1.1 REVISION
Definition 1.1.1: A group is a pair G, consisting of a non-emptey set G and a binary operation defined on G ,satisfying the following four properties: 1. G is closed under the operation . 2. The operation is associative. 3. G contains an identity element e for . 4. Each a ∈ G has an inverse a−1 ∈ G.
Remark 1.1.2: If it happens that the group operation satisfies the commutativity property, then G, is called a commutative group or an abelian group.
Example 1.1.3:
Let K4 e, a , b , c .Define on K4 by
e a a e a , e b b e b e c c e c , c b b c a a b b a c , a2 b2 c2 e
Construct the group table?
Solution :
eabc eeabc aaecb bbcea ccbae
Thus K4 is an abelian group which is called Klien -4-group.
Definition 1.1.4:
1 Let H be a non-empty subset of the group G. Then H is said to be a subgroup of G if H itself a group under the same operation as that of G,we denote it by H ≤ G.
Theorem 1.1.5: Let G be a group and ∅≠H ⊆ G. Then H ≤ G a,b ∈ H ab−1 ∈ G.
Definition 1.1.6: The center of a group G , denoted by ZG, is the set ZG a ∈ G : xa ax ∀ x ∈ G.
Theorem 1.1.7: Let G be a group. Then ZG ≤ G.
Theorem 1.1.8: let H be a non-empty subset of a group G. Then
H ≤ G i H is closed ab ∈ H ∀ a,b ∈ H ii a−1 ∈ H for each a ∈ H.
______
Exercise 1.1.9: 1. Let G be a group and a ∈ G .DefineCa x ∈ G : xax−1 a.Show that Ca ≤ G which is called the centralizer of an element a.
2. Assume that H, K are subgroups of the commutative group G.Let HK g ∈ G : g hk h ∈ H, k ∈ K.Prove that HK ≤ G. ______Definition 1.1.10: Let n be a fixed positive integer. Two integers a and b are said to be congruent modulo n, written
a ≡ b mod n a − b is divisible by n.
2 Definition 1.1.11: For an arbitrary integer a,let
a a x ∈ Z : x ≡ a mod n x ∈ Z : x a qn q ∈ Z
We call a a the congruenss class modulo n, determined by a and refer to a as a representative of this class.
Theorem 1.1.12:
(1) For each ve integer n, the mathematical system Zn,⊕ forms a commutative group, known as the group of integers modulo n.
(2) Zn,⊙ is a commutative semi-group, with identity 1 ∈ Zn.
Definition 1.1.13: Let G be any non-empty set. A permutation on a set G is a bijection from
G to itself. Denote SG the set of all permutations on G.Take G N 1,2,3,...,n. Then the set of all permutations on N will be denoted by Sn.
Theorem 1.1.14:
1. The set SG is a group under function composition ∘. This group is called symmetric group on the set G. 2. The system Sn,∘ forms a group known as the symmetric group on n symbols,whichis non-commutative for n ≥ 3. 3. An,∘ form a group of order n!/2 which is called the alternating group of n-letters, where An ∈ Sn : is an even permutation.
Definition 1.1.15: A group G is called a cyclic group if there exist at least one element a ∈ G such that
G an : n ∈ Z , if the operation is . G na : n ∈ Z , if the operation is .
The element a is called the generator of G. The cyclic group generated by a is denoted by 〈a.
3 Example 1.1.16: (Infinite cyclic) Consider the group Z,.Then Z, is a cyclic group generated by 1, − 1.
Example 1.1.17: (Finite cyclic)
The additive group Zn,⊕ is a cyclic group with generator 1 .
Definition 1.1.18: (i) Let G be a group . The order of a ∈ G is the smallest ve integer n such that an e na e if the operation is () .We denote it by oa n 0. (ii) If there is no such integer we say that a has infinite order denoted by oa .
Definition 1.1.19: If G is a group such that for a ∈ G,G am : m ∈ Z,then G is a cyclic group. Denote it by G a .
Theorem 1.1.20: Let G be a cyclic group. Then (i) G is commutative (every cyclic group is commutative) (ii) If H ≤ G H is cyclic. (every subgroup of a cyclic group is cyclic).
Theorem 1.1.21: Let a be an element in a group G .Ifn is the least ve integer such that an e, i.e., oa n. Then (i) a has order n,and a e, a, a2, ..., an−1. (ii) If am e n ∣ m. (iii) If ak as k ≡ s mod n.
Definition 1.1.22: The order of an element a oa of the group G is the order of the subgroup generated by a. That is oa o a .
Theorem 1.1.23: Let G be a finite cyclic group of order n i.e., G a , |G| n with generator a ∈ G. Then (1) If d m,n is the greatest common divisor for the integer m,n, then the subgroup generated by am is the same as the subgroup generated by ad.
4 i.e., ad am da ma
and am contains n ∣ d elements.
(2) The distinct subgroups of G are those subgroups ad da if the operation is (), where d is a ve divisor of n.
(3) If 1 m,n i.e., m,n are relatively prime, then am is a generator of G.
am if the operation is . i.e., G . ma if the operation is ()
5 Theorem 1.1.24: Let G a be an infinte cyclic group. i.e., oa . Then 1. If n ≠ m ∈ Z an ≠ am. 2. a, a−1 are the only generators of G.
Definition 1.1.25: Let H ≤ G and a ∈ G. The left coset of H in G is defined as follows: aH ah : h ∈ H if the operation is . a H a h : h ∈ H if the operation is .
Theorem 1.1.26: Let H ≤ G. Then there is a one-one correspondence between H and aH, for a ∈ G. i.e., |H| |aH|.
Definition 1.1.27: Let H ≤ G, a,b ∈ G. Then a and b are left (or right )congruent modulo H denote it by a ≡l b ≡r
−1 −1 a ≡l b mod H ab ∈ H a ≡r b mod H ba ∈ H.
Theorem 1.1.28: Let H ≤ G. (i) Then the relation ≡l mod H is an equivalence relation on G. (ii) The equivalence classes are left cosets of H in G.
a b ∈ G : a ≡l b mod H b ∈ G : a−1b ∈ H b ∈ G : b ∈ aH aH.
Remark 1.1.29: (1) The set of left cosets of H in G partition G. i.e., G is the union of all left cosets of H in G and distinct left coets are disjoint.
6 (2) If b ∈ a a b. i.e., if b ∈ aH aH bH.
(3) Let H ≤ G . Then aH bH a−1b ∈ H.
Definition 1.1.30: Let H ≤ G. Then the index of H in G is defined to be the number of distinct left cosets of H, which is denote it by G : H.
Theorem 1.1.31 (Lagrange Theorem): Let G be a group of finite order n, i.e., |G| n, H be a subgroup of G of order k. Then every subgroup of G is a divisor of |G|. i.e., |H| ∣ |G| and |G| |H|G : H.
Proof : Let |G| n and |H| k. Then by theorem 1.1.26, if a ∈ G |H| |aH| k. G is the union of all disjoint left cosets of H in G. i.e.,
G a1H a2H a3H ... arH ,ai ∈ G where aiH ∩ ajH ∅ for i ≠ j.
i.e., G : H r |G| |a1H| |a2H| ... |arH| n kr k ∣ n |H| ∣ |G|.
Corollary 1.1.32: (i) Let |G| n. Then the order of any element a ∈ G divides n,andan e. (ii) Any finite group of prime order has no-non-trivial subgroup. (iii) Every group G of prime order is cyclic. (iv) Any non-commutative group has at least six elements.
Definition 1.1.33: A subgroup H of a group G is said to be normal in G if every left coset of H in G is also a right coset of H in G.i.e., H G aH Ha ∀ a ∈ G.
Remark 1.1.34:
7 1. G , e are normal. (Trivial normal subgroups} 2. Every subgroup of a commutative group is normal. 3. Every subgroup of a cyclic group is normal. 4. aH Ha it does not require that ah ha ∀ h ∈ H.
Theorem 1.1.44: The subgroup H is normal in G aHa−1 ⊆ H for each element a ∈ G. i.e., aH Ha aHa−1 ⊆ H ∀ a ∈ G.
Theorem 1.1.45: Let G be a group. Then (1) If H ≤ G s.t. H ⊆ ZG H G. (2) ZG G. (3) Every subgroup of a commutative group is normal.
Quotient Group: Let H G. Denote the set of all left (right) distinct cosets of H in G,by:
a aH : a ∈ G a Ha a ∈ G if the operation is . G/H or a a H : a ∈ G a H a : a ∈ G if the operation is
Theorem 1.1.46: 1. Let H G. Then G/H forms a group with respect to the coset multiplication,which is called the quotient group of G by H. 2. If G is commutative then G/H is commutative.
Definition 1.1.47: 1. Given a group G and a pair a,b ∈ G the commutator of a and b is defined to be the product aba−1b−1 ,wedenoteitbya,b. 2. Denote G,G to be the subgroup generated by the set of all commutators (derived subroup) i.e.,G,G ai, bi : ai, bi ∈ G (finite product with one or more terms).
8 Theorem 1.1.48 :(Important for the application of the sylow theorem) Let G be a group, and G′ be the set of all commutators aba−1b−1 of a group G. Then 1. G′ G. 2. G/G′ is commutative group. 3. If H G, then G/H is commutative G′ ⊆ H. ______Exercise 1.1.49: 1. Let H, K be subgroups of the group G ,and H ≤ K ≤ G. Show that G : H G : KK : H.
2. If G : H 2 H G.
3. Show that An Sn.
4. Given H, K are subgroups of G. Prove the following : (i) If K G HK ≤ G. (ii) If H, K are normal of G HK G.
5. Show if G a and H G G/H is cyclic. ______Definition 1.1.50: Let G,∗ and G′,∘ be two groups. 1. A homomorphism from G,∗ into G′,∘ is a mapping f : G G′ such that fa ∗ b fa ∘ fb. 2. If f is one - one and onto , then f is said to be isomorphism,denote it by G G′. 3. A homomorphism from G into itself is called endomorphism. 4. An isomorphism from G onto itself is called automorphism. 5. Denote HomG,G′ The set of all homomorphisms from the group G into G′. 6. AutG{The set of all isomorphisms from G onto itself}. 7. Let f : G G′ be a homomorphism. Then ImG fG fa : a ∈ Gand kerf a ∈ G : fa e′ f−1e′.
Theorem 1.1.51: 1. HomG,G,∘ forms a semi-group with identity. (∘ is a composition) 2. AutG,∘ is a subgroup of the symmetric group . i.e., AutG ≤ SG. 3. Let f : G G′ be a homomorphism. Then fG ≤ G′ while kerf G.
9 Theorem 1.1.52: Let H G. Then the mapping : G G/H , a aH where a ∈ G is an onto homomorphism with ker H. is called a natural homomorphism).
Theorem 1.1.53:
(i) Let G be a finite cyclic group of order n 0. Then G Zn . (ii) Let G be an infinite cyclic group . Then G Z . (iii) Any two cyclic groups of the same order are isomorphic.
10 Theorem 1.1.54 (Cayley):
(i) Any group G is isomorphic to a subgroup of the symmetric group SG. (ii) When G is finite .Then any finite group of order n is isomorphic to a subgroup of the symmetric group Sn.
______Exercise 1.1.55: 1. Let f : G G′ be a homomorphism. Prove the following: (i) fZG ⊆ ZfG. (ii) If G′ is commutative , then G,G ⊆ kerf.
2. Let f : G G′ be a homomorphism. Prove the following: (i) If G is commutative fG is commutative. (ii) If G is cyclic fG is cyclic. ______
Theorem 1.1.56 (1st homomorphism theorem): If f : G G′ be a homomorphism from a group G into a group G′, then G/kerf fG.If f is onto, then G/kerf G′.
Theorem 1.1.57 (2ed homomorphism theorem): Let H,K be subgroups of the groupp G.andK G. Then (a) H ∩ K H (b) HK ≤ G (c) K HK. Moreover, H/H ∩ K HK/K .
Theorem 1.1.58 (3ed homomorphism theorem): Let H,K be two normal subgroups of the group G with K ⊆ H. Then (i) H/K G/K. ii G/K╱H/K G/H.
______
Exercise 1.1.59 : 1. Assume H,K are subgroups of the group G, with K G and G HK. Show that G/K H H ∩ K e.
11 12 2. Let H, K1, K2 ≤ G and K1,K2 G.IfH ∩ K1 H ∩ K2, prove that
HK1/K1 HK2/K2.
3. Let H1, H2, K ≤ G and K G.IfH1K H2K, prove that
H1/H1 ∩ K H2/H2 ∩ K. ______
Number Theory:
Theorem 1.1.60:
k1 k2 kr 1. Any ve a 1 can be written uniquely in a standared form a p1 p2 ...pr where ki i 1,...,r is a ve integer such that each pi is a prime with 1 p1 p2 ... pr.
n1 n2 nr 2. Any divisor b of a is of the form b p1 p2 ...pr where 0 ≤ ni ≤ ki for each i.
3. Every ve integer n 1 can be expressed as a product of primes numbers, this representation is unique.
13 1.2 DIRECT PRODUCT OF GROUPS
In this section we study products in the category of groups. The product is important not only as a means of constructing new groups from old , but also for describing the structure of certain groups in terms of particular subgroups (whose structure for instance, may already be known). In this section, we pay attention to finite groups.
Definition 1.2.1: Let H and K be subgroups of the group G. Then G is said to be the internal direct product (G is decomposable into internal direct product) of H and K, written G H ⊗ K,if 1. H and K are normal subgroups of G. 2. G HK. 3. H ∩ K e.
Example 1.2.2: 2 2 Consider the Klein 4-group K4 e,a,b,ab where a b e and ab ba.SinceK4 is commutative , then all its subgroups are normal. If H is a cyclic subgroup generated by a , i.e., H a and K is a cyclic subgroup generated by b , i.e., K b . Then H ∩ K e and G HK K4 H ⊗ K. Thus K4 is the internal direct product of two cyclic subgroups of order 2.
Remark 1.2.3: Some groups cannot be expressed as the internal direct product of two non-trivial normal subgroups (such groups are referred to as indecomposable). For let G be an infinite cyclic group generated by a, G a , and suppose G H ⊗ K,whereH,K are non-trivial subgroups of G H,K are cyclic subgroups (since G is cyclic ) generated by an, am respectively. H an , K am anm ∈ H ∩ K H ∩ K ≠ e for this would imply that G is finite.
14 There exist several criteria for a group to be an internal direct product of its subgroups . We establish one such condition below.
15 Theorem 1.2.4: The group G is an internal direct product of H and K iff (i) each element x ∈ G can be uniquely expressed in the form x hk, h ∈ H, k ∈ K,and (ii) any element of H commutes with any element of K, i.e., hk kh.
Proof : Suppose G H ⊗ K, we have to prove (i) and (ii). Since G HK, H ∩ K eany element x ∈ G can be written as x hk, h ∈ H, k ∈ K.
To show that x is uniquely expressiable, let x h1k1,where h1 ∈ H, k1 ∈ K.
hk h1k1.
−1 −1 Multiplying this equation on the left by h and on the right by k1 ,weget
−1 −1 −1 −1 kk1 h h kk1 , h h1 ∈ H,K.
