {e} = G0 E G1 E G2 E ··· E Gn = G in G is a composition series if for every i in 0,1,...,n the group Gi is maximally normal in Gi+1. Another consequence of Theorem 44 is Theorem 46. (Jordan H¨older) Let G be a group with composition series G, then every other compositon series of G is equivalent to G. Example 9. A few examples: (1) Let n be a positive integer, the cyclic group Z/nZ is a ﬁnite group and hence it has a composition series. Let n = m1m2 ...mk be a decomposition of n into positive factors. MATH 328 39

Then there is a corresponding normal series

{e} = m1m2 ...mkZ/nZ E m2 ...mkZ/nZ E ··· E mk−1mkZ/nZ E mkZ/nZ E Z/nZ.

The factors are isomorphic to Z/miZ with 1 ≤ i ≤ k. If we choose all mi to be prime, then this normal series is a composition series since all factors Z/pZ for p prime are simple. The equivalence of different composition series corresponds to the uniqueness of prime factor decomposition up to order. (2) If we specialize the previous example to n = pr for some positive integer r, then there is a unique composition series as there is a unique prime factor decomposition (no matter which order). (3) the group Z of the integers with operation addition is inﬁnite dimensional. It has no composition series, as every non-trivial subgroup is an inﬁnite cyclic group and hence not simple. Thus the smallest term not equal to {0} of a normal series is not simple and hence neither is the corresponding factor. (4) For n = 3 is {e} E A3 E S3 a composition series in S3, since the factors S3/A3 =∼ Z/2Z and A3 =∼ Z/3Z are simple. S4 has more than just one composition series. For n ≥ 5, we have proven that An is simple and hence {e}EAn ESn is a composition series (since Sn/An =∼ Z/2Z is also simple).

Deﬁnition 16. Let G be a group. A normal series

G : {e} = G0 E G1 E G2 E ··· E Gn = G is called abelian normal series, if all its factors are abelian groups. G is called a solvable group if G has at least one abelian normal series.

Example 10. Some examples are: (1) Every abelian group is solvable (2) The group a b G = ∈ GL (k)|c = 0 c d 2 is solvable (k is a ﬁeld). The product of two group elements is a b a b a a a b + b d 1 1 2 2 = 1 2 1 2 1 2 0 d 0 d 0 d d 1 2 1 2 so that we see that the map a b α : G −→ k∗ × k∗, 7→ (a,d) 0 d is a group homomorphism (k∗ is the multiplicative subgroup of k consisting of all invertible elements in k). The kernel is 1 b G := ker α = 1 0 1 It is a normal subgroup of G and it is isomorphic to the abelian additive group of k. Hence {1} E G1 E G is an abelian normal series. 40 T CREUTZIG

(3) Thegroup S4 is solvable. An abelian normal series is given by G1 = {e,(12)(34),(13)(24), (14)(23)} (the Klein four group), G2 = A4 and G3 = S4.

Theorem 47. A subgroup of a solvablegroup is solvableand a homomorphicimage of a solvable group is also solvable.

Proof. Let G be a solvable group and let

G : {e} = G0 E G1 E G2 E ··· E Gn = G be an abelian normal series. Let U be a subgroup of G. Then each Ui := U ∩ Gi is normal in Ui+1. The ﬁrst isomorphosm theorem tells us that ∼ U ∩ Gi+1/(U ∩ Gi) = U ∩ Gi+1/(U ∩ Gi+1 ∩ Gi) = (U ∩ Gi+1)Gi/Gi ⊂ Gi+1/Gi and hence as a subgroup of an abelian group it must be abelian itself. Let ϕ : G −→ G′ be a surjective homomorphism. We have to show that G′ is solvable. For each i, the subgroup ϕ(Gi) is normal in ϕ(Gi+1) since the homomorphic image of a normal subgroup is normal in the image. The thus deﬁned normal series is abelian, since the map g 7→ ϕ(g)ϕ(Gi) induces a surjective group homomorphism Gi+1/Gi −→ ϕ(Gi+1)/ϕ(Gi) and hence the factors must be abelian as the homomorphic image of an abelian group is abelian.

Theorem 48. Let G be a ﬁnite solvable group, then G has an abelian normal series

G : {e} = G0 E G1 E G2 E ··· E Gn = G such that the factors Gi+1/Gi are cyclic of prime order.

Proof. Let G : {e} = G0 E G1 E G2 E ··· E Gn = G be a composition series of G. That means all factors are simple groups. By the previous theorem all factors are solvable. We thus have to prove that all solvable simple groups are cyclic of prime order. Let H be a simple solvable group. Since the only normal subgroups of H are {e} and H the only (and hence abelian) normal series of H is {e}EH. Hence, H is abelian. For every h in H is < h > a proper normal subgroup of H. Hence H =< h > and the order |H| must be prime.