−1 −1 Since H ∩ K ekk1 e and hh1 e k k1, h h1.
Now, we have to prove that any element of H commutes with any element of K. Consider the commutator hkh−1k−1, ∵ H G hkh−1k−1 ∈ H ∵ K G hkh−1k−1 ∈ K. Thus hkh−1k−1 ∈ H ∩ K ehk kh.
Conversely, suppose that the conditions (i) and (ii) hold, we have to prove that G H ⊗ K,andH,K G ??. From (i) we get, G HK. To prove that H G,pickh ∈ H, then −1 −1 −1 xhx h1 k1h h1k1 h1hh1 ∈ H. Similarly, K G. To complete the proof, we must show that H ∩ K e. Let x ∈ H ∩ K x ∈ H, x ∈ K.Nowx can be expressed as
16 x xe x ∈ K, e ∈ H x ex e ∈ H, x ∈ K x has two representations , x e.
17 The next theorem exhibits the interplay between the notions of internal direct product and the quotient group.
Theorem 1.2.5: Let G H ⊗ K , then G/H K and G/K H.
Proof : Define the mapping f : G K by fx k where x hk, h ∈ H, k ∈ K. 1. f is well defined. Since G H ⊗ K , then by theorem 1.2.4, x has unique representation, which implies that f is well defined.
2. f is onto. ∀ k ∈ K ∃ x ∈ G s.t. fx k.
3. f is homomorphism.
Let x, y ∈ G x hk, y h1k1 where h,h1 ∈ H, k,k1 ∈ K. Consider
fxy fhkh1k1 fhh1kk1 kk1 fxfy.
Thus by 1st homomorphism theorem, G/kerf K. kerf x ∈ G : fx e H
G/H K.
Theorem 1.2.6: If H, K are finite subgroups of the group G, then |H||K| |HK| . |H ∩ K|
Proof : Let D H ∩ K. D is a subgroup of G and , since D ⊆ K, D is a subgroup of K. Then we can decompose the subgroup K into the union of finite number
18 t of disjoint right cosets of D in K, i.e., K Dki. i1
K Dk1 Dk2 ... Dkt ki ∈ K
|K| |Dk1 | |Dk2 | ... |Dkt | |K| |D| |D| ...... |D| |K| t|D| |K| t . |D| Since HD H for D ⊂ H,weget t t t HK H Dki HDki Hki. i1 i1 i1
The right cosets Hk1, Hk2,...,Hkt are pair wise distinct for let
Hki Hkj for some i,j
−1 −1 −1 Hkikj H kikj ∈ H kikj ∈ H ∩ K D Dki Dkj. the two right cosets of D in K are equal , (since these cosets are disjoint) |H||K| Hence Hki ≠ Hkj |HK| |H|t |HK| . |H ∩ K|
Corollary 1.2.7: If G is a finite group and G H ⊗ K |G| |H||K|.
Theorem 1.2.8: Let H and K be normal subgroups of the finite group G with |G| |H||K|. If either (i) G HK or (ii) H ∩ K e, then G H ⊗ K.
Proof : (i) Suppose G HK with |G| |H||K|, we want to show that G H ⊗ K. Since G HK
|H||K| |G| |HK| . |H ∩ K|
19 But |G| |H||K| |H ∩ K| 1 H ∩ K e G H ⊗ K.
20 (ii) Suppose H ∩ K e |H||K| |HK| |H||K|. |H ∩ K| |HK| |G| HK G G H ⊗ K.
Example 1.2.9:
Consider the commutative group Z10,⊕. Then the only two non-trivial normal subgroups are those H 5 0̄,5̄ , K 2 0̄,2̄,4̄,6̄,8̄.
Since |Z10 | |H|.|K| and H ∩ K 0̄, then Z10 2̄ ⊕ 5̄ .
We have restricted our attention to the internal direct product of two subgroups. The definition lends itself to the following generalization:
Definition 1.2.10: A group G i said to be the internal direct product of the subgroups H1,H2,...Hn, denoted by G H1 ⊗ ... ⊗ Hn iff 1. Hi G ∀ 1 ≤ i ≤ n. n 2. G Hi i1 3. Hi ∩ Hj e. i≠j
Theorem 1.2.11:
The group G is the internal direct product of the subgroups H1,....Hn iff (i) Every element x ∈ G is uniquely expressible in the form x h1h2...hn, with hi ∈ Hi. (ii) hi ∈ Hi and hj ∈ Hj i ≠ jhihj hjhi.
21 In the opposite direction : Given two subgroups G1 and G2 , is there a group G which is the internal direct product of H and K isomorphic to G1 and G2? For an answer, we take G to be the cartesian product of G1 G2.
Definition 1.2.12: The cartesian product of the subgroups H,K is the set of all ordered pair h,k where h ∈ H, k ∈ K, which denoted by G H K h,k : h ∈ H, k ∈ K. Define a binary operation on H K by
h1,k1.h2,k2 h1h2,k1k2. One can easily verfiy that H K,. makes a group called external direct product. The identity element of H K is eH,eK and the inverse of h,k is h−1,k−1.
The answer of the question above is given in the following theorem:
Theorem 1.2.13: Let G be a group with normal subgroups H and K such that G H ⊗ K. Then G H K.
Proof : Define f : G H K by fhk h,k , h ∈ H, k ∈ K. 1. f is well defined: Since every element in G is uniquely expressible in the form hk, then f is well defined.
2. f is homomorphism:
For let x hk, y h1k1 where h,h1 ∈ H, k,k1 ∈ K, then
fxy fhkh1k1 hh1kk1 hh1,kk1 h,kh1,k1 fxfy.
3. f is 1-1: Let x ∈ kerf fx fhk e,eh,k e,e h e,k e x e.
22 4. f is onto: ∀ h,k ∈ H K ∃ hk ∈ H ⊗ K s.t. fhk h,k.
Thus G H K.
The notion of external product can easily be broadened to allow for more factors.
Definition 1.2.14:
Let G1, G2,..., Gn be a finite collection of groups, and consider their cartesian product
G G1 G2 ... Gn a1,...,an : ai ∈ Gi. Multiplication is defined in G by
a1,a2,...,an.b1,b2,...,bn a1b1,a2b2,...,anbn : aibi ∈ Gi. Under this binary operation G is a group, known as the external product of the set Gi where e e1,e2,...,en and the inverse of a1,...,an is −1 −1 a1 ,...,an .
Theorem 1.2.15:
Let G1,G2, ...,Gn be a finite groups, and a1,a2,...,an ∈ G1 G2 ... Gn. Then 1. |G1 G2 ... Gn | |G1 ||G2 |...|Gn |. 2. If oa1 r1, oa2 r2,...,oan rn, then oa1,a2,...,an least common multiple of all the ri.
Theorem 1.2.16:
Let G be a group with normal subgroups H1, H2,...,Hn such that G H1 ⊗ H2 ⊗ ... ⊗ Hn. Then
G H1 ⊗ H2 ⊗ ... ⊗ Hn H1 H2 ... Hn.
Proof : The proof is similar to the proof of theorem 1.2.13.
Example 1.2.17:
List the elements of the group Z2 Z3. Show that this group is cyclic and find the generator.
23 Solution :
Z2 Z3 0̄,0̄,0̄,1̄,0̄,2̄,1̄,0̄,1̄,1̄,1̄,2̄. To find the generator, we start by the element 0,0 :
0̄,0̄ n.0̄,0̄ : n ∈ Z 0̄,0̄ 0̄,1̄ n.0̄,1̄ : n ∈ Z 0̄,0̄,0̄,1̄,0̄,2̄ 0̄,2̄ n.0̄,2̄ : n ∈ Z 0̄,0̄,0̄,2̄,0̄,1̄ 1̄,0̄ n.1̄,0̄ : n ∈ Z 0̄,0̄,1̄,0̄ 1̄,1̄ n.1̄,1̄ : n ∈ Z 0̄,0̄,1̄,1̄,0̄,2̄,1̄,0̄,0̄,1̄,1̄,2̄
Thus Z2 Z3 is a cyclic group generated by 1̄,1̄. Since there is only one cyclic group of order 6 ,we have Z2 Z3 Z6.
Example 1.2.18:
Consider Z3 Z3. This is a group of order 9 elements. Since Z3 Z3 is not a cyclic group therefore Z3 Z3 ≇ Z9.AlsoZ2 Z2 is a group of order 4 which is not cyclic, therefore Z2 Z2 K4.
The preceding examples illustrate the following theorem which shows how we decompose the group Zn,⊕.
Theorem 1.2.19:
If gcdm,n 1, then Zmn Zm Zn.
Proof : Since the integers n ∣ |Zmn |,andm ∣ |Zmn |, then these integers generates the following two cyclic subgroups respectively: H n.1̄ n̄0̄,n̄,2n,...,m − 1n of order m, i.e., |H| m. K m.1̄ m̄0̄,m̄ ,2m,...,n − 1m of order n, i.e., |K| n. From Cayley’s theorem H Zm,andK Zn. Moreover, H, K are normal subgroups of Zmn such that
24 |Zmn | mn |H||K|. We claim that H ∩ K e, for let ā ∈ H ∩ K,
ā ∈ H and ā ∈ K oā ∣ |H| and oā ∣ |K| oā ∣ m and oā ∣ n n ∣ m or m ∣ n. But m,n 1 oā 1 ā1 e 0̄ ā 0̄. Hence using theorem 1.2.8, we get Zmn H ⊗ K Zm Zn.
Theorem 1.2.20:
k1 k2 kr If the integer n 1 has the prime factorization n p1 p2 ...pr (distinct primes), then
k k kr Zn Z 1 Z 2 ... Zpr . p1 p2
Example 1.2.21: 3 2 Consider the cyclic group Z72,⊕. Then 72 2 .3 ,therefore
Z72 Z23 Z32 Z8 Z9.
25 PROBLEMS 1.2
1. Let G S3,∘ , 123 , 12 .IsG the internal direct product of ,. Justify your answer?
2. Use another method to prove theorem 1.2.5.
∗ 3. Show that G G1 ⊗ G2 where G R ,., G1 R ,., G2 −1,1,..
4. Let G Z12,⊕ and H1 3̄ , H2 2̄ , K1 4̄ . (i) Is Z12 H1 ⊕ H2 justify your answer? (ii) Is Z12 H1 ⊕ K1 justify your answer?
5. Given the groups Z4,⊕, Z12,⊕, Z18,⊕. Answer the following questions: (i) Find |Z4 Z12 | ,and|Z12 Z18 |. (ii) Find the order of o2̄,6̄ ∈ Z4 Z12 and o8̄,10 ∈ Z12 Z18. (iii) Find the order of o3̄,6̄,12,16 ∈ Z4 Z12 Z20 Z24.
6. Prove Theorem 1.2.20, using induction on r.
7. List the elements of Z2 Z4. Find the order of each of the elements. Is this group cyclic?
8. Find all proper non-trivial subgroups of Z2 Z2.
9. Let G1 Z4,⊕ and G2 S3,∘. (i) List the elements of Z4 S3. (ii) Is this group commutative?
2 4 10. Let G1 a : a 1 and G2 b : b 1 . (i) List the elements of G G1 G2. 2 3 (ii) Let A1 1,1,a,1, A2 1,1,1,b,1,b ,1,b . Show that G A1 ⊗ A2.
11. Prove that G1 G2 G2 G1 for any groups G1, G2.
12. Prove that an external direct product of commutative groups is commutative.
13. Show that a finite commutative group G ≠ e is cyclic iff G Z k1 Z k2 ... Zpkr p1 p2 r ki where pi are distinct primes.
26 14. Find two normal subgroups of Z15 and prove that Z15 is the internal direct product of these two normal subgroups.
15. Let G be a group , H G,K G, H ∩ K e. Show that G G/H G/K.
27 1.3 FINITE COMMUTATIVE GROUPS
The task of this section is to generalize theorem 1.2.20 to arbitrary finite commutative groups. That is, we intend to prove that any commutative k1 k2 kr group of order n p1 p2 ...pr splits into a direct product of r −cyclic k1 k2 kr subgroups of orders p1 p2 ...pr . As a step toward this goal, it is expedient to introduce a new class of groups , the p −groups, which are essential in this section and in the next section when we develop the Sylow theorems.
Definition 1.3.1: Let p be a prime number. A group G is said to be ap −group if the order of each element of G is some power of p (not necessarily the same power).
Example 1.3.2: 2 Consider the group Z9,⊕, then |Z9 | 3 . By Lagrange’s theorem , every element in Z9 must divides the order of Z9 . Thus if a ∈ Z9, then oa 1or 2 3or3 Z9,⊕ is a 3 −group.
Example 1.3.3:
If p is a prime number, then Zpn ,⊕ is a p −group for any n 0. Since by Lagrange’s theorem, every element of Zpn must have order dividing pn. Hence every element has order a power of p.
It may happen that G is not a p −group, yet some of its subgroups are p −groups. In connection with that possibility, we make the following defintion:
Definition 1.3.4: Let G be a finite commutative group that has order divisible by the prime number p.Define Gp to be the set of all elements of G that have orders that are powers of p.
n Gp a ∈ G : oa p ,forsomen.
Theorem 1.3.5:
28 The set Gp is a subgroup of G.
Proof : 0 0 (i) Since e p oe p e ∈ Gp Gp ≠∅. r s (ii) Let a,b ∈ Gp oa p , ob p for some ve integers r,s. rs rs rs r ps s −pr Since G is commutative , then ab−1 p a p b−1 p a p b p e. Hence by therem 1.1.21, oab−1 ∣ prs oab−1 will be a power of p. Thus −1 ab ∈ Gp.
Remark 1.3.6: 1. Without assumption that G is commutative , theorem 1.3.5, does not hold, i.e., this result false for non-commutative groups, for instance, in the symmetric groups S3 , the elements having order some power of 2 do not comprise a group.
2. Gp is the largest p −subgroup of G.ForifH is an arbitrary p −subgroup, then every element of H has order a power of p, hence by the definition of Gp, each element of H must lie in Gp, that is H ⊆ Gp.
Example 1.3.7:
Consider the commutative group Z6,⊕. Then o0̄ 1, o1̄ 6, o2̄ 3, o3̄ 2, o4̄ 3, o5̄ 6.
For p 2orp 3, the subgroups Gp are given by
G2 0̄,3̄, G3 0̄,2̄,4̄.
The group Z6 is not a p −group, but G2 is a 2-subgroup and G3 is a 3-subgroup.
In this section , we shall characterize finite p −groups as those groups having p −power order. It will be convenient to pause and drive the commutative case now. For this , we need the following lemma.
Lemma 1.3.8: Let H be a normal subgroup of the group G. If G is a p − group H, G/H are p − groups.
Proof : Clearly, H is a p −group even H is not a normal subgroup of G.
29 To prove that G/H is a p −group, let ā ∈ G/H ā aH : a ∈ G. But G is a p −group, oa is some power of p, i.e., oā pr. Consider r r r ā p aH p a p H H oā ∣ pr ā has order power of p.