Deﬁnition 17. Let G be a group, then [a,b] := aba−1b−1 is called the commutator of a and b in G. The commutator subgroup or derived subgroup is D(G) = aba−1b−1|a,b ∈ G .

The commutator group is normal, since −1 −1 gaba−1b−1g−1 = gag−1 gbg−1 gag−1 gbg−1 . The factor group is abelian since ghD(G) = ghh−1g−1hgD(G) = hgD(G). On the other hand, if N is a normal subgroup of G, such that the factor group is abelian, then abN = baN MATH 328 41 and hence aba−1b−1 in N. In other words, D(G) ⊂ N. Deﬁne D0(G) = G and inductively Dn(G) = D(Dn−1(G)), then we get the derived series of G G = D0(G) D D1(G) D D2(G) D D3(G) D .... All factor groups are abelian. m Theorem 49. A group G is solvable if and only if there exists m in Z≥0 with D (G) = {e}. m m−i Proof. If D (G) = {e}, then Gi = D (G) form an abelian normal series. Let G : {e} = G0 E G1 E G2 E ··· E Gn = G be an abelian normal series. Then for all r > 0 it holds that D(Gr) ⊂ Gr−1, since Gr−1 is a i normal subgroup of Gr with abelian factor group. We claim that D (G) ⊂ Gn−i for all i ≤ n. 0 we prove this claim by induction. The case n = 0 is clear as D (G) = G = Gn. Assume that i D (G) ⊂ Gn−i then i+1 i D (G) = D(D (G)) ⊂ D(Gn−i) ⊂ Gn−i−1 = Gn−(i+1). So the claim follows by induction. Especially it follows that n D (G) ⊂ Gn−n = Go = {e}.

Exercise 31. Let p be a prime number and let G be a finite p-group. Let H be a proper subgroup of G. Prove that H is a proper subgroup of its normalizer NG(H) in G. (Hint: Consider the two cases Z(G) 6⊂ H and Z(G) ⊂ H). Solution. If Z(G) 6⊂ H, then there exists at least one g ∈ Z(G) and G 6∈ H. But g must be in the normalizer, since ghg−1 = h for every h in H. If Z(G) ⊂ H, then let the group order of G be pn and we prove the statement by induction for n. If n = 1, then G is cyclic of prime order and has no non-trivial subgroups. Now, let the statement be true for all p-groups of order at most pn−1. Since Z(G) ⊂ H and the center is normal in both G and H, the factor group H/Z(G) is a proper subgroup of G/Z(G). By Theorem 28 its order is at most pn−1. By induction hypothesis, there exists g in G with gZ(G) 6∈ HZ(G) and hence g 6∈ H with gHg−1Z(G) = HZ(G). This can only be true if g is in the normalizer of H. Exercise 32. Let p be a prime number and let G be a finite p-group of order pn. Show that a composition series of G has n distinct subgroups 6= {e}. Show that G is solvable. Solution. By Sylow’s theorem a group of order pn has a subgroup of order pn−1. By Lemma 39 this subgroup is normal. Set G = Gn and this subgroup call it Gn−1. Then we can inductively n−i by the same argument define subgroups Gi such that Gi is normal in Gi+1 and has order p . This gives a normal series since G0 = {e}. It has the desired number of subgroups and it is a composition series since each factor is cyclic of order p hence simple. Especially, we see that every factor is abelian. Hence G is solvable. Exercise 33. Let p be a prime number and let G be a finite p-group of order pn. Let H be a subgroup of G. Show that there is a composition series that contains H. Solution. We construct a normal series of G that contains H. For this let

{e} = H0 E H1 E ··· E Hm = H 42 T CREUTZIG be a chosen normal series of H. Define Gi = Hi. We extend this normal series to a normal series for the normalizer of H in G. By Exercise 31 H is a proper subgroup in NG(H). Define Gm+1 = NG(H) and more generally define Gm+r = NG(Gm+r−1). By Exercise 31 in each of these steps, we extend the normal series to a larger subgroup of G, hence this procedure has to terminate after at most n steps, and we get a normal series of G containing H. Any refinement of this normal series to a compositon series will contain H .

13. ABELIAN GROUPS Usually, we have denoted group operation by multiplication. In this section, we aim to classify ﬁnitely generated abelian groups. We will work with the integers, their subgroups and factor groups. It is thus more natural to denote group operation by addition. Especially, we will write g + g + ···+ g = ng for adding g to itself (n − 1)-times. This means the group generated by an element g in a group G is hgi = {ng|n ∈ Z} = Zg.

Definition 18. An abelian group is called finitely generated if there is a finite subset of elements g1,...,gn in G such that they generate G:

G = hg1,...,gni = Zg1 + ··· + Zgn. The map n Z −→ G, (m1,...,mn) 7→ m1g1 + ··· + mngn is then a surjective group homomorphism. Let K be the kernel of this map. K must then be a subgroup of Zn and we have the isomorphism G =∼ Zn/K. On the other hand, if ϕ : G −→ G′ is a surjective group homomorphism. Then every element in G” can be expressed in terms of ϕ(g1),...,ϕ(gn) if g1,...,gn are a generating set of G. In other words, if ϕ : Zn −→ G is a surjective group homomorphism, then G must be ﬁnitely generated. Our strategy to classify ﬁnitely generated abelian groups is thus to understand subgroups K of Zn. Let us start with an example that illustrates the strategy.

Example 11. Consider the case n = 2 and the subgroup K = {(2n,2n)|n ∈ Z} of Z2. The group Z2 is generated by the two elements (1,1) and (0,1). Consider the automorphism of Z2 induced by the map (1,1) 7→ (1,0), (0,1) 7→ (0,1). Under this automorphism the subgroup K is mapped to the subgroup L = {(2n,0)|n ∈ Z} = 2Z ⊕{0}. Hence Z2/K =∼ Z2/L =∼ Z/2Z ⊕ Z. The symbol ⊕ is called the direct sum. It denotes the direct product of two groups and indicates that we choose the symbol + for the group operation.

We aim to generalize this example. MATH 328 43

Proposition 50. Every subgroup of Zn is finitely generated by at most n generators. Proof. We proof this statement by induction on n. The case n = 1 is clear as every subgroup of Z is either isomorphic to Z or {0}. Let now n > 1 and the statement be true for all subgroups m n of Z if m < n. Let K be a subgroup of Z . Every element of K is of the form (m1,...,mn) for some integers mi. Let F be the set of first components m1 of elements of K. Then F must be a subgroup of Z. Define f to be either f = 0 if F = {0} or f the smallest positive element of F if F 6= {0}. Choose an element ( f ,m2,...,mn) in K. Then every other element (r1,...,rn) in K must satisfy r1 = sf for some integer s (if not then by the Euclidean algorithm there exists 1 ≤ r < f and t in Z with r1 = tf + r and hence r in F. Contradicting the minimality of f ). It follows that (r1,...,rn) = s( f ,m2,...,mn)+(0,r2 − sm2,...,rn − 2mn). ′ n−1 The elements of the form (0,r2 − sm2,...,rn − 2mn) form a subgroup K of Z . By induction hypothesis K′ is finitely generated by at most n − 1 elements and hence K is finitely generated by at most (n − 1) + 1 = n elements. Theorem 51. Let K be a subgroup of Zn. Then

K =∼ d1Z ⊕···⊕ drZ for some positive integers di with di|di+1 and r ≤ n. Especially n Z /K =∼ Z/d1Z ⊕···⊕ Z/drZ ⊕ Z ⊕···⊕ Z. Here, the summand Z appears (n − r) times.

Proof. We have just proven that K is ﬁnitely generated. Let k1,...,km be a set of such genera- n tors. Each ki is an element of Z and so it is of the form ki =(ki1,...,kin). We can thus use the components of the generators to deﬁne a matrix

k11 ... k1n . .. . A :=  . . .  k ... kmn  m1  with integer entries. Applying row and column operations in Z to this matrix now corresponds to changing generators of K and Zn as follows (1) Interchanging two rows i and j corresponds to reordering the generators of K. The i-th generator becomes k j and vice versa. (2) Multiplying row number i by minus one corresponds to changing the generator ki to −ki. (3) Adding row i to row j replaces k j by k j + ki. The analogous column operations change the generators of Zn in the analogous way. We will prove in a moment that these row and column operations allow one to bring the matrix A into Smith Normal Form S, that is

d1 ... 0 0 ... 0 ......  ......  0 ... dr 0 ... 0 S =    0 ... 0 0 ... 0    ......   ......     0 ... 0 0 ... 0     44 T CREUTZIG with di|di+1.

We thus have classiﬁed all ﬁnitely generated abelian groups up to isomorphism.

Corollary 52. Let G be a ﬁnitely generated abelian group. Then there exist positive integers di with di|di+1 and 0 ≤ r ≤ n such that

G =∼ Z/d1Z ⊕···⊕ Z/drZ ⊕ Z ⊕···⊕ Z.