30 Theorem 1.3.9: A finite commutative group G is a p −group |G| is a power of p.
Proof : Suppose |G| is a power of p, i.e., |G| pr. Then by Lagrange’s theorem every element a ∈ G must divide order of G every element a ∈ G is of order some power of p,i.e., G is a p −group.
Conversely, let G be a p −group, we want to prove that |G| is a power of p,i.e., if |G| n we want to prove that n px??. We are going to prove this by using induction on |G|. (i) For n 1, |G| p0, so the result is true for all finite groups of order 1.
(ii) Suppose that the theorem is true for all finite commutative p −subgroups of order less than |G|,thatis,ifH ≤ G such that |H| n |H| ps.
(iii) We have to prove that the theorem is true for |G| n. Let H be a subgroup of G generated by any element a ∈ G, i.e., H a .Thenwehave two cases: H a ≤ G ↙↘ H a G H G G is a cyclic group generated by a. ∵ H is a non-trivial subgroup ∵ G is a p − group , |H|,|G/H| are less than |G|. every element has p −power order ∵ H, G/H are finite commutative group oa pr by lemma 1.3.8, H,G/H are p −groups. o a oa pr Hence by supposition, |H| ps1 , |G/H| |G| pr. But |G| |H|.|G/H| ps1 .ps2 ps1s2
It follows in both cases that |G| is necessarily a power of p and the proof is completed.
We are now ready to deal with the central topic of the present section, to wit, the description of all finite commutative groups. The first step in this program ties the structure of such groups to that of p −groups.
31 Theorem 1.3.10:
k1 k2 kr Let e ≠ G be a finite commutative group of order n p 1 p 2 ...p r (distinct primes), then 1. G is the internal direct product of p −groups, i.e.,
G Gp 1 ⊗ Gp 2 ⊗ ...... ⊗ Gp r
s where Gp i a ∈ G : oa p i for some s.
2. ki Gp i pi for every i.
Proof :
1. G Gp 1 ⊗ Gp 2 ⊗ ... ⊗ Gp r . To prove this we have to prove that every element a ∈ G is of the form
a a1a2.....ar where ai ∈ Gp i (I) and G G G ....G G ...G e (II) p i ∩ p 1 p 2 p i−1 p i1 p r (I) Let oa m, m ∣ n, by theorem 1.1.60,
k1 k2 kr n1 n2 nr m ∣ p 1 p 2 ...p r m p 1 p 2 ...p r where 0 ≤ ni ≤ ki m Put mi ni .Themi are relatively prime and so, there exist integers ui for which p i
u1m1 u2m2 .... urmr 1.
a au1m1u2m2....urmr a au1m1 .au2m2 ....aurmr
u m Setting ai a i i for i 1,...,r,weget
a a1.a2...... ar
uimi It remains to show that ai a ∈ Gpi ?? i.e., oai is a power of pi ??
ni Let vi pi , then
v u vi uimi i uim m i ui ai a a a e e.
32 ni pi ai e ai ∈ Gpi .
(II) G G G ....G G ...G e .Let p i ∩ p 1 p 2 p i−1 p i1 p r a G G G ....G G ...G ∈ p i ∩ p 1 p 2 p i−1 p i1 p r
a G and a G G ....G G ...G ∈ pi ∈ p 1 p 2 p i−1 p i1 p r
s oa pi and a b1b2....bi−1bi1...br where bi ∈ Gpi
s t1 t2 ti−1 ti1 oa pi , ob1 p1 , ob2 p2 ,..., obi−1 pi−1 , obi1 pi1 , ..., obr
t t t1 i−1 i1 tr s p p p pr pi 1 i−1 i1 a e, b1 e, ....,bi−1 e, bi1 e, ..., br e.
t1 t2 ti−1 ti1 tr t t Set t p1 p2 ...pi−1 pi1 ...pr b1...bi−1bi1...br e a e .
s Therefore api e and at e . s But pi , t are relatively prime, so there exist integers u, v such that
s ps u t v 1 upi vt a a i .a e.
2. ki Gp i pi for every i.
Since Gp1 , Gp2 , ..., Gpr are p-groups, then by theorem 1.3.9,
|Gp1 |, ...,|Gpr | are some power of p. i.e.,
ji |Gpi | pi for some non-negative integer ji.
But |G| |Gp1 |.|Gp2 |.....|Gpr |
k1 k2 kr j1 j2 jr p 1 p 2 ...p r p 1 p 2 ...p r .
The uniqueness of the prime factorazation of |G| implies that ji ki for every i 1,...,r. ki Hence Gp i pi for every i.
33 The preceding proof establishes that a finite commutative group G is the direct product of its subgroups Gp for all primes p such that Gp ≠ e; which is simply to say, for all primes p dividing |G|. This reduces the study of arbitrary finite commutative groups to the study of finte commutative p −groups. The basic result on p −groups from which the whole structure theory can be pinned down is derived below:
Theorem 1.3.11: A finite commutative p −group G is the internal direct product of finite number of cyclic p −subgroups.
Proof : The proof is not required.
Theorems 1.3.10 and 1.3.11 may be brought together to give a complete description of the groups under consideration. The following theorem is sometimes called the Fundamental Theorem for Commutative Groups:
Theorem 1.3.12 (Frobenius) Every finite commutative group is the internal direct product of finitely many cyclic p −subgroups.
Switching to external direct products, and using the fact that any cyclic group of order k is isomorphic to Zk, one could just express Theorem 1.3.12 as follows:
Corollary 1.3.13: If G is a finite commutative group, then
G Z n1 Z n2 ... Zpnr (1) p1 p2 r
for suitable primes pi, not necessarily distinct, and integers ni ∈ Z .
So far, we have learned that any finite commutative group is a direct product of cyclic groups, each of which is a p −group, There can be no factorization, since a finite cyclic group, say G a ,ofp −power order is indecomposable. We clarify this point in the following theorem:
Definition 1.3.14
34 A group G is decomposable if it is isomorphic to a dircet product of two proper non-trivial normal subgroups. Otherwise G is indecomposable.
35 Theorem 1.3.15: A finite cyclic group G of order pn is indecomposable.
Proof : Suppose G is a cyclic group of order pn such that G is decomposable, i.e., G a , |G| pn and G H ⊗ K.
|H| pr, |K| ps s.t. r s n,1≤ r,s ≤ n. Let m maxr,s, then m n . Writing the generator a ∈ G as a hk where h ∈ H, k ∈ K, it follows that apm hkpm hpm kpm e |G| ≤ pm. If |G| pm ,since|G| pn. If |G| pm ,sincepm pn. Therefore G ≠ H ⊗ K.
The above Theorem 1.3.14 allow us to rephrase Theorem 1.3.12 in the following form, which we state as a corollary:
Corollary 1.3.16: Any finite commutative group is the direct product of a finite number of indecomposable cyclic groups.
Remark 1.3.17: 1. We will now identify, up to isomorphism, all possible commutative groups of a given order. By Frobenuis’ theorem any finite commutative group G is the direct product of prime power order cyclic groups. Thus, it will be sufficient to consider the case wherein |G| pn, p is a prime. Theorem 1.3.13, assert that
G Zpn1 Zpn2 ... Zpnr
n n n n p p 1 .p 2 ....p r n n1 n2 ... nr.
2. The phrase up to isomorphism signfies that any commutative group of order pn should structurally identical (isomorphic) to one of the groups of order pn exhibited.
3. For any ve integer n, a summation n n1 n2 ... nr, where 0 ni ≤ nj for i j ,is
36 called a partition of n.
37 4. The first remark indicate that each commutative group of order pn give rise to a partition of n. From the opposite point of view, starting with a partition n n1 n2 ... nr ,we can always produce a commutative group G of order pn; simply take G to be the external direct product of Zpn1 Zpn2 ...Zpnr .
5. Briefly, there are just as many non-isomorphic commutative groups of order pn as there are ways to express n as a sum of positive integers.
2 6. As an example, there are only two commutative groups of order p , either Zp2 or Zp Zp. 2 These groups are not isomorphic, for Zp2 contains an element of order p , but every element of Zp Zp other than the identity has order p.
7. The number of non-isomorphic commutative groups of order n pk1 .pk2 ....pkr (distinct primes) is
m k1.k2...kr,
where ki denotes the number of partitions of the integer ki.
Example 1.3.18: Find all commutative groups up to isomorphism (all non-isomorphic) of order 24.
Solution : |G| 24 23.31. m 3.1 3.1 3. Hence there are 3 − commutative groups up to isomorphism of order 24 namely, 1. G Z2 Z2 Z2 Z3.
2. G Z22 Z2 Z3. 3. G Z23 Z3.
Example 1.3.20: Find all non-isomorphic commutative groups of order 243.
Solution : |G| 243 35 5 5 0 4 1 3 2 3 1 1 2 1 1 1 1 2 2 1 1 1 1 m 5 7, therefore there are 7 non-isomorphic commutative
38 groups namely:
1. G Z35 2. G Z34 Z3 3. G Z33 Z32 4. G Z33 Z3 Z3 5. G Z32 Z3 Z3 Z3 6. G Z3 Z32 Z32 7. G Z3 Z3 Z3 Z3 Z3.
Remark 1.3.21: There are other theorems which give insight into the structure of finite commutative groups. A seconed basic result states that any finite commutative group G is isomorphic to a direct product
Zm1 Zm2 .... Zmr where mi ∣ mi1 for 1 ≤ i ≤ r − 1.(2)
Theorem 1.3.22 (Fundamental Th. of Finitely Generated Com. Groups) Let G be a finite commutative group, then G is isomorphic to two different types of direct products of cyclic groups as follows:
G Z n1 Z n2 ... Zpnr (1) p1 p2 r
where the pi are primes and are not necessarily distinct, this direct product of cyclic groups of prime power order isomorphic to G is unique except for a rearrangement of the factors.
G Zm1 Zm2 ....Zmr (2)
where mi ∣ mi1 . The numbers mi are called The torsion coefficients of G , and they are unique except for a rearrangement of the factors.
Remark 1.3.23: The following, is describing the method of finding a group expressed in (2) which is isomorphic to a given direct product of cyclic groups of prime power order given in (1): For each prime appearing in the order of the group , write the subscripts in the direct product involving that prime in a row in order of increasing magnitude. Keep the right-hand ends of the rows aligned. Thus starting with Z2 Z22 Z3 Z3 Z5,we form the array
39 222 33 5
Then take the product of the numbers in each column getting
Z2 Z22 Z3 Z3 Z5 Z6 Z60.
Example 1.3.24: Find all commutative groups up to isomorphism of order 360 in both types.
Solution : |G| 360 23325. m 3.2.1 3.2.1 6. Using theorem 1.3.22 ,we get:
1. G Z2 Z2 Z2 Z3 Z3 Z5 Z2 Z6 Z30
2. G Z2 Z22 Z3 Z3 Z5 Z6 Z60
3. G Z23 Z3 Z3 Z5 Z3 Z120
4. G Z2 Z2 Z2 Z32 Z5 Z2 Z2 Z90
5. G Z2 Z22 Z32 Z5 Z2 Z180
6. G Z23 Z32 Z5 Z360
Our final theorem provides an application of the results of this section:
Theorem 1.3.25: If n 1 is a square-free integer (that is, n is not divisible by the square of any prime), then any commutative group of order n is isomorphic to Zn.
Proof : Since n is a square-free integer, we have
|G| n p1p2...pr where the pi are distinct primes.
40 Now Theorem 1.3.10, asserts that
G Gp1 ⊗ Gp2 ⊗ ... ⊗ Gpr with |Gpi | pi.
Since Gp1 , Gp2 , ..., Gpr are primes order , then
Gp1 Zp1 , Gp2 Zp2 , ..., Gpr Zpr . Hence
G Zp1 ⊗ Zp2 ⊗ ... ⊗ Zpr Zp1p2...pr Zn.
Corollary 1.3.26: If n is a square-free integer, then all commutative groups of order n are isomorphic.
41 PROBLEMS 1.3
1. The homomorphic image of p − group is a p −group. i.e., if f : G G′ is a homomorphism and H ≤ G such that H is a p −group, show that fH is a p-group.
2. Find all commutative groups up to isomorphism of order 720, 1089. Express them in both forms (1) and (2) of theorem 1.3.22 and pair up isomorphic groups of forms (1) and (2).
3. How many groups up to isomorphism are there of order (a) 24 (b) 25 (c) of order (24)(25).
4. Find all non-isomorphic commutative groups of order 400.
5. Given G G1 G2 ... Gn,whereGi are groups. Show that a. ZG ZG1 ZG2 ... ZGn. b. G,G G1,G1 G2,G2 ... Gn,Gn.
6. Find the non-isomorphic commutative groups of order p3,p4,p5.
42 1.4 THE CLASS-EQUATION AND CAUCHY THEOREM
We are going in this section to develop the class-equation of a finite group G and this equation is another way for counting the order of any finite group. To find the class-equation it depends on defining an equivalence relation on a set G, measures the size of the equivelance classes under this relation, and then equates the number of elements in the set to the sum of the orders of these equivalence classes. This kind of an approach will be illustrated in this section. We shall introduce a relation, prove it is an equivalence relation, and then find a neat algebraic description for the size of each equivalence class. From this simple description we will get a powerful results about finite groups. The development of the class-equation of a finite group, needs some preparation. We begin by introducing a notion of conjugacy for group elements.
Definition 1.4.1: Let G be a group and a,b ∈ G. The element b is said to be a conjugate of a, if there exists some x ∈ G for which b xax−1.
Theorem 1.4.2: Given a,b ∈ G we may define a relation in G by requiring that a b b is a conjugate of a. This relation is an equivalence relation.
Proof : 1. Since a eae−1 a is conjugate to itself a a. 2. Let a b b xax−1 x−1bx a x−1bx−1−1 a b a. 3. Let a b, b c b xax−1, c yby−1 c yxax−1y−1 yxayx−1
a c.
The relation induces a partition of the set G into equivalence classes,
43 usually referred to as the conjugacy classes of G.
44 Definition 1.4.3: The equivalence class containing an element a of a group G under the relation of conjugacy is called the conjugate class of a and denoted by Ca. Ca b xax−1 : x ∈ G.
Remark 1.4.4: 1. The conjugacy class of an element a consists of all the elements in G that are conjugate to a.
2. Two conjugacy classes are either disjoint or identical
3. The group G is the union of its disjoint conjugacy classes.
Example 1.4.5: Let G be a commutative group. Then Ca xax−1 : x ∈ G axx−1 : x ∈ G a.
Example 1.4.6 (commutative group):
Find all distinct conjugacy classes for Z4.
Solution :
Z4 0̄,1̄,2̄,3̄. −1 −1 −1 C0̄ x0̄x : x ∈ Z4 0̄ ⊕ 0̄ ⊕ 0̄ ,... , 3̄ ⊕ 0̄ ⊕ 3̄ 0̄. −1 C1̄ x1̄x : x ∈ Z4 1̄. −1 C2̄ x2̄x : x ∈ Z4 2̄. −1 C3̄ x3̄x : x ∈ Z4 3̄.
Example 1.4.6(non-commutative group):
Find all distinct conjugacy classes for S3.