It remains to explain the Smith Normal Form. Proving its existence is constructive and pro- vides an algorithm. It goes as follows: We will present a method to use elementary row and column operations (in Z) to reduce A to a matrix of form d 0 0 A 2 where d divides every entry of A2. Then applying the same procedure to A2 gives a smaller matrix A3 and so on, so that we can inductively bring A into Smith Normal Form. We call the absolut value of a non-zero entry its size, and we call the upper left entry the pivot. If A is the zero matrix, then we are done. Otherwise: In each step, If there is an element of smaller but non-zero size than the pivot, we move that element to the pivot by changing rows and columns. Since every entry is integer this has to be done only a finite number of times. Step 1: Repeatedly add or substract the pivot from all other entries in the first row or first column in such a way that the size of these entries gets reduced. Each entry either becomes zero or smaller than the pivot. In the latter case move this entry to the pivot. Since the size of the pivot can only be reduced a finite number of times. After a finite number of steps all entries except for the pivot in the first row and in the first column must be zero. Step 2: If the pivot divides every entry of A2, then we are done. Otherwise choose an entry of A2 that is not divided by the pivot. Add the corresponding row to the first row and go back to step one. Performing step one will again redice the size of the pivot. So that after a fintie number of performing Step 1 and Step 2 we must terminate.

Example 12. Let Z3 −→ G be a group homomorphismwith kernel K = h(6,3,3),(4,5,7),(3,2,2)i. Show that G =∼ Z/6Z. We have to bring the matrix 6 3 3 A = 4 5 7   3 2 2   MATH 328 45 in Smith Normal Form. We apply the algorithm step by step (Ci denotes the i-th row, Ri the i-th column) 6 3 3 3 2 2 2 3 2 R ↔ R C ↔ C C → C −C A = 4 5 7 −−−−−→1 3 4 5 7 −−−−−→1 2 5 4 7 −−−−−−−→3 3 1       3 2 2 6 3 3 3 6 3 2 3 0   2 1 0  1 20 C → C −C C ↔ C C → C −2C 5 4 2 −−−−−−−→2 2 1 5 −1 2 −−−−−→1 2 −1 5 2 −−−−−−−−→2 2 1       3 6 0 3 3 0 3 30       1 00 R2 → R2+R1 1 0 0 10 0 R → R −3R C ↔ C C → C −3C −1 7 2 −−−−−−−−→3 3 1 0 7 2 −−−−−→2 3 02 7 −−−−−−−−→3 3 2       3 −3 0 0 −3 0 0 0 −3 10 0  1 0 0  1 0 0  C ↔ C C → C −2C R → R +3R 02 1 −−−−−→2 3 0 1 2 −−−−−−−−→3 3 2 0 1 0 −−−−−−−−→3 3 2       0 0 −3 0 −3 0 0 −3 6 1 0 0      0 1 0   0 0 6 It follows that K =∼Z ⊕ Z ⊕ 6Z and hence G/K =∼ Z/Z ⊕ Z/Z ⊕ Z/6Z =∼ Z/6Z.

Exercise 34. Let K = h(21,63),(105,0)i be a subgroup of Z2. Show that Z2/K =∼ Z/21Z ⊕ Z/315Z.

Solution. In analogy to the previous example we compute

21 63 C → C −3C 21 0 R → R −5R 21 0 A = −−−−−−−−→2 2 1 −−−−−−−−→2 2 1 , 105 0 105 −315 0 −315 so that K =∼ 21Z ⊕ 315Z and the claim follows.

Exercise 35. Let K = h(12,18,0),(−6,0,36),(−6,24,30)i be a subgroup of Z3. Show that Z3/K =∼ Z/6Z ⊕ Z/6Z ⊕ Z/306Z.

Solution. Again, in analogy to the previous example we compute

12 18 0 −6 0 36 −60 0 R3 → R3−R1 R ↔ R C → C +6C R → R +2R A = −6 0 36 −−−−−→2 1 12 18 0 −−−−−−−−→3 3 1 12 18 72 −−−−−−−−→2 2 1       −6 24 30 −6 24 30 −6 24 −6 −60 0  −60 0  −6 0 0  R ↔R C ↔C C →C +4C 0 18 72 −−−−→3 2 0 24 −6 −−−−→3 2 0 −6 24 −−−−−−−→3 3 2       0 24 −6 0 18 72 0 72 18 −60 0  −60 0   C →R +12R 0 −6 0 −−−−−−−−→3 3 2 0 −6 0 ,     0 72 306 0 0 306 so thatK =∼ 6Z ⊕ 6Z ⊕306Z and the claim follows.  46 T CREUTZIG

14. APPLICATIONS As a fun end to this course I’d like to give a few applications of group theory that at the same time reviews important aspects of this course. Group Actions and Number Theory Lemma 53. Let p be a prime number and let G be a ﬁnite group, then |G|p−1 = |{g ∈ G|gp = e} mod p. Proof. We consider the set

X := (g1,...,gp)|g ∈ G, g1g2 ...gp = e .