Solution :
S3 i, 23, 13, 12, 123, 132
45 −1 Ci i : ∈ S3 i −1 C23 23 : ∈ S3 23, 12, 13 −1 C123 123 : ∈ S3 123, 132
The conjugate classes are Ci, C23, C123.
Remark 1.4.7: 1. For a finite group an immediate question presents itself: How large is Ca? Of course, this depends strongly on the element a. For instance, if a ∈ ZG, then ax xa for all x ∈ G, hence xax−1 a; in other words, the conjugacy class of a in this case consists merely of the element a itself. On the other hand, if Ca consists only of the element a, then xax−1 a for all x ∈ G; this gives us that xa ax for all x ∈ G,hence a ∈ ZG.So ZG is characterized as the set of those elements, a,inG whose conjugacy class has only one element , a, itself.
2. For a commutative group G,SinceG ZG, two elements are conjugate iff they are equal. So conjugacy is not an interesting relation for commutative groups; however, for non-commutative it is highly interesting notion. (see ex. 1.4.5, 1.4.6).
3. Example 1.4.6, illustrates that the conjugate classes of a group need not contain the same number of elements.
4. The number of elements in each conjugate class divide the order of G (finite group).
5. Given a ∈ G, Ca consists of all xax−1 as x runs over G.Sotodeterminewhicharethe distinct conjugates of a, we need to know when two conjugates of a coincide, which is the same as asking : When xax−1 yay−1? In this case, transposing we obtain ax−1y x−1ya;inotherwords,x−1y must commute with a. This brings us to introduce the following definition which is called the centralizer of a in G.
Definition 1.4.8: Let G be a group and a ∈ G. The centeralizer of a in G is defined to be the set of all elements in G that commute with a, and denoted by Ca. i.e., Ca x ∈ G : xa ax.
Theorem 1.4.9: For a ∈ G, Ca is a subgroup of G.
46 Proof : Exercise.
47 Example 1.4.10:
In S3 , evaluate the centralizer of every element. What is the ZS3 ?
Solution :
S3 i, 23, 13, 12, 123, 132.
1. Ci ∈ S3 : i i S3
2. C12 ∈ S3 : 12 12 i, 12
3. C13 ∈ S3 : 13 13 i, 13
4. C23 ∈ S3 : 23 23 i, 23
5. C123 ∈ S3 : 123 123 i, 123, 132
6. C132 ∈ S3 : 132 132 i, 123, 132
ZS3 ∈ S3 : ∀ ∈ S3 i.
Hence the centralizer is different than ZS3.
Remark 1.4.11: 1. The centralizer of an element is closely related to the center of a group. These concepts are not identical.
2. ZG Ca. a∈G
3. a ∈ ZGG Ca.
4. G is commutative its conjugacy classes is a singleton.
5. a ∈ ZGCa a.
The centralizer leads to a very useful fact about the size of conjugacy classes:
Theorem 1.4.12: Let G be a finite group, a ∈ G. 1. Then the numbers of conjugate elements of a in G equals to the number of all distinct left cosets of Ca in G. i.e., |Ca| G : Ca.
48 2. The number of Ca is a divisor of |G|. i.e., |Ca| ∣ |G|.
49 Proof : 1. For arbitrary elements x,y ∈ G, the following two conjugates are equal iff:
xax−1 yay−1 a x−1yay−1x a x−1yax−1y−1 ax−1y x−1ya x−1y ∈ Ca xCa yCa
So x, y give rise to the same conjugate of a if and only if x and y are in the same left coset of of Ca in G. Thus there are as many conjugates of a in G as there are left cosets of Ca in G. Hence |Ca| G : Ca.
2. By Lagrange’s Theorem: |G| |Ca|.G : Ca G : Ca ∣ |G| |Ca| ∣ |G|.
Conjugacy class-equation: Let G be a finite group, as each element of G lies in one and only one conjugacy class, we may determine the number of elements in G by counting them class by class as in the following theorem:
Theorem 1.4.13 (The Class Equation I): For any finite group G,wehave |G| |G| ∑G : Ca ∑ ,(I) a∈G a∈G |Ca| where the sum runs over one a from each conjugacy class.
Proof : Let Ca1, Ca2, ..., Cat be all distinct conjugate classes of G. Hence
G Ca1 Ca2 ... Cat
and Cai ∩ Caj ∅ if i ≠ j. Hence
|G| |Ca1| |Ca2| ... |Cat| Using Theorem 1.4.12, we get
50 |G| G : Ca1 G : Ca2 ... G : Cat |G| ∑G : Ca. a∈G
Theorem 1.4.14 (The Class Equation II): Let G be a finite group . Then the class equation in Theorem 1.4.13, can be written as : |G| |ZG| ∑ G : Ca (II) a∉ZG where |ZG| is the number of elements in G whose conjugacy class has only one-element.
Proof : Observe that by remark 1.4.11, a ∈ ZG Ca a. Hence |ZG| equal to the union of all conjugacy classes that contain one-element, i.e.,
|ZG| |Ca| ( disjoint classes) a∈ZG ∑ |Ca| a∈ZG ∑ G : Ca a∈ZG
Therefore as a ∈ G a ∈ ZG or a ∉ ZG. Therefore the class equation can be count as: |G| ∑G : Ca a∈G ∑ G : Ca ∑ G : Ca a∈ZG a∉ZG ∑ |Ca| ∑ G : Ca a∈ZG a∉ZG |ZG| ∑ G : Ca. a∉ZG As an application of what we have talking about here, we shall prove the Cauchy’s Theorem. First we are going to prove the Cauchy’s Theorem for only commutaive groups. Next, we are going to establish Theorem 1.4.15
51 without the hypothesis of commutativity, that is we are going to prove that Cauchy’s Theorem is true for all groups. The class equation provides the key to the proof of Cauchy’s Theorem 1.4.16.
52 Theorem 1.4.15 (Cauchy Theorem For Commutative Group): Let G be a finite commutative group of order n.Ifp is a prime number such that p ∣ |G|, then G contains an element of order p.i.e., ∃ a ∈ G : e ≠ a, ap e.
Proof : We prove this theorem by induction on |G|. If |G| 1, there is no such p and the theorem is trivially true. Suppose that the theorem is true for all commutative groups of order less than |G|. Now, we have to prove that the theorem is true for G. G ↙↘ 1 G has no-proper subgroup (2) G has a proper subgroup e ≠ H ≠ G ↙↘ Then any a ≠ 0inG must be (i) p ∣ |H| generator of G,G a . Since H G, then H is a (ii) p ∤ |H| G is cyclic. finite commutative group of Since p ∣ |G| order less than |G|.Then by |G| p assumption there would be an any element a ≠ e ∈ G element a ∈ H ⊂ G s.t.ap e. satisfies ap e. Therefore it is true for G.
Case (ii) p ∤ |H| : Let |H| m,sincep ∤ |H| p ∤ m. ∵ G is commutative H G and we can form G/H. Moreover, G/H is a commutative group. Since p ∤ |H| and p ∣ |G| p ∣ |G/H|. ∵ H ≠ e |H| 1and|G/H| |G|. Thus, by induction since G/H is a finite commutative group of order less than |G|, there exists an element in G/H such that the order of this element is p. i.e., ∃ 0̄ ≠ ā ∈ G/H s.t. āp ē aHp H ap ∈ H.
53 By Lagrange’s Theorem ap|H| e apm amp e. Set b am ∈ G. We prove that b am ≠ e. If b e am e amH H aHm H ām ē p ∣ m . Hence b ≠ e, therefore there is an element b ∈ G such that bp e.
Theorem 1.4.16 (Cauchy’s Theorem For All Groups): Let G be a finite group of order n.Ifp is a prime number such that p ∣ |G|, then G contains an element of order p.i.e., ∃ a ∈ G : e ≠ a, ap e.
Proof : We prove this theorem by induction on |G|. If |G| 1, there is no such p and the theorem is trivially true. Suppose that the theorem is true for all groups of order less than |G|. Now, we have to prove that the theorem is true for G. Consider the class equation for G : |G| |ZG| ∑ G : Ca. a∉ZG G ↙↘ (1) G ZG (2) G ≠ ZG
∃ a ∈ G s.t.a ∉ ZG. G is commutative. For such an a, G : Ca 1, Hence by theorem |G| |Ca|.G : Ca |Ca|. 1.4.15,G contains an Since |Ca| |G| and the theorem is assumed element of order p. to hold for he subgroup Ca, therefore we have two cases : Ca ↙↘ (i) p ∣ |Ca| (ii) p ∤ |Ca|, a ∉ ZG
54 In this case the Since p ∣ |G| p ∣ |Ca|G : Ca theorem holds p ∣ G : Ca p divides by assumption. each term of ∑G : Ca p ∣ ∑G : Ca. Using the class equation p ∣ |ZG|,sinceZG is a commutative group, by theorem 1.4.15, ZG contains an element of order p, and hence G contains an element of order p.
Example 1.4.17:
Consider the commutative group Z8,⊕ and the non-commutative group S3,∘. Since 2 ∣ |Z8 | and 2 ∣ |S3 |, then by theorem 1.4.15, 1.4.16, respectively, Z8, S3 contain an element of order 2, these elements are 4̄ ∈ Z8 and 12 ∈ S3.
We now have the background to derive a partial converse of Lagrange’s theorem, to the effect that any finite group of order divisible by a prime number p , has a subgroup of order p.
Corollary 1.4.18: If the prime number p divides the order of the finite group G , then G contains at least a subgroup of order p.
Proof : By Cauchy’s Theorem there exists an element a ∈ G such that ap e. Consider the cyclic subgroup generated by an element a. i.e., H a e, a, a2, ..., ap−1 Then H is a subgroup of order p.
Cauchy’s Theorem has many consequences. We shall present one of
55 these, in which we determine completely the nature of certain groups of order pq ,wherep and q are distinct primes.
Theorem 1.4.19: Let G be a group of order pq,wherep and q are prime numbers and p q. Then 1. G has one and only one subgroup of G say A of order p. 2. A G.
Proof : 1. G has one and only one subgroup of G say A. By Cauchy’s theorem, G has a subgroup of order p. Assume that we have two distinct subgroups of order p,sayA and B. i.e., |A| p |B|,andA ≠ B. Consider the set AB x ab : a ∈ A, b ∈ B.
2 We claim that AB has p distinct elements, for let x a1b1,y a2b2 ∈ AB s.t. x y
a1b1 a2b2,wherea1,a2 ∈ A, b1,b2 ∈ B −1 −1 a2 a1 b2b1 ∈ A,B −1 −1 a2 a1 ∈ A ∩ B, b2b1 ∈ A ∩ B. But A ∩ B ≤ A |A ∩ B| ∣ |A| p. |A ∩ B| 1or|A ∩ B| p.
If |A ∩ B| p A B . Therefore |A ∩ B| 1 A ∩ B e
−1 −1 a2 a1 e and b2b1 e
a2 a1 and b2 b1 Thus the number of distinct elements in AB is p2. But all these elements are in G,which has only pq p2 elements(since p q. With this contradiction we see that A B and A is the only subgroup of order p.
2. A G. Consider the two left and right cosets, then
56 |xA| |Ax| |A| p. Since A is the only subgroup of order p xA Ax A A G.
Another application of Cauchy’s Theorem is given in the following Corollary:
Notation : We have been proved the following corollary 1.4.20, in section 1.3, theorem 1.3.9, but for only finite commutative group.
Corollary 1.4.20: G ≠ e is a finite p − group iff |G| pk,forsomek 0.
Proof : ( Let |G| pk, then by Lagrange’s theorem , each element of G has a power of p as its order, This G is a p −group.
57 ( Suppose G is a finite p −group such that |G| n ≠ pk. By Fundamental theorem for arithmetic 1.1.60:
k |G| p1p2...pr s.t. pi ≠ p ∀ i
p1 ∣ |G|, p2 ∣ |G|, ..., pr ∣ |G|
Therefore by Cauchy’s theorem , G contains an elements of order p1,of order p2, ..., of order pr, hence we arrive at contradiction , since G is a p −group. Thus p is the only divisors of |G|. |G| pk.
While we have seen examples of groups that have a trivial center (for instance, the symmetric groups Sn with n 2, finite p −group possess non-trivial centers. This is a content of the following theorem due to Burnside.This theorem is another application of the class equation.
Theorem 1.4.21 (Burnside): If G is a finite p-group with more than one element then ZG ≠ e. i.e., |ZG| 1.
Proof : Since G is a finite p −group then |G| pn. The proof follows directly from the class equation |G| |ZG| ∑ G : Ca. a∉ZG We have two cases: ↙↘ (1) a ∈ ZG (2) a ∉ ZG
Ca ≠ G |Ca| ∣ |G| If a ∈ ZG |Ca| pr, r n ZG has e,anda Using the class equation, we get |ZG| 1. as p ∣ |G|, p ∣ ∑G : Ca p ∣ |ZG| |ZG| 1
58 This last theorem has the most interesting application, which you may have seen in solving problem 2 of section 1.4.This is
59 Theorem 1.4.22: If G is a group of order p2,wherep is a prime, then G is commutative.
Definition 1.4.23: A group G is said to be a simple group , if it has only trivial normal subgroups.
Corollary 1.4.24: A non-commutative group G of order pk where p is a prime number and k 1 is never simple.
Proof : From Theorem 1.4.21, |ZG| 1. Hence the center of G, which is normal in G, forms a non-trivial subgroup.
60 PROBLEMS 1.4
1. Suppose G is any group. Show that if ZG is cyclic G is commutative.
2. Prove that if |G| p2, then G is commutative. (use problem 1)
3. If |G| p3,andG is not commutative |ZG| p.
4. Let N G. Then any subgroup H̄ H/N where H ≤ G, containing N. Show that if H̄ G/N then H G.
5. If |G| pn, n ≥ 1. Then G has a normal subgroup of order pn−1.
6. Let G A4,∘. Find all distinct conjugate classes and then verify theorem 1.4.13.
∗ 7. Find all conjugate classes of Z7 ,⊙.
8. Let G be a group of order 9. Show that G is either a cyclic group or oa 3 ∀ a ∈ G.
9. Let G a s.t. |G| 22. Show that there is an element a ∈ G s.t. a2 e and 10 elements of order 11.
61 1.5 THE SYLOW THEOREMS
Non-commutative finite groups are vastly more complicated than finite commutative groups which were completely classified up to isomorphism in section 1.3.The Sylow Theorems are a basic first step in understanding the structure of an arbitrary finite group.Our motivation is the question: If a positive integer m divides the order of a group G, does G have a subgroup of order m? This is the converse of Lagrange’s Theorem. The answer to this question is no. For example, the alternating group A4 is of order 12, but has no subgroup of order 6. Cauchy’s Theorem gives a weak version of the converse (for prime divisors). The major results in this section are due to Sylow.
Example 1.5.1:
Show that the alternating group A4 has no subgroup of order 6.
Solution :
1 2 3 4 5 6 7 8
A4 i, 234, 243, 134, 143, 124, 142, 123, 132,
1 2 3 1324, 1423, 1234 .