Let g1,...,gp−1 be p − 1 arbitrary group elements. Then there is a unique gp, namely gp = −1 p−1 (g1 ...gp−1) , such that (g1,...,gp) in X. Hence the cardinality of X is |G| . We also have −1 that g1 =(g2 ...gp) and hence also (g2,...,gp,g1) is in X. In other words a cyclic permutation of the entries maps an element of X to another element of X. This gives an action of the group Z/pZ on X. The ﬁxed-points of this action are exactly the elements of the form {g ∈ G|gp = e}. Hence the Lemma follows from Lemma 27. Theorem 54. (Fermat) Let p be prime and n 6= 0 mod p, then np−1 = 1 mod p. Proof. First, we observe that (−1)p−1 = 1 mod p, since for p odd, p − 1 is even and hence (−1)p−1 = 1. While for p = 2, we have (−1)2−1 = −1 = 1 mod 2. We can thus restrict to the case n > 0. Let G = Z/nZ, then the previous Lemma tells us that np−1 = |{m ∈ Z/nZ| pm = 0}| mod p, but since by assumption n is co-prime to p the only element in Z/nZ with this propery is the neutral element 0. Theorem 55. (Wilson) Let p be prime, then (p − 1)! = −1 mod p.

Proof. Consider the previous Lemma with G = Sp, then p 0 = |{σ ∈ Sp |σ = e}| mod p. But all elements of order divisible by p are the p-cycles (order p) and the neutral element e (order one). These are (p − 1)! + 1 elements, so that the theorem follows. Cryptography using Cyclic Groups This is an application that your fellow student Stephen Romansky explained to me (and the example is coming from the correspondig wikipedia page). Assume that two person need a shared secret key to encrypt information. Call these people Ann and Bob. They decide for a ﬁnite cyclic group Z/23Z an element of the group 5 that has a multiplicative inverse, 14. They want to use this information to share a common key that even though all information is transmitted in public, the key won’t. For this each of the two chooses an integer. Ann chooses MATH 328 47 a = 6 and computes 56 = 8 mod 23. Bob chooses b = 15 and computes 515 = 19 mod 23. They exchange the numbers 8 and 19, and compute 815 = 2 mod 23 (Bob) respectively 196 = 2 mod 23 (Ann). Both of them now share the secret number 2 which they can use for sharing secret information. This is actually a secret information, as only the numbers g = 5, p = 23,ga = 8 mod 23, gb = 19 mod 5 are publicly known. In practice this secret is of course easy to encode for a cyclic group of small order. However, if the group order is prime p, with p ∼ 10300 and a and b are of order 10100, then even knowing g, p and ga,gb mod p the computer time for solving for a or b is too long. Lie Groups Lie groups are a type of groups that appear in many applications, a famous one being quantum physics. There name is due to the mathematician Sophus Lie.

Deﬁnition 19. Let G be a n-dimensional real manifold, and let {Uα } be a set of open subsets of G that cover G, such that each open subset looks like Rn in the sense that there is a bijective map n φα : Uα → R that is bi-continuous. In other words both φα and its inverse are continuous maps. A transition map is then a map −1 φαβ = φβ ◦ φα φα (Uα ∩Uβ ) from φ U ∩U to φ U ∩U . A manifold is called smooth if all transition maps are α α β β α β smooth, that is inﬁnitely differentiable. A Lie group G (over R) is a smooth manifold and also a group, such that group multiplication and inversion are smooth maps.

The group GLn(C) is a Lie group, the classical ones are: Example 13. Interesting subgroups of GL(n,C) are (1) The orthogonal (compact) group O(n), t O(n) = g ∈ GLn(R)|gg = I (2) The unitary (compact) group U(n), ∗ t U(n) = g ∈ GLn(C)|g(g ) = I (3) The special (compact) unitary group SU(n), SU(n) = {g ∈ U(n)|detg = 1} (4) The symplectic group (compact) Sp(2n) t Sp(2n) = g ∈ GL2n(R)|gJg = J for 0 −I J = n I 0 n Deﬁnition 20. A Lie algebra g is a vector space g together with a bilinear operation [ , ] : g × g → g satisfying [U,T] = −[T,U] (antisymmetry) 48 T CREUTZIG and [U,[S,T]]+[T,[U,S]]+[S,[T,U]] = 0 Jacobiidentity for all U,S,T in g.

Lie groups and Lie algebras are very closely related. The important map is the exponential map: 1 1 exp : Mat (C) → GL(n,C), exp(X) = I + X + X 2 + X 3 + ... (14.1) n 2 6 This series converges for every X (using the existence of Jordan normal form). The classical Lie algebras given in the next example are mapped to the classical Lie groups via the exponential map.