Suppose A4 has a subgroup H of order 6. Note that
2 2 2 3 1 2 3 i and i i for i 1,...8
o1 o2 o3 2andoi 3 ∀ i 1,...,8. Since |H| 6, and 2,3 ∣ |H|, then by corollary 1.4.18, H contains a subgroup of order 3 and a subgroup of order 2
i, j ∈ H for some i,j
say 1, 1 ∈ H
62 2 i, 1, 1 2, 11 4, 11 8 ∈ H 2 2 4 3 ∈ H and 8 7 ∈ H. This means that H has at least 8 distinct elements, which contradicts the assumption that |H| 6. Thus 1 ∉ H. A similar argument shows 2,3 ∉ H. This means that H does not contain subgroups of order 2 . Therefore A4 does not contain a subgroup of order 6.
The answer of our question will be in the following theorem which is called The First Sylow Theorem.
Definition 1.5.2: Let G be a finite group and p be a prime number. If pr ∣ |G| and pr1 ∤ |G| then a subgroup P of order pr is called a Sylow p −subgroup of G. i.e., P is a Sylow p −subgroup if P is a p −group and is not properly contained in any other p −subgroup of G.
Theorem 1.5.3 (First Sylow Theorem): Let G be a finite group , p be a prime such that pr ∣ |G| and pr1 ∤ |G|. Then G contains a Sylow p −subgroup of order pr.
Proof : Let |G| n prm where p ∤ m. We will prove the theorem by induction on n. The theorem is certainly true for n 1, and so we assume that it holds for all groups of order less than n. Consider the class equation |G| |ZG| ∑ G : Ca. a∉ZG We will consider two cases: ↙↘ (1) pr ∣ |ZG| (2) pr ∤ |ZG|, pr ∣ |G|
63 p ∣ |ZG|, by Cauchy’s theorem ZG has an element a of order p. Let N be a subgroup generated by a By Lagrange’s Theorem |G| |Ca|G : C |N| p.Since ZG is commutative ∵ pr ∣ |G| pr ∣ |Ca| or pr ∣ G : Ca N ZG G.Consider G/N : If pr ∣ G : Ca pr ∣ ∑G : Ca r n p .m r−1 r |G/N| p p p m n. By class equation p ∣ |ZG| . r pr−1 ∣ |G/N|, by assumption Hence p ∣ |Ca| and |Ca| |G|, G/N has a subgroup of order pr−1, by induction assumption, Ca has a subgrou r say, H̄ H/N of order pr−1. H of order p ,consequently so does G. |H̄ | |H|/|N| |H| |N|.pr−1 pr. ∃ a subgroup of order pr.
The basic tools needed to unterstand the last two Sylow Theorems are very similar to those used in section 1.4, except that we will now deal with conjugate subgroups rather than conjugate elements. More precisely, let H be a fixed subgroup of a group G and A, B be any subgroups of G.
Definition 1.5.4: Let G be a group, H be a fixed subgroup and A, B be two subgroups of G. We say that A is H −conjugate B, if there exists an x ∈ H such that B xAx−1 xax−1 : a ∈ A, when G H,thenwesayA is conjugate to B.
Example 1.5.5:
Consider the symmetric group S3 i, 12,13,23,123,132. ∵ |S3 | 6 and 2 ∣ |S3 |, then by Theroem 1.5.3, S3 has a Sylow 2-subgroup of order 2.Namely,
P1 i, 12, P2 i, 23, i, 13. Thus we have 3 −Sylow 2 −subgroup of order 2. (The number of the Sylow p −subgroups will be given in the Third Sylow’s Theorem). These subgroups are conjugate since ∃ 132, 123 ∈ S3 s.t.,
64 −1 −1 −1 P2 132i132 , 13223132 i, 12 P1 −1 −1 −1 P3 123i123 , 12313123 i, 12 P1
−1 −1 P1 P2 P3 . Note the number of Sylow 2 −subgroups must divides |G|.
Theorem 1.5.6: If H ≤ G and a ∈ G, then aHa−1 aha−1 : h ∈ H ≤ G, which is called the conjugate subgroup of H induced by the element a.
Proof : Let x,y ∈ aHa−1
−1 −1 x ah1a , y ah2a where h1,h2 ∈ H. Consider
−1 −1 −1 −1 −1 xy ah1a ah2a ah3a ∈ aHa .
Theorem 1.5.7: The conjugate subgroups of H are all isomorphic to H.
Proof : Define f : H aHa−1 by fh aha−1 h ∈ H. It is easy to show that this mapping is an isomorphism H aHa−1 |H| |aHa−1 |.
Notation : A Sylow p −subgroup of a group G is a maximal p −subgroup of G,thatis, a p −subgroup contained in no larger p −subgroup of G.
Theorem 1.5.8: Let P be a Sylow p −subgroup of the group G.Iff is any automorphism of G, then fP is a Sylow p −subgroup of G.
Proof :
65 Let f : G G be an automorphism with P be a Sylow p −subgroup of G. P is a p −group fP is a p −group of G. We have to show that fP is a Sylow p −subgroup of G ?? Assume that fP is not a Sylow p −subgroup of G. Then fP ⊂ P′ (in some p − subgroup).(*) Since f−1 : G G is an automorphism with P′ is a p-group f−1P′ is a p − group containing P i.e., P ⊂ f−1P′. This inclusion is proper, for from (*) ∃ a ∈ P′ s.t. a ∉ fP
f−1a ∈ f−1P′ s.t. f−1a ∉ P. HencewearrivethatP is contained in a p −subgroup of G ,sinceP is a Sylow p −subgroup of G. Hence fP must be a Sylow p −subgroup of G.
Corollary 1.5.9: Any conjugate of a P Sylow p −subgroup of G is again a Sylow p −subgroup of G.
Proof : −1 Define fa : G G by faP aPa , a ∈ G. Then clearly, fa is an 1-1, onto, and homomorphism. Hence fa is an −1 automorphism, then by theorem 1.5.8, faP aPa is a Sylow p −subgroup of G.
Theorem 1.5.10 (Second Sylow’s Theorem)
If P1 and P2 are Sylow p −subgroups of a group G, then there exists −1 x ∈ G s.t. P1 xP2x . i.e., that is any Sylow p −subgroups of a group G are conjugate.
Remark 1.5.11: Any two Sylow p −subgroups of a group G are isomorphic.
Theorem 1.5.12 (Third Sylow’s Theorem) The number of Sylow p −subgroups of a finite group G divides |G| and is
66 of the form np 1 pm ∣ |G| where m 0,1,2,......
Applications:
Example 1.5.13:
Find all Sylow subgroups of a group S3.
Solution :
|S3 | 3! 6 2.3 By 1st Sylow’s Theorem, there are : Sylow 2 −subgroups of order 2 and Sylow 3 −subgroups of order 3. To find the number of these subgroups, we use 3ed Sylow’s Theorem:
Let n2 the number of Sylow 2-subgroups 1 2mm 0,1,......
if m 0 n2 1 ∣ 6 ✓
if m 1 n2 3 ∣ 6 ✓
if m 1 n2 5 ∤ 6 Hencewehave3−Sylow 2 −subgroups of order 2 which are :
P1 i,12, P2 i, 13, P3 i, 23.
Let n3 the number of Sylow 3 −subgroups 1 3mm 0,1,......
if m 0 n3 1 ∣ 6 ✓
if m 1 n3 4 ∤ 6
if m 1 n2 7 6 Hence there is only one Sylow 3-subgroup of order 3 which is
A3 i, 123, 132 S3.
Example 1.5.14: Find all Sylow subgroups of the group of order 36.
Solution : |G| 36 22.32. Let n2 be the number of Sylow 2 −subgroups of order 4. Then
67 n2 1 2mm 0,1,.... if m 0 n2 1 ∣ 36 ✓ if m 1 n2 3 ∣ 36 ✓ if m 2 n2 5 ∤ 36 if m 3 n2 7 ∤ 36 if m 4 n2 9 ∣ 36 ✓ if m 5 n2 11 ∤ 36 if m 6 n2 13 ∤ 36 13 3 39 elements |G| Thus we have 9 −Sylow 2 −subgroups of order 4.
Let n3 be the number of Sylow 3 −subgroups of order 9. Then n3 1 3mm 0,1,.... if m 0 n3 1 ∣ 36 ✓ if m 1 n3 4 ∣ 36 ✓ if m 2 n3 7 ∤ 36 7 8 56 elements |G| Thus we have 4 −Sylow 3 −subgroups of order 9.
Example 1.5.15: Find all Sylow subgroups of the group of order 56.
Solution : |G| 56 23.7 Let n2 be the number of Sylow 2 −subgroups of order 8. Then n2 1 2mm 0,1,.... if m 0 n2 1 ∣ 56 ✓ if m 1 n2 3 ∤ 56 if m 2 n2 5 ∤ 56 if m 3 n2 7 ∣ 56 ✓ if m 4 n2 9 ∤ 56 9 7 63 elements |G| Thus we may have 7 −Sylow 2 −subgroups of order 8.
Let n7 be the number of Sylow 7 −subgroups of order 7. Then n7 1 7mm 0,1,.... if m 0 n7 1 ∣ 56 ✓ if m 1 n7 8 ∣ 56 ✓ if m 2 n7 15 ∤ 56 7 8 56 elements |G| Thus we may have 8 −Sylow 7 −subgroups of order 7.
Theorem 1.5.16:
68 A Sylow p −subgroup of a finite group G is normal iff it is a unique Sylow p −subgroup.
Proof : ( Let H and K be two Sylow p −subgroups such that H G. From the 2ed Sylow’s Theorem, H and K are conjugate. i.e., K gHg−1 for some g ∈ G. But H G gH Hg ∀ g ∈ G gHg−1 H H K.
Let H be the only Sylow p −subgroup of a group G. Then for g ∈ G the conjugate gHg−1 is a subgroup of G such that |gHg−1 | |H|. By corollary 1.5.9, the conjugate of H is again a Sylow p −subgroup, gHg−1 is a Sylow p −subgroup.But we have only one therefore gHg−1 H H G.
Example 1.5.17: Show that there is no simple groups of order 56.
Solution : From example 1.5.15, The number of Sylow 7 − subgroups is either 1 or 8. 1. If we have 1-Sylow 7 −subgroup of order 7, then by theorem 1.5.16, this subgroup is normal and hence G is not simple.
2. If we have 8 −Sylow 7 −subgroups of order 7, then each subgroup has 6 elements except the identity so 8 6 48 elements. So it remains 56 − 48 8 elements. Thus there is only 1 Sylow 2 −subgroup of order 8 which is again normal by theorem 1.5.16.Hence G is not simple.
On the other side, the number of Sylow 2 − subgroups is either 1 or 7. 1. If we have 1 Sylow 2 −subgroup of order 8, then by theorem 1.5.16, this subgroup is normal and hence G is not simple.
2. If we have 7 −Sylow 2 −subgroups of order 8, then each subgroup has 7 elements except the identity so 7 7 49 elements.
69 So it remains 56 − 49 7 elements. Thus there is only 1 Sylow 7 −subgroup of order 7 which is again normal by theorem 1.5.16.Hence G is not simple. More Applications: Theorem 1.5.18: Let G be a group of order pq where p and q are distinct prime numbers such that p ∤ q − 1andq ∤ p − 1. Then the following is true: 1. G has a normal subgroup P of order p. 2. G has a normal subgroup Q of order q. 3. G is not a simple group. 4. G is a cyclic group.
Proof : 1. G has a normal subgroup P of order p. From 1st Sylow’s Theorem, G has a Sylow p −subgroup of order p,sayP. By 3ed Sylow’s Theorem the number of Sylow p −subgroups of order p are
np 1 mp m 0,1,.....
There are 4 −possibilities: np ↙↙↘↘ 1 np 1 2 np p 3 np q 4 np pq there is only 1 mp p 1 mp q one Sylow subgroup mp p − 1 mp q − 1 of order p. p ∣ p − 1 p ∣ q − 1 P G. . . 1 mp pq mp pq − 1 p ∣ pq − 1 .
2. G has a normal subgroup Q of order q. From 1st Sylow’s Theorem, G has a Sylow q −subgroup of order q,sayQ. By 3ed Sylow’s Theorem the number of Sylow q −subgroups of order q are
nq 1 mq m 0,1,.....
70 There are 4 −possibilities: nq ↙↙↘↘ 1 nq 1 2 nq q 3 nq p 4 nq pq there is only 1 mq q 1 mq p one Sylow subgroup mq q − 1 mq p − 1 of order q. q ∣ q − 1 q ∣ p − 1 Q G. . . 1 mq pq mq pq − 1 q ∣ pq − 1 .
3. G is not a simple group. From (1), and (2) G is not simple.
4. G is a cyclic group. From (1) and (2) G has a normal subgroup P of order p and a normal subgroup Q of order q. i.e., P G , Q G and |P| p , |Q| q. Therefore P and Q are cyclic subgroups. Hence
P Zp and Q Zq.
Consider the subgroup P ∩ Q .ByLagrange’sTheorem |P ∩ Q| ∣ pq |P ∩ Q| ∣ p or |P ∩ Q| ∣ q
|P ∩ Q| 1 P ∩ Q e. Moreover, |P||Q| |P||Q| |PQ| pq |G|. |P ∩ Q| 1 Therefore, G PQ. Hence
71 G P ⊗ Q P Q Zp Zq Zpq. so that G is cyclic.
Example 1.5.19: Show that any group of order 15 must be commutative.
Solution : From Theorem 1.5.18, |G| 3.5 such that 3 ∤ 5 − 1 and 5 ∤ 3 − 1. Hence G is cyclic G is commutative.
72 The following concepts are necessary for examples 1.5.24, 25, 26.
Definition 1.5.20: (i) Let H be a subgroup of G and a ∈ G.Then the conjugate subgroup of H induced by an element a is a set aHa−1 aha−1 a ∈ G.
(ii) If aHa−1 H for some a ∈ G, we say that the subgroup H is self-conjugate under a.
Remark 1.5.21: 1. Clearly,aHa−1 is a subgroup of G.
2. Any subgroup is self-conjugate under each one of its own elements, since hHh−1 H.
3. The concept of a normal subgroup can be arrived at in an alternative way: H G aHa−1 H ∀ a ∈ G. H is self -conjugate under every element of G.
4. The conjugate subgroups of H are all isomorphic to H, hence isomorphic to each other.Since we can define f : H aHa−1 fh aha−1. Clearly, f is an isomorphism.
5. The conjugate subgroups have the same order. i.e., |H| |aHa−1 |.
Definition 1.5.22:
(i) The normalizer of H in G ,denoted by NGH, is defined to be the set of all elements under which H is self-conjugate, i.e.,
−1 NGH g ∈ G : gHg H.
(ii) If K is a subgroup of G, then NKH is defined as :
−1 NKH k ∈ K : kHk H.
73 Remark 1.5.23:
1. NGH is a subgroup of G. −1 −1 Let a,b ∈ NGH aHa H and bHb H. −1 −1 −1 −1 −1 −1 Consider ab Hab ab Hba aHa H NGH ≤ G.
2. H is not necessarily normal in G.
3. H NGH. Since ∀ a ∈ NGH : aH Ha.
4. NGH is the largest subgroup of G in which H is normal in NGH. For let M be another subgroup of G such that H M aH Ha ∀ a ∈ M −1 aHa H ∀ a ∈ M a ∈ NGH M ≤ NGH.