Example 14. The important examples are

(1) TheLiealgebraof O(n,R) is so(n,R). It consists of all matrices X satisfying X +Xt = 0. (2) The unitary Lie algebra is denoted by u(n,R) and consists of all complex matrices X satisfying X ∗ + Xt = 0. (3) The special unitary Lie algebra is denoted by su(n,R) and consists of all traceless unitary matrices. (4) The symplectic Lie algebra is denoted by sp(2n,R) and consists of all matrices X satis- 0 −I fying XJ + JXt = 0 with J = n . I 0 n

The importance of Lie groups and Lie algebras in nature, is their action as the symmetry group of a physical system. These actions are called representations, and you already see a few obvious examples, as n × n matrices naturally act on n-dimensional vector space (this is called the standard or fundamental representation for above classical Lie groups and algebras). Further matrices act naturally on itself. The adjoint representation is the action via conjugation (for Lie groups) respectively commutator (for Lie algebras). Let us discuss their appearance in physics.

Quantum Physics

Consider the problem of angular momentum in quantum mechanics. In physics, the angular momentum of an object in three-dimensional space is L = x × p, where x is a vector indicating the position of the object and p the momentum vector, × is the cross-product. In quantum mechanics, observables are replaced by operators. x is replaced by the operator that acts by multiplication with x (we will still denote it by x) and p is replaced by −ih¯∇. In order to compute the angular momentum operator, one prefers polar coordinates, that is x = r sinϑ cosϕ, y = r sinϑ sinϕ, z = r cosϑ.

The following picture illustrates the coordinate change, and er,eϑ ,eϕ are then the standard orthonormal unit vectors in polar coordinates. MATH 328 49 z

er e ϑ ′ ϕ ϑ ϕ′

eϑ ϕ y

x The standard unit vectors in the old and new coordinates are related as er = sinϑ cosϕex + sinϑ sinϕey + cosϑez,

eϑ = cosϑ cosϕex + cosϑ sinϕey − sinϑez,

eϕ = −sinϕex + cosϕey. The quantum mechanical angular momentum operator is then computed from space and momentum operators in polar coordinates:

x = rer, p = −ih¯∇, L = x × p = Lxex + Lyey + Lzez d cosϑ d d cosϑ d L = −ih¯ −sinϕ − cosϕ , L = −ih¯ cosϕ − sinϕ x dϑ sinϑ dϕ y dϑ sinϑ dϕ 2 d 2 2 1 d d 1 d Lz = −ih¯ , L = −h¯ sinϑ + . dϕ sinϑ dϑ dϑ (sinϑ)2 dϕ2 Having computed this objects, one can next compute their commutators:

[Lx,Ly] = Lz, [Ly,Lz] = Lx, [Lz,Lx] = Ly, these are exactly the commutation relations of the Lie algebra so(3). One can also compute that 2 the total angular momentum L commutes with all three Lx,Ly and Lz. Finding now the states of the quantum mechanical system amounts to study the Schr¨odinger equation 2 L ψλ,∆(ϕ,ϑ) = ∆ψλ,∆(ϕ,ϑ). 2 Since Lz commutes with L one can diagonalize the action (there is a unitarity condition that guarantees that) and so one additionally requires