5. H G NGH G. If NGH G ,then by (4), NGH is the largest subgroup of G in which H is normal in NGH H G. Conversely, let H G gH Hg ∀ g ∈ G gHg−1 H ∀ g ∈ G g ∈ NGH G NGH.
Example 1.5.24: Let |G| 24. Show that G has a normal subgroup of order 8 or 4.
Solution : |G| 24 23.3 By 1st Sylow’s Theorem G has a Sylow 2 −subgroup of order 8. Let n2 be the number of Sylow 2 −subgroups of order 8. Then
n2 1 2mm 0,1,....
if m 0 n2 1 ∣ 24 ✓ if m 1 n2 3 ∣ 24 ✓ if m 2 n2 5 ∤ 24 5 7 35 elements |G|
Hence n2 1orn2 3.
If n2 1
74 ∃ 1 − Sylow 2 − subgroup of order 8 which is normal in G.
75 If n2 3 ∃ 3 − Sylow 2 − subgroups of order 8.
Let H and K be two of them. i.e., |H| |K| 8. Then |H|.|K| |HK| 8.8 |G| |H ∩ K| |H ∩ K|
|H ∩ K| 8.8 2.66(*) 24 But |H ∩ K| ∣ |H| |H ∩ K| 1or2or4or8. From (*) , we get |H ∩ K| 4or8. If |H ∩ K| 8,then |H|.|K| |HK| 8.8 8.8 8 |H ∩ K| |H ∩ K| 8 HK K (since H and K are distinct).
Thus |H ∩ K| 4. It remains to show that H ∩ K G ?? |H| |K| H : H ∩ K 8 2andK : H ∩ K 2 |H ∩ K| 4 |H ∩ K|
H ∩ K H and H ∩ K K
But from remark 1.5.23, NGH ∩ K is the largest subgroup of G in which H ∩ K is normal. Hence
H ⊂ NGH ∩ K and K ⊂ NGH ∩ K
HK ⊂ NGH ∩ K |HK| |NGH ∩ K|
|H|.|K| 8.8 |NGH ∩ K| |HK| 16 |H ∩ K| 4
76 But |NGH ∩ K| ∣ 24 and 16
|NGH ∩ K| 24 G NGH ∩ K Hence from remark 1.5.23, H ∩ K G. Therefore there exists a normal subgroup of order 4.
Example 1.5.25: Let |G| 108. Show that G has a normal subgroup of order 27 or 9.
Solution : |G| 108 22.33 By 1st Sylow’s Theorem G has a Sylow 3 −subgroup of order 27. Let n3 be the number of Sylow 3 −subgroups of order 27. Then
n3 1 3mm 0,1,....
if m 0 n3 1 ∣ 108 ✓ if m 1 n3 4 ∣ 108 ✓ if m 2 n3 7 ∤ 108 7 26 182 elements |G|
Hence n3 1orn3 4.
If n3 1 ∃ 1 − Sylow 3 − subgroup of order 27 which is normal in G.
If n3 4 ∃ 4 − Sylow 3 − subgroups of order 27.
Let H and K be two of them. i.e., |H| |K| 27 and H ≠ K. Then |H|.|K| |HK| 27.27 |G| |H ∩ K| |H ∩ K|
|H ∩ K| 27.27 6.7(*) 108 But |H ∩ K| ∣ |H|
77 |H ∩ K| 1or3or9or27. From (*) , we get |H ∩ K| 9 or 27. If |H ∩ K| 27 , then |H|.|K| |HK| 27.27 27.27 27 |H ∩ K| |H ∩ K| 27 HK K (since H and K are distinct).
Thus |H ∩ K| 9. It remains to show that H ∩ K G ?? |H ∩ K| 32 , |H| 33 and |H ∩ K| 32, |K| 33
H ∩ K H and H ∩ K K
But from remark 1.5.23, NGH ∩ K is the largest subgroup of G in which H ∩ K is normal. Hence
H ⊂ NGH ∩ K and K ⊂ NGH ∩ K
HK ⊂ NGH ∩ K |HK| |NGH ∩ K|
|H|.|K| 27.27 |NGH ∩ K| |HK| 81 |H ∩ K| 9
But |NGH ∩ K| ∣ 108 and 81
|NGH ∩ K| 108 G NGH ∩ K Hence from remark 1.5.23, H ∩ K G. Therefore there exists a normal subgroup of order 27.
Example 1.5.26: Let |G| 45 5.32. Show that G is a commutative group.
Solution : By 1st Sylow’s Theorem G has a Sylow 5 −subgroup of order 5 . Let n5 be the number of Sylow 5 −subgroups of order 5. Then
78 n5 1 5mm 0,1,....
if m 0 n5 1 ∣ 45 ✓ if m 1 n5 6 ∤ 45 if m 2 n5 11 ∤ 45 if m 3 n5 16 ∤ 45 16 4 64 elements |G|
Hence there is only 1 −Sylow 5-subgroup of order 5 say, H, H G.
Consider the quotient group G/H : we have |G/H| 32 which implies, by problem (2) in section (1.4), that G/H is a commutative group . Now, using Theorem 1.1.48 in section 1.1 “ G/H is commutative G′ ⊆ H ”
we get, G′ ⊆ H |G′ | 1or |G′ | 5(1)
Similarly , G has a Sylow 3 −subgroup of order 9.
Let n3 be the number of Sylow 3 −subgroups of order 9. Then
n3 1 3mm 0,1,....
if m 0 n3 1 ∣ 45 ✓ if m 1 n3 4 ∤ 45 if m 2 n3 7 ∤ 45 (7 8 56 elements |G|
Hence there is only 1 −Sylow 3-subgroup of order 9 say, K, K G.
Consider the quotient group G/K : We have |G/K| 5 which implies, that G/K is a cyclic group G/K is a commutative group .
79 AgainusingTheorem1.1.48insection1.1weget, G′ ⊆ K |G′ | 1or |G′ | 3or |G′ | 9(2) From 1 and 2 we get , |G′ | 1 G′ e G is a commutative group.
80 PROBLEMS 1.5
1. Show that any group of order 2p has a normal subgroup of order p.
2. Show that there are no simple groups of orders 20, 30, 96,and 56.
3. Show that there is exactly one Sylow 13 −subgroup in a group of order 130.
4. Show that any group of order 12 must contain a normal subgroup of order 4, 3.
5. Show that any group of order 48 must have a normal subgroup of order 16 or of order 8.
6. Show that any group of order 36 must have a normal subgroup of order 9 or of order 3.
7. Prove that any group of order 1225 must be a commutative group.
8. If P is a Sylow p −subgroup of G, prove that P is a Sylow p −subgroup of NGP and is the only p −Sylow subgroup of NGP.
9. Let H and K be subgroups of a group G. Show that
H K H ⊆ K ⊆ NGH.
81 CHAPTER (II) RINGS THEORY
Group theory gives an information concerning addition in the integers Z the rational Q ,the real numbers R , and the complex numbers C,since additively each of the above number systems is a group. One can also apply group theory to multiplication and division in Q, R,andC (but not in Z)by noting that the non-zero elements of any one of these form a multiplicative group.
However, group theory is not sufficient to deal with all the algebraic properties of the number systems and their generalization. For example, group theory alone does not enable us to derive consequences of the distributive law since this axiom involves both addition and multiplication. Therefore, we now begin a general study of algebraic structures among which are Z, Q, R and C - that satisfy a set of axioms involving addition, and multiplication.
82 2.1 RINGS AND SUBRINGS
This section is devoted to a study of algebriac structures having two binary operations, denoted by S,∗,∘ . Obvious examples are Z,,., Q,,. and PA,∩,. We shall be concerned with the abstract system which is known as a ring. Since a ring is a combination of commutative group and a semi-group so many of the important concepts of a group theory have natural generalizations to systems with two operations.
Definition 2.1.1: A ring is an ordered triple R,,. consisting of a non-empty set R and two binary operations , . defined on R such that : 1. R, is a commutative group . 2. R,. is a semi-group. 3. The operation . is distributive over the operation , i.e. (i) a.b c a.b a.c. (ii) b c.a b.a c.a ∀ a,b,c ∈ R.
Definition 2.1.2: 1. AringR is said to be a commutative ring if ∀ a,b ∈ R : a.b b.a.
2. AringR is said to be a ring with identity (unity),ifR has an element 1 ∈ R s.t. a.1 1.a a ∀ a ∈ R. 3. Let R. ,. be a ring with unity 1. An element 0 ≠ a ∈ R is called a unit of R,ifithasa multiplicative inverse in R. i.e., ∃ b ∈ Rs.t. ab ba 1.
Example 2.1.3: 1. Z,,., Q,,.,R,,.,andC,,. are rings with usual addition and multiplication. 2. Zn,⊕,⊙ is a commutative ring with identity 1 ∈ Zn. 3. N,,., Z,,. and D,,. are not a ring, where D denote the set of odd integers. 4. 0,,. is a commutative ring which is called trivial ring. 5. In Q,,.,andR,,. every non-zero number is a unit. 6. In Z,,. the units are 1, − 1. 7. nZ,,. is a commutative ring without unity.
83 Example 2.1.4: Let Z 2 a b 2 : a,b ∈ Z. Then Z 2 ,,. is a commutative ring where , . are defined by: If x a b 2 , y c d 2 ∈ Z2 where a,b,c,d ∈ Z, then x y a c b d 2 ∈ Z 2 x.y ac 2bd ad bc 2 ∈ Z 2 . It is easy to check that Z 2 is a commutative ring.
Example 2.1.5:
Let R be a ring with unity 1R ∈ R and S be a ring with unity 1S ∈ S. Consider the cartesian product R S r,s : r ∈ R, s ∈ S.
Let x r1,s1, y r2,s2. Define the addition and multiplication by
x y r1,s1 r2,s2 r1 r2,s1 s2 ∈ R S
x.y r1,s1.r2,s2 r1.r2,s1.s2 ∈ R S.
Then R S is a ring with unity 1 1R,1S and if R, S are commutative , then R S is commutative. Thus if R1,R2,...,Rn are rings, then we define
R1 R2 .... Rn r1,r2,....,rn : ri ∈ Ri ,i 1,...,n and addition and multiplication by
r1,...,rn s1,...,sn r1 s1,...,rn sn
r1,...,rn.s1,...,sn r1s1,...... ,rnsn.
Thus R1 R2 ... Rn is a ring.
Example 2.1.6: Let R,,. be a ring. Define
MnR aij nn : aij ∈ R, i,j 1,...,n the set of all n n matrices with entries from R.
Then MnR,,. is a ring , where , . are the usual addition and
84 multiplication of matrices. Moreover, MnR is a non-commutative ring even if R is commutative (why ?).
85 If R has unity, then MnR also has a unity and denoted by
100...0 010...0 In ...... 000...1
MnR is called the ring of all n n −matrices. For example,
ab M2Z : a,b,c,d ∈ Z cd
is a ring.
Example 2.1.7: Let
ab T : a,b,c,d ∈ Q 0 d
and let
ab uv x , y ∈ T 0 d 0 w
Define
ab uv a ub v x y ∈ T 0 d 0 w 0 c w
ab uv au av bw x.y . ∈ T. 0 d 0 w 0 cw
Then T,,. is a ring.
86 Definition 2.1.8: If R is a ring and ∅≠S ⊆ R, we say that S is a subring of R denoted by S ≤ R if S itself is a ring over the same operations defined on R.
Theorem 2.1.9: A non-empty subset S of a ring R is a subring iff ∀ a,b ∈ S we have 1. a − b ∈ S 2. ab ∈ S.
Proof : Suppose S,,. is a subring of R,,.. Then S,,. is a ring itself. Therefore S, is a commutative group and S,. is a semi-group. Thus a − b ∈ S and ab ∈ S.
Conversely, suppose that a − b ∈ S and ab ∈ S. Then 1. S, is a commutative group,for a − a ∈ S 0 ∈ S 0 − a ∈ S −a ∈ S . Moreover, ∀ a,b ∈ S : a − −b ∈ S a b ∈ S . Lastly, is commutative S, is a commutative group. 2. S,. is a semi-group,forab ∈ S and . is associative. 3. Clearly . is distributive over , therefore S,,. is a ring.
Example : The set of even integers Ze forms a subring of a ring Z.
Definition 2.1.10: Let R,,. be a ring. Then the center of R denoted by ZR is defined by ZR a ∈ R : ax xa ∀ x ∈ R.
Theorem 2.1.11: The center of R is a subring of R.
Proof : 1. ZR ≠∅ for 0.x x.0 0 ∈ ZR. 2. Let a,b ∈ ZRax xa and bx xb ∀ x ∈ R. Consider a − bx ax − bx xa − xb xa − b a − b ∈ ZR.
87 3. Consider abx abx axb axb xab xab ab ∈ ZR.Thus ZR ≤ R.
Remark : R is commutative ZR R.
88 Theorem 2.1.12: Let R,,. bearing.Then ∀ a,b,c ∈ R,wehave 1. 0.a a.0 0. 2. a−b −ab −ab. 3. ab − c ab − ac. 4. a − bc ac − bc. 5. −a−b ab.
Proof : 1. 0.a a.0 0. Consider a.0 a.0 0 a.0 a.0 by cancellation law a.0 0. Similarly, for the other side.
2. a−b −ab −ab. a−b ab a−b b a.0 0 a−b −ab. −ab ab −a ab 0.b 0 −ab −ab. Thus a−b −ab −ab.
3. ab − c ab − ac. ab − c a.b −c a.b a.−c a.b − a.c.
4. a − bc ac − bc. a − b.c a −b.c a.c −b.c ac − bc.
5. −a−b ab −a−b −−ab −−ab ab.
89 PROBLEMS 2.1
1. Show that S is a subring of M2R ,where
a 2b S : a,b ∈ R . ba
2. Suppose that G is a group with respect to addition , with identity 0. Define a multiplication in G by a.b 0 ∀ a,b ∈ G. Show that G forms a ring with respect to these operations.
3. If R1 and R2 are subrings of the ring R, show that R1 ∩ R2 is a subring of R.
4. Find subrings R1, R2 of Z such that R1 R2 is not a subring of Z.
5. For a fixed element a ∈ R where R is a ring. Prove that R′ x ∈ R : ax 0 is a subring of R.
6. Let R and S be arbitrary rings. In the cartesian product R S of R and S,define r,s r′,s′r r′, s s′ and
r1,s1 r2,s2 r1 r2,s1 s2
r1,s1.r2,s2 r1r2,s1s2. Prove that R S is a ring . This ring is called the direct sum of R and S and denoted by R ⊕ S.
7. Suppose R is a ring in which all elements x satisfy x2 x. Prove that (a) x −x for each x ∈ R. (Hint : consider x x2 (b) R is commutative. (Hint : consider x y2.
90 8. Show that a ba − b a2 − b2 in a ring R iff R is commutative.
9. Show that a b2 a2 2ab b2 in a ring R iff R is commutative.
10. Decide if each of the following sets is a ring w.r.t. usual addition and multiplication. If it is not a ring state at least one condition that fails to hold. (a) R1 5n : n ∈ Z}. (b) R2 a b 2 : a,b ∈ Q} (c) R3 a b 3 : a,b ∈ Z} 3 3 (d) R4 a b 2 c 4 : a,b,c ∈ Q}
91 2.2 IDEALS AND INTEGRAL DOMAINS
In this section , we develop some theory of rings that paralles the theory of groups. We shall see that the concept of an ideal in a ring is analogous to that of a normal subgroup in a group.
Definition 2.2.1: (1) A non-empty subset I of a ring R is called a right ideal of R if (i) a − b ∈ I ∀ a,b ∈ I. (ii) ra ∈ I ∀ a ∈ I ,andr ∈ R.
(2) A non empty subset I is called a left ideal of a ring R if (i) a − b ∈ I ∀ a,b ∈ I. (ii) ar ∈ I ∀ a ∈ I and r ∈ R.
(3) A non-empty subset I of a ring R is called an ideal of R if (i) a − b ∈ I ∀ a,b ∈ I. (ii) ra ∈ I ∀ a ∈ I ,andr ∈ R (iii) ar ∈ I ∀ a ∈ I ,andr ∈ R.
Example 2.2.2: (1) The subring I 0 and I R are always ideals of a ring R. These ideals are called trivial ideals. (2) The set E of all even integers is a subring of Z whichisanidealofZ.
Example 2.2.3: It is easy to show that
ab T : a,b,c ∈ Z 0 c
forms a non-commutative subring of M2Z with respect to the operations of matrix addition and multiplication. In this subring consider the subset
92 0 b I : b ∈ Z 00
which is clearly non-empty. Since 0 x 0 y 0 x − y (i) − ∈ I , x − y ∈ Z. 00 00 00
(ii) Let r ∈ T, a′ ∈ I, i.e.,
ab 0 x r and a′ , then 0 c 00
0 x ab 0 xc a′r ∈ I , xc ∈ Z 00 0 c 00
ab 0 x 0 ax (iii) ra′ ∈ I, ax ∈ Z 0 c 00 00
Thus I is an ideal of T.
Example 2.2.4: Let R be a commutative ring with unity 1 ∈ R. For any fixed a ∈ R the set a ar : r ∈ R is an ideal of R (which is called ideal generated by one element).
Solution : (i) Clearly, a ≠∅,fora.1 1.a a ∈ a . (ii) Let x,y ∈ a and r,s ∈ R, then x ar ,and y as x − y ar − as ar − s ∈ a r − s ∈ R.
93 (iii) Let t ∈ R and x ar ∈ a . Then tx xt art art ∈ a rt ∈ R. Thus a is an ideal of R.
94 Example 2.2.5: Let R bearinganda ∈ R be a fixed element of R.Define aR ax : x ∈ R Ra xa : x ∈ R. Then Ra is a left ideal of R,andaR is a right ideal of R.
Solution : (i) Clearly, Ra ≠∅, for 0 0.a ∈ Ra. (ii) Let y,z ∈ Ra y x1a and z x2a where x1,x2 ∈ R, then
y − z x1a − x2a x1 − x2a ∈ Ra for x1 − x2 ∈ R.
(iii) Let r ∈ R, y ∈ Ra , then
ry rx1a rx1a ∈ Ra for rx1 ∈ R. Therefore Ra is a left ideal.
Definition 2.2.6: Let R, R′ be rings . A function f : R R′ is called a ring homomorphism if (i) fa b fa fb. (ii) fab fafb, a,b ∈ R.
Example 2.2.9: Study the following mapping from being ring homomorphism, 1-1, and onto ??
: Z̄ 6 Z̄ 6 defined by ā 4ā.
Solution : (i) is a ring homomorphism: ā ⊕ b̄ 4ā ⊕ b̄ 4ā ⊕ 4b̄ ā ⊕ b̄. ā ⊙ b̄ 4ā ⊙ 4b̄ 4a4b 16ab 4ab 4ab ab ā ⊙ b̄
(ii) Since Z̄ 6 0̄,2̄,4̄ is not onto.
95 (iii) Sinse kerf 0̄,3̄f is not 1 − 1.
96 Definition 2.2.10: An element 0 ≠ a ∈ R (ring) is called a left (right) zero divisor if ∃ 0 ≠ b ∈ R s.t. ab 0 ba 0. An element is called a zero divisor if it is both right and left zero divisor.
Example 2.2.11:
1. In Z̄ 6 0̄,1̄,2̄,3̄,4̄,5̄ the zero divisors are 2̄,3̄,4̄. 2. The ring Z has no zero divisors.
Definition 2.2.12: AringR is called an integral domain if (i) R is a commutative ring. (ii) R has unity 1. (iii) ∀ a,b ∈ R :ifab 0 a 0orb 0 (it has no zero divisors).
Example 2.2.13: 1. Z,,. is an integral domain. 2. Z̄ 10, ,. is not an integeral domain. 3. E,,. is not an integral domain. 4. R,,., Q,,., C,,. are integral domains.
Theorem 2.2.14:(cancellation law) If a,b, c are elements of an integral domain R such that a ≠ 0and ab ac then b c.
Proof : Suppose a,b,andc are elements in R such that a ≠ 0andab ac. ab − ac 0 ab − c 0 b − c 0 since a ≠ 0, R is an integral domain) b c.
97 Theorem 2.2.15:
For n 1,Z̄ n,⊕,⊙ is an integral domain iff n is a prime number.
Proof : Since Z̄ n,⊕,⊙ is an integral domain Z̄ n is a commutative ring which has unity. It remains to show that Z̄ n has no zero divisors iff n is a prime number.
Suppose that n is a prime number, and ā ≠ 0̄ in Z̄ n such that āb̄ 0̄ for some b̄ ∈ Z̄ n. ab 0̄ ab − 0 qn , q ∈ Z ab qn n ∣ ab. Since n is a prime number, then n ∣ a or n ∣ b. But ā ≠ 0̄ n ∤ a n ∣ b b̄ 0̄.
Therefore Z̄ n has no zero divisors and hence is an integral domain.
Suppose n is not a prime. Then n has divisors other than n ,so there are integers a and b such that n ab,1 a n,1 b n.
ā ≠ 0̄ and b̄ ≠ 0̄. But ā.b̄ ab n̄0̄. Therefore ā is a zero divisor in Z̄ n, which implies that Z̄ n is not an integral domain. Hence we arrive to a contradiction since Z̄ n is an integral domain. Hence n must be a prime number.
A field is another special type of ring, and we shall examine the relationship between a field and an integral domain. We begin with the following definition.
Definition 2.2.16: 1. AringisR,,. called a division ring or skew field if every non-zero element of R is
98 a unit in R, i.e, (if its non-zero elements form a group under multiplication). 2. A field is a commutative division ring.
99 Example 2.2.17:
Q,,.,R,,., C,,.,andZp,,. are examples of field.
Notation :Every integral domain and every division ring has at least two elements.
The relation between fields and integral domains is stated in the following theorem:
Theorem 2.2.18: Every field is an integral domain . The converse is not true.
Proof : Since F is a field F,,. is a commutative ring with unity. It remains to show that F has no zero divisors. i.e., if x,y ∈ Fs.t. xy 0 x 0ory 0?? Suppose x,y ∈ F s.t. xy 0. If x ≠ 0, then every non-zero element has inverse. x1 ∈ F x−1xy x−1.0 1.y 0 y 0. Similarly if y ≠ 0 x 0. Therefore F is an integeral domain.
Notation The converse is not true, for Z,,. is an integral domain but it is not a field.
Theorem 2.2.20: Every finite integral domain is a field.
Proof : Let R be a finite integral domain. |R| n, i.e., R r1,r2,...,rn. Let a ∈ R s.t. a ≠ 0. Then
aR ar1,ar2,...,arn.
The elements of aR are distinct for if ars ard rs rd (cancellation
100 low). aR R. But 1 ∈ R aR 1 ars for some rs ∈ R. Hence a is a unit. Thus every non-zero element has inverse R is a field.
Corollary 2.2.21:
Z̄ n is a field n is a prime.
Definition 2.2.22: (1) AringR with 1∈ R is called a Boolean ring if x2 x ∀ x ∈ R. (2) Let R be a ring and 0 ≠ a ∈ R then a is called idempotent if a2 a. (3) If there are positive integers n such that nx 0 ∀ x ∈ R . Then the smallest positive integer m such that mx 0 ∀ x ∈ R is called the characteristic of R. If no such positive integer exists , then R is said to be characteristic zero.
Example 2.2.23:
1. The charateristic of Z̄ 6 is 6. 2. The characteristic of Z, R, C is zero.
Exercise : Suppose R is a Boolean ring. Prove the following: (a) x −x ∀ x ∈ R. (b) R is a commutative ring. (c) chR 2.
101 PROBLEMS 2.1, 2.2
1. If I1, I2 are two ideals of the ring R. Prove that I1 ∩ I2 is also an ideal of R.
2. Find two ideals I1 and I2 of the ring Z such that: (a) I1 I2 is not ideal of Z. (b) I1 I2 is an ideal of Z.
3. If I1 and I2 are two ideals of the ring R.Prove that
I1 I2 x y : x ∈ I1, y ∈ I2
is an ideal of R that contains each I1 and I2.
4. Let I be an ideal of a ring R with 1 ∈ R. Show that if 1 ∈ I I R.
5. If R is a ring with 1 ∈ R,andI is an ideal of R containing a unit. Show that I R.
6. Let R be a ring. Define
rs T : r,s,t ∈ R ⊂ M2R. 0 t
(a) Show that T is an ideal of R. (b) Show that S is an ideal of T where
0 r S : r ∈ R ⊂ M2R. 00
8. If A is an ideal of a ring R and B is an ideal of a ring S. Show that A B is an ideal of R S.
9. Let R and S be rings. Define
102 1 : R S R by 1r,s r
2 : R S S by 2r,s s.
Show that 1, 2 are onto homomorphism, and evaluate ker1,ker2.
10. Let f : Z Z defined by fx nx.Show that f is not a homomorphism. onto 11. Let f : R R′ be a homomorphism and onto function. Show that : (a) If I is an ideal of R , fI is an ideal of R′. (b) If I′ is an ideal of R′, then f−1I′ is an ideal of R.
11. Decide which of the following are integral domains and which are fields with respect to the usual operations of addition and multiplication. State a reason for each that fails to be an integral domain or a field. (a) S m n 2 : m,n ∈ Z. (b) M a b 2 : a,b ∈ Q . (c) R a b 3 2 : a,b ∈ Q . (c) The set of all complex numbers of the form a bi, a,b ∈ Z.
12. Show that if R is an integral domain then R has no nilpotent element a ∈ R such that a ≠ 0.
13. Prove that if the element a ∈ R has a multiplicative inverse in R then a is not a zero divisor in R.
14. Find all zero divisors in Z̄ n for the following values of n. (a) n 6(b)n 5(c)n 8.
15. Consider the set S 0̄,2̄,4̄,6̄,8̄,10,12,14,16 ⊆ Z̄ 18. (a) Is S an integral domain. (b) Is S a field.
103 2.3 POLYNOMIALS OVER A RING
This section presents the facts about polynomials that are necessary for studying field and polynomial equations. Polynomials with real numbers as coefficients will be familiar from elementary algebra and calculus; now we must allow for the possibility that the coefficients are from some ring other than R (real numbers) .
Definition 2.3.1: 1. Let R be a commutative ring with 1 ∈ R and let x be an indeterminate. A polynomial in x with coefficients in R,orapolynomial in x over R,is an expression of the form
n 0 1 2 n i a0x a1x a2x ... anx ∑ aix , i0
where n is a non-negative integer and each ai is an element of R.Thesetofall polynomials in x over R is denoted by Rx.
2. If n is the largest non-negative integer such that an ≠ 0, then we say that the polynomial 0 1 2 n fx a0x a1x a2x ... anx has degree n, written degfx n,andan is called the leading coefficient of fx.
3. If the leading coefficient is 1, then fx is said to be monic.
4. a0 is called the constant term of fx,therefore elements of R are referred to as constant polynomial.
Remark 2.3.2: 1. The degree of the zero polynomial is not defined. 2. The non-zero elements of R are of degree zero.
Definition 2.3.3: Suppose that R is a commutative ring with unity, x is an indeterminate, and that n m i i fx ∑ aix and gx ∑ bix i0 i0
104 are polynomials in x over R. Then
fx gx iff ai bi for all i.
Definition 2.3.4: n i Let R be a commutative ring with unity. For any fx ∑ aix and i0 m i gx ∑ bix in Rx,wedefineAddition and Multiplication as follows: i0 k i 1. fx gx ∑ai bix ,wherek larger of the two integers n,m. i0 nm i i 2. fxgx ∑ cix where ci ∑ ajbi−j i0 j0 Theorem 2.3.5: Rx,,. forms a commutative ring with unity, which is called the ring of polynomials over R.
Example 2.3.6:
Let fx, gx ∈ Z6x such that fx 1x0 5x1 3x3 and gx 4x0 2x1.
1. fx gx 5x0 1x1 3x3 5 x 3x3 2. fxgx 4x0 4x1 4x2 4 4x 4x2. 3. The constant term of fx is 1 and the leading coefficient of fx is 3. 4. The polynomial gx has constant term 4 and leading coefficient 2. 5. degfx 3, and deggx 1. 6. degfxgx 2 ≠ degfx deggx.
Theorem 2.3.7: If R is an integral domain and fx, gx are non-zero polynomials of Rx,then degfxgx degfx deggx.(2.1)
Proof : Let R be an integral domain, and suppose that
105 n m i i fx ∑ aix has degree n and gx ∑ bix has degree m i0 i0
in Rx. Then an ≠ 0andbm ≠ 0 anbm ≠ 0sinceR is an integral domain. But anbm is the leading coefficient in fxgx since
nm fxgx a0b0 ..... anbmx . Therefore degfxgx n m degfx deggx.
Theorem 2.3.8: Rx is an integral domain iff R is an integral domain.
Proof : ( Assume that R is an integral domain. If fx and gx are non-zero polynomials (not constant) in Rx, then both fx and gx have degrees.i.e., degfx ≠ 0 and deggx ≠ 0 By theorem 2.3.7, degfxgx degfx deggx degfxgx ≠ 0 fxgx is not a zero polynomial. i.e., fxgx ≠ 0, and this shows that Rx is an integral domain.
If Rx is an integral domain, then R must be an integral domain since R is a commutative ring with unity and R ⊆ Rx.
Notation 2.3.9: The equation (2.1), cannot hold when one of the factors is a zero-polynomials, for the degree of zero polynomials is undefined and the product of any zero polynomial and any polynomial yields the zero polynomial.
106 DIVISIBILTY AND GREATEST COMMON DIVISOR
If a ring R is not an integral domain, the division of polynomials over R is not very satisfactory subject for study because of the possible presence of zero divisors. In order to obtain the results that we need on division on polynomials, the ring of coefficients actually needs to be a field. For this reason, we confine our attention for the rest of this chapter to rings of polynomials Fx where F is a field. This assures us that Fx is an integral domain, and that every non-zero element of F has a multiplicative inverse.