Lzψλ,∆(ϕ,ϑ) = λψλ,∆(ϕ,ϑ). It turns out that the solutions to these two equations are quantized by the quantum numbers: 2 ∆ = h¯ ℓ(ℓ − 1), λ = hm¯ , ℓ ∈ Z≥0, m ∈ Z ∩ [−ℓ,ℓ], and the solutions are the standard spherical harmonics on the 2-sphere. The numbers ∆ and Lz are the quantities measurable by experiment. Let me explain the Lie group meaning of all this. The 2-sphere itself is not a group, however its points are identified with elements of a left coset, 50 T CREUTZIG that is S2 =∼ SU(2)/U(1) of the Lie group SU(2) by the subgroup U(1) (not normal). While we can act on a group itself both from the left and from the right, the set of left cosets only carries an action from the left. Hence, S2 carries an action of SU(2). In Lie theory parlor, the differential operators Lx,Ly,Lz are called invariant vector fields and they describe infinitesimal translations tangent to the coset. The solutions to above differential equation is the harmonic analysis on S2 whose solutionsform a basis of the space of square integrable functions on S2 with respect to a measure (sinϑdϑ ∧ dϕ) that is left-invariant under the action of SU(2). These functions carry an action of the Lie algebra su(2), but they are actually exactly half of those representations of su(2) that can be integrated, that is representations that are also representations of the Lie group SU(2). Integrability of representations is a severe constraint and it is somehow the mathematical reason for quantization of physical observables. Further, the total angular momentum is the Laplacian on S2 and it is obtained from the unique invariant degree two element in the universal envelopping algebra (think of it as polynomials in the generators of su(2)) of su(2). Proving that the two-sphere is a coset of SU(2) is a bit of work, see [S]. One does it as follows. SU(2) acts on the Lie algebra su(2) via conjugation. But su(2) is a three-dimensional Lie algebra, so that this action defines a representation of SU(2) on R3. The corresponding group homomorphism is given by Re a2 − b2 Im a2 + b2 −Re(2ab) a b 2 2 2 2 φ : SU(2) −→ GL3(R), 7→ −Im a − b Re a + b Im(2ab) . −b∗ a∗   ab∗ + a∗b −i(ab∗ − a∗b) |a|2 −|b|2   The image of this group homomorphismis the Lie group SO(3), and its kernel is {±I2} =∼ Z/2Z. The action of SO(3) on a vector in R3 preserves the norm (this is exactly the condition ggt = I), but S2 is exactly the subspace of R3 of norm one vectors. Hence we have constructed an action of SU(2) on S2. This action is transitive, i.e. S2 is one single orbit and hence by Theorem 23 the two-sphere is isomorphic to the quotient of SU(2) by the isotropy subgroup of any point of S2. Take p =(0,0,1) then the subgroup a 0 U(1) =∼ a ∈ C,|a|2 = 1 0 a∗ is the isotropy subgroup of this point. Note, that we could have taken any points isotropy subgroup, they are all isomorphic (in this case).

Another physics example in which Lie groups are the main ingredient are gauge theories. The standard model of particle physics is for example a gauge theory for the Lie group SU(3)× SU(2) × U(1). Here group theory really meets geometry, and all the physics objects have a precise meaning. For example. quarks and other fermions like electrons are vector bundles, gauge fields like the photon or the gluons are connections and curvature is the field strength. All these objects carry an action of the gauge group, a compact Lie group. If this action depends on the position of space-time of the field (which it usually does), then this is called a local gauge group. The physics problem is described by a Lagrangian, which is a special polynomial in the fields and their derivatives, namely this polynomial is invariant under the group action. This is the meaning of symmetry: a group is symmetry group of the problem if it acts on the physical objects in such a manner that the Lagrangian of the problem is invariant. I am not MATH 328 51 able to go into details here, but let me give you the group action in the case of the standard model. The gauge group here is the direct product of unitary 3 × 3 matrices (SU(3)), unitary 2 × 2-matrices (SU(2)) and the circle U(1). These correspond to quantum chromodynamic and the electro-weak interaction. A matrix acts naturally on column vectors and row vectors. The former is called the standard representation (or fundamental representation in physics) while the latter is the conjugate or anti-fundamental representation. For SU(2), these two are naturally isomorphic. Fermions are spin 1/2 fields, meaning that they carry the standard representation of SU(2). Examples are electrons, neutrinos and quarks. The quarks also carry the standard representation of SU(3), while electrons are invariant under that group. This means that quarks interact under the strong interaction and elcetrons don’t. The second type of fields are gauge bosons. These are fields in the adjoint representation of the Lie group. For example, SU(3) is 8- dimensional and there are 8 gluons, the gauge bosons of the strong interaction. Similarly, there is the gauge boson of electro-dynamics, the photon and there are three W-bosons, the gauge bosons of the weak interaction. Further, there are then particles that are composed out of the elementary particles. For example, the proton is composed out of three quarks, meaning that the proton field is an element of the threefold exterior product of the standard representation of SU(3).

15. PRACTICE QUESTIONS A few exercises, I only sketch the proofs.

Exercise 36. Prove that Z/nZ ⊕ Z/mZ =∼ Z/(nm)Z if and only if n and m are co-prime.

Solution. Let g and h be generators of Z/nZ and Z/mZ. Then (g,h) generates Z/nZ ⊕ Z/mZ if and only if they are co-prime.

Exercise 37. Which of the following groups are isomorphic: Z/3Z ⊕ Z/3Z ⊕ Z/2Z ⊕ Z/2Z, Z/9Z ⊕ Z/2Z ⊕ Z/2Z, Z/18Z ⊕ Z/2Z, Z/6Z ⊕ Z/6Z?

Solution. By the previous exercise these are the ﬁrst and last one. Also the middle two are isomorphic.

Exercise 38. Which of the following groups are isomorphic:

(Q,+),(R,+),R+,·)?

Solution. A linear map ϕ from Q to R has image {qϕ(1)|q ∈ Q} (you might like to write a detailed argument on this). This can never be surjective. The map x 7→ ex is a group homomorphism it is both injective and surjective, since the logarithm is its inverse.