Definition 2.3.10: Let gx ≠ 0andfx are in Fx , then gx divides fx [or gx is a factor of fx if there exists hx in Fx such that fx gxhx.Ifgx divides fx , we write gx ∣ fx.Ifgx does not divide fx , we write gx ∤ fx.
Remark 2.3.11: Polynomials of degree zero have two special properties: 1. Any non-zero element a ∈ F is a factor of every fx ∈ Fx since a−1fx ∈ Fx and fx aa−1fx.
2. If gx ∣ fx, then agx ∣ fx for all non-zero elements a ∈ F since fx gxhxfx agxa−1hx.
Example 2.3.12: Since 6x2 − x − 2 2x 13x − 2 in Qx2x 1 ∣ 6x2 − x − 2. Moreover, a2x 1 ∣ 6x2 − x − 2 ∀ 0 ≠ a ∈ F.ex: 52x 1 ∣ 6x2 − x − 2 because 6x2 − x − 2 52x 11/53x − 2
The Division Algorithm for integers has the following analogue in Fx.
Theorem 2.3.13: (The Division Algorithm) Let 0 ≠ gx and fx be elements of Fx . There exist unique elements
107 qx (quotient)andrx (remainder)inFx such that fx gxqx rx, with either rx 0ordegrx deggx.
108 Example 2.3.14: Divide fx 2x5 3x4 7x3 x2 4x 1bygx x3 2x2 3x 1in Z11x.
2x2 10x 3 x3 2x2 3x 1∣ 2x5 3x4 7x3 x2 4x 1 ∓2x5 ∓ 4x4 ∓ 6x3 ∓ 2x2 − x4 x3 − x2 4x 1 −1 ≡ 10 ∈ Z11 10x4 x3 10x2 4x 1 ∓10x4 ∓ 9x3 ∓ 8x2 ∓ 10x 20 ≡ 9, 30 ≡ 8 3x3 2x2 5x 1 −8 ≡ 3, − 6 ≡ 5 ∓3x3 ∓ 6x2 ∓ 9x ∓ 3 7x2 7x 9 −4 ≡ 7, − 2 ≡ 9 2x5 3x4 7x3 x2 4x 1 2x2 10x 3x3 2x2 3x 1 7x2
Our next objective in this section is to show that any two non-zero polynomials over F have a greatest common divisor in Fx. We saw that if gx is a divisor of fx , then agx is also a divisor of fx, for every non-zero a ∈ F. By choosing a to be the multiplicative inverse of the leading coefficient gx, the leading coefficient in agx can be made equal to 1 (monic polynomial). So one conditions that we place on a greatest common divisor of two polynomials is that it be monic. Without this condition, the greatest common divisor of two polynomials would not be unique.
Definition 2.3.15: Let fx and gx be non-zero polynomials in Fx. A polynomial dx in Fx is the greatest common divisor of fx and gx if these conditions are satisfied: 1. dx is a monic polynomial. 2. dx ∣ fx and dx ∣ gx. 3. hx ∣ fx and hx ∣ gx imply that hx ∣ dx. The greatest common divisor of fx and gx is usually denoted by gcdfx,gx.
The following remark gives a systematic way of finding a greatest
109 common divisor of any two polynomials.
Remark 2.3.16: 1. The greatest common divisor of two polynomials can be computed using a procedure known as the Euclidean algorithm. 2. If fx gxqx rx,thengcdfx,gx gcdgx,rx. 3. Given non zero polynomials fx, gx, the Euclidean algorithm uses the division algorithm repeatedly to obtain
fx gxq1x r1x with 0 r1x or degr1x deggx
gx r1xq2x r2x with 0 r2x or degr2x degr1x
r1x r2xq3x r3x with 0 r3x or degr3x degr2x etc.
Since degr1x degr2x ...., the remainders get smaller and smaller, and after a finite number of steps we obtain a remainder rn1x 0. The algorithm ends with the equation
rn−1x rnxqn1x 0. This gives us the greatest common divisor:
gcdfx,gx gcdgx,r1x gcdr1x,r2x ... gcdrn−1x,rnx
Theorem 2.3.17: Let fx and gx be non-zero polynomials in Fx.Then there exists a unique greatest common divisor dx of fx and gx in Fx. Moreover, dx can be expressed as dx fxsx gxtx for sx and tx in Fx,anddx is a monic polynomial of the least degree that can be written in this form.
Example 2.3.18: 3 1 2 1 1 2 2 1 Find the g.c.d of fx x 3 x − 4 x − 12 , gx x − 3 x − 3 in Qx, and then find sx, tx s.t. dx fxsx gxtx.
Solution :
110 Using the long division we get:
r1x
3 1 2 1 1 1 3 x 3 x − 4 x − 12 4 4 x 2 2 1 x 1 2 2 1 x − 3 x − 3 x − 3 x − 3 2 2 1 3 1 fx x − 3 x − 3 x 1 4 x 4 s.t. degr1x deggx. g x Next ,we compute ,toget r1x x2 − 2 x − 1 3 3 4 x − 4 3 1 3 3 4 x 4
x2 − 2 x − 1 4 x − 4 3 x 1 0. 3 3 3 3 4 4
3 1 Thus the last non-zero remainder 4 x 4 is a common divisor of highest degree and so the greatest common divisor is the monic polynomial
dx −4 3 x 1 x 1 . 3 4 4 3 To find sx and tx,wehave dx fxsx gxtx,
fx gxx 1 3 x 1 4 4 gxx 1 3 x 1 4 3 gxx 1 3 dx 4
3 dx −gxx 1 fx 4 dx − 4 x 1gx 4 fx 3 3 sx 4 and tx − 4 x 1 3 3
Example 2.3.19: 5 3 2 4 3 2 Let fx x 2x x 2x,gx x x x ∈ Z3x.Evaluate gcdfx,gx?
111 Solution : Using the long division we get:
r1x x5 2x3 x2 2x x4 x3 x2x 2 2x3 2x2 2x
such that degr1x deggx. g x Next, divide we get: r1x x4 x3 x2 2x2x3 2x2 2x 0 So the last non-zero remainder 2x3 2x2 2x is common divisor of highest degree and so the greatest common divisor is the monic polynomial dx 2−12x3 2x2 2x x3 x2 x.
112 Example 2.3.20: Let fx x3 1, gx x2 1 ∈ Rx. Evaluate gcdfx,gx?
Solution : Using the long division we get:
r1x x3 1 x2 1x −x 1
such that degr1x deggx. g x Next, divide we get: r1x x2 1 −x 1−x − 1 2 Thus gcdfx,gx 1. i.e., fx and gx are relatively prime.
113 IRREDUCIBLES AND UNIQUE FACTORIZATION THEOREM
In the factorization of polynomials over a field F, the concept of an irreducible polynomial is analogous to the concept of prime integer in the factorization of integers.
Definition 2.3.21: Let fx be a polynomial over the field F.Anelementc ∈ F is called a root or (zero) of the polynomial fx ∈ Fx if fc 0.
Example 2.3.22: 1. 1, − 1 are roots of fx x2 − 1 ∈ Zx. 2. fx x2 1 has no roots in Rx. 2 3. 1, 3 are zeros of fx x 2x 3 ∈ Z6x.
Theorem 2.3.23: (The Remainder Theorem) If fx ∈ Fx,andc ∈ F, then the remainder in the division of fx by x − c is fc. fx x − cqx fc.
Proof : Divide fx by x − c we get fx x − cqx rx where degrx degx − c Since x − c has degree one, the remainder rx has dgree zero. i.e., rx r is constant. fx x − cqx r fc 0 r r fc.
Example 2.3.24: Divide fx x4 − 2x3 x2 − x 1 ∈ Qx by x − 2weget, fx x − 2x3 x 1 3 Also, f2 3.
114 Theorem 2.3.25: (Factor Theorem) If fx ∈ Fx and c ∈ F, then x − c is a factor of fx iff fc 0. Proof : This is an immediate corollary of the Remainder Theorem.
Theorem 2.3.26: Let F be a field. Then fx is a unit in Fx iff fx is a non-zero constant polynomial.
Proof : Suppose fx is a unit fxgx 1. By theorem 2.3.7, degfxgx degfx deggx 0 degfx 0 and deggx 0 fx and gx are non zero constant polynomial.
Suppose fx b is a non-zero constant polynomial in Fx.SinceF is a field and b ∈ F then every element is a unit. ∃ b−1 ∈ F s.t. bb−1 1. Take b−1 gxfxgx 1. But gx b−1 is a constant polynomial, i.e., gx ∈ Fx. Hence fx is a unit.
Remark 2.3.27: 1. Qx, Rx, Cx has infinitely many units. 2. The theorem above may be false if F is not a field. For example the constant polynomial 2 ∈ Zx is not a unit, and a non-constant polynomial 3x 1 in Z9x is a unit since 3x 16x 1 1.
Definition 2.3.28: A polynomial fx ∈ Fx is said to be an associate of gx ∈ Fx if fx c.gx for some non-zero elements c ∈ F.
Remark 2.3.29: 1. The associates of fx are just the products of gx with a unit in Fx. 2. If fx cgx gx c−1fx fx is an associate of gx gx is an associate of fx. 3. Each polynomial of degree at least one has two obvious sets of divisors: its associates and the polynomial of degree zero.
115 Example 2.3.30: 1. 2x2 − 1, 6x2 − 3 are associates over Q. 2. The associate of 4 ∈ Z10 are4,2,8,and6since4.1 4, 4.3 2, 7.4 8, 9.4 6 where1,3,7,9areunitsofZ10. 3. The only associate of 26 ∈ Z are 26, − 26.
116 Definition 2.3.31: 1. A non constant polynomial px ∈ Fx is said to be an irreducible (or prime) if its only divisors are its associates and the non zero constant polynomials (units). i.e., if px gxhx then one of gx and hx is of degree zero and the other is an associate of px. Or another definition A polynomial px ∈ Fx is said to be an irreducible (or prime)ifpx has a ve degree and px cannot expressed as a product px gxhx with both gx and hx of ve degreein Fx. 2. A non-constant polynomial that is not irreducible is said to be reducible.
Example 2.3.32: The polynomial fx x 2 is irreducible in Qx. Since the divisors of fx must have degree 0 or 1. Divisors of degree 0 are non-zero constant polynomial. Suppose gx ∣ fx such that deggx 1
fx gxhx deghx 0 hx constant fx cgxgx c−1fx i.e., gx is an associate of x 2.
Remark 2.3.33: A similar argument in the general case shows that every polynomial of degree 1 in Fx is irreducible of Fx.
Example 2.3.34: The polynomial 3x4 − 3x2 − 6 can be factored as
3x2 − 2x2 1 in Qx 3x 2 x − 2 x2 1 in Rx 3x 2 x − 2 x ix − i in Cx
Each factor is irreducible in its context.
It is usually not easy to decide whether or not a given polynomial is irreducible over a certain field. However, the following theorem is sometimes helpful for polynomials with degree less than 4.
117 Theorem 2.3.35: If fx is a polynomial of degree 2 or 3 over the field F, then fx is irreducible over F iff fx has no roots (zeros) in F.
Example 2.3.36: Determine whether or not each of the following polynomials is irreducible over Z5. 1. fx x3 2x2 − 3x 4. 2. gx x2 3x 4.
Solution : Routine computations show that f0 4, f1 4, f2 4, f3 0, f4 3.
Thus 3 is a root of fx in Z5,andfx is reducible over Z5. However, gx is irreducible over Z5 since gx has no roots in Z5 : g0 4, g1 3, g2 4, g3 2, g4 2.
Irreducible polynomials play a role in the factorization of polynomials corresponding to the role prime integers play in the factorization of integers. This is illuatrated by the follwing theorem:
Theorem 2.3.37: 1. If px is an irreducible polynomial over a field F and px ∣ fxgx in Fx, then either px ∣ fx or px ∣ gx in Fx. 2. If px is an irreducible polynomial over a field F and px ∣ f1x...fnx in Fx, then px divides some fjx.
Theorem 2.3.38: (Unique Factorization Theorem) Every polynomial of ve degree over the field F can be expressed as a product of monic irreducible polynomials, the irreducible polynomials being unique excpet for the order and for the unit (non-zero constant) factors in F. m fx a pix,0≠ a ∈ F i1
where a is a leading coefficient , and pix are monic irreducible
118 polynomials over F.
Example 2.3.39: Find the factorization using Theorem 2.3.41, for
4 3 fx 2x x 2x 4 over Z5.
Solution : First we determine the roots of fx : f0 4, f1 2, f2 0, f3 1, f4 1.
Thus 2 is the only zero of fx in Z5, and the Factor Thorem assures that x − 2 is a factor of fx. Dividing by x − 2weget fx x − 2 2x3 3x 3
gx Similarlythezeroofgx is 2. Hence gx x − 22x2 4x 1 and fx x − 222x2 4x 1
2 2 Since 2x 4x 1 has no zeros in Z5, then 2x 4x 1 is irreducible over Z5. To arrive at the desired factorization, we need only to factor the leading coefficient of fx from the factor 2x2 4x 1 : fx x − 222x2 4x 1 x − 222x2 4x 2.3 2x − 22x2 2x 3
Theorem 2.3.40: (Eisenstien’s Irreducibility Criterion) n Let fx a0 a1x a2x ... anx be a polynomial of positive degree in Zx. If there exists a prime integer p such that
2 p ∣ a0, p ∣ a2, ..., p ∣ an−1 and p ∤ an, p ∤ a0
119 then fx is irreducible over Q.
Example 2.3.41: Consider fx 25x5 − 9x4 3x2 − 12 ∈ Zx.Takep 3. 3 ∣−12 , 3 ∣ 3, 3 ∣−9, and 3 ∤ 25, 32 ∤−12. Hence fx is irreducible over Q.
120 PROBLEMS 2.3
1. Consider the following polynomials over Z8 fx 2x3 7x 4, gx 4x2 4x 6, hx 6x2 3. Evaluate the following: a. fxgx b. fxhx gxhx.
2. For fx, gx,andZnx given in the following, find qx and rx in Zn that satisfy the conditions in the Division Algorithm. 3 2 a. fx 2x 3x 4x 1, gx 3x 1, in Z5x. 4 2 2 b. fx x 5x 2x 2, gx 3x 2inZ7x.
3. For fx, gx,andZnx given in the following, find the greatest common divisor dx of fx and gx in Zn and find sx, tx such that dx fxsx gxtx. 4 2 3 2 a. fx x 2x x 1, gx x x 2x 2, in Z3x. 4 3 2 2 b. fx 4x 2x 6x 4x 5, gx 3x 2, in Z7x.
4. Determine whether or not each of the following polynomials is irrducible over each indicated fields. a. x2 2x 2 over Q, R and C. 4 2 b. x 2x 1 over Z3, Z5 and Z7.
5. Write each of the following polynomials as a product of its leading coefficients and a finite number of monic irreducible polynomials over Z5. a. fx 2x3 1 b. fx 3x3 2x2 x 2. c. fx x4 x3 2x2 3x 2.
6. Show that each of the following polynomials is irreducible over the field of rational numbers. a. fx x4 − x3 − 2x2 6x − 4 b. gx 6x4 x3 3x2 − 14x − 8
121