Exercise 39. Show that (Q,+) is not ﬁnitely generated, i.e., if X ∈ Q is a ﬁnite set, then the group generated by X is not Q. 52 T CREUTZIG

Solution. Let X = {p1/q1,..., pk/qk} be a ﬁnite subset of Q where pi and qi are coprime integers. Then, the group generated by X is given by

hXi = ∑mi pi/qi mi ∈ Z = ∑ mi ∏q j pi/q mi ∈ Z,q = ∏q j ( i ) ( i j i ! j ) 6=

In particular we see that every element in hXi is an integer multiple of 1/p, hence this cannot be all Q. Exercise 40. Let p be prime and X a set with less than p elements. Show that the only action of Z/pZ is the trivial one. Solution. The orbit length has to divide the group order. Since X has less than p elements there is no orbit of size p and hence every orbit has to have length one, that means Z/pZ acts trivially. Exercise 41. Let G be a finite group, show that g|G| = e for all g in G. Solution. The order of a group element has to divide the group order. Let g be in G, d its order and dn = |G|. Then g|G| = gdn =(gd)n = en = e. Exercise 42. Let G be a finite group and H a subgroup of G of index n. Show that gn! in H for every g in G. Solution. G acts on G/H. The latter has n elements, hence this action defines a group homo- n! n! morphism ϕ from G to Sn. Hence by the previous exercise ϕ(g ) = ϕ(g) = e and hence the action of gn! is trivial on G/H but this means it is an element of H. Exercise 43. Let G be a group with a conjugacy class C that has two elements. Show that G is not simple. Solution. The conjugacy class of the identity is {e}, so G has at least three elements G acts on C by conjugation. This defines a group homomorphoshm from G to S2 =∼ Z/2Z. This map cannot be injective as G has more than two elements. It is surjective since the action is transitive. So the kernel is neither G nor the identity and hence must be a non-trivial nomral subgroup.

Exercise 44. Prove that the only normal subgroup of Sn is An if n ≥ 5. Use that An is simple.

Solution. Since An is simple and since its index is two in Sn the normal series {e} E An E Sn is a composition series. Let N be another normal subgroup of Sn, then {e} E N E Sn is a normal series of same length as the composition series just mentioned. Hence this must also be a composition series, thus N has either index two in Sn or order two. Order two subgroups are never normal except in S2, and An is clearly the only index two subgroup of Sn. (since such a group must contain at least one transpositon if it is not equal to An, but since it also must be normal it then has to contain all transpositions, but those generate Sn) Exercise 45. Let G be a simple group of order 168. Show that G has either 7 or 28 different 3-Sylowsubgroups. Solution. By Sylows theorem the possibilities for the number of 3-Sylowgroups is 1,4,7,28. Simple makes 1 impossible, but also 4. The reason is that G acts transitively on the set of 3- Sylowgroups via conjugation. If there were 4, then this would deﬁne a group homomorphism to S4. This cannot be injective as the order of S4 is too small, but it is surjective hence the kernel is a non-trivial subgroup. MATH 328 53

Exercise 46. Let G be a simple group of order prm, where r ≥ 1,m > 1 and p and m are co-prime and p is prime. Let n be the number of p-Sylowsubgroups of G. Let S be such a p-Sylowsubgroup and let H = NG(S) be its normalizer in G. Then [G : H] = n (we have proven this in step 3 of the proof of Sylow’s theorem) and S is not normal in G. Conclude that |G| must divide n!. Solution. G acts transitively on the set of p-Sylowgroups deﬁning a group homomorphism to Sn, since G is simple this map must be injective and hence G is isomorphic to a subgroup of Sn. Exercise 47. If G is a group of order 250000 = 2456 then G is not simple. Solution. Assume that G is simple. Let n be the number of 5-Sylowgroups, then n = 16 is the only possibility. By the previous exercise 250000 must divide 16!, but it doesnot.

REFERENCES [A] M Artin. Algebra, 2011 Pearson Education [Co] K Conrad. Expository Papers, http://www.math.uconn.edu/ kconrad/blurbs/ [Cr] T Creutzig. Ma Ph 451 – Mathematical Methods for Physics I. Winter 2014. [G] F. M. Goodman. Algebra - Abstract and concrete, see http://www.math.uiowa.edu/ goodman/algebrabook.dir/algebrabook.html. [H] W. Holzmann. Classiﬁaction of Finitely Generated Abelian Groups, http://www.cs.uleth.ca/ holzmann [JS] J C Jantzen, J Schwermer. Algebra, Springer. [S] F Spinu. Lie Groups,Lecture Notes.

(T Creutzig) 573 CAB, UNIVERSITY OF ALBERTA E-mail address: [email protected]