DOCSLIB.ORG

School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet IV: Composition Series and the Jordan–Hölder Theor

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet IV: Composition Series and the Jordan–Hölder Theor CMRD 2010 School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet IV: Composition series and the Jordan–H¨older Theorem (Solutions) 1. Let G be a group and N be a normal subgroup of G. If G/N and N both have composition series, show that G has a composition series and that the set of composition factors of G equals the union of those of G/N and those of N. [Use the Correspondence Theorem to lift the terms of a composition series to a chain of subgroups between G and N.] Solution: Let G/N = G /N > G /N > >G/N = N1 = 1 0 1 ··· r { } be a composition series for G/N. By the Correspondence Theorem, each sub- group in this series has the form Gi/N , where Gi is a subgroup of G containing N. Furthermore, Gi/N Gi 1/N for all i, so the Correspondence Theorem (applied − to the correspondence between subgroups of Gi 1 containing N and subgroups − of Gi 1/N ) shows that Gi Gi 1. Therefore − − G = G >G > >G = N, 0 1 ··· r and Gi Gi 1 for all i. − Let N = N >N > >N = 1 0 1 ··· s be a composition series for N. Putting this together with the above chain of subgroups, we obtain G = G >G > >G = N >N > >N = 1. 0 1 ··· r 0 1 ··· s Here Gi 1/N Gi 1/Gi = − − ∼ Gi/N (by the Third Isomorphism Theorem) is a simple group since the Gi/N form a composition series for G/N, while Nj 1/Nj is simple since these are the factors in − a composition series for N. Hence this last series is a composition series for G. 1 2. Let p, q and r be prime numbers with p<q<r.LetG be a group of order pqr. (a) Suppose that G does not have a unique Sylow r-subgroup. Show that it has a unique Sylow q-subgroup Q. [Hint: If there is no normal Sylow q-subgroup, then how many elements are there of order r and how many of order q?] (b) Show that G/Q has a unique Sylow r-subgroup of the form K/Q,whereK G and K = qr. | | (c) Show that K has a unique Sylow r-subgroup and deduce that, in fact, G has a unique Sylow r-subgroup R (contrary to the assumption in (a)). [Hint: Why is a Sylow r-subgroup of K also a Sylow r-subgroup of G?Rememberthatall the Sylow r-subgroups of G are conjugate.] (d) Show that G/R has a unique Sylow q-subgroup. (e) Deduce that G has a composition series G = G0 >G1 >G2 >G3 =1 where G = qr and G = r.Uptoisomorphism,whatarethecompositionfactorsofG? | 1| | 2| Solution: (a) Let G have nr Sylow r-subgroups. By Sylow’s Theorem, n 1 (mod r) and n pq. r ≡ r | Now p<q<r,sop, q 1 (mod r). Thus if n = 1 then n = pq. ≡ r r The Sylow r-subgroups of G are cyclic of order r, so any distinct two intersect in the identity. Hence the Sylow r-subgroups contain a total of n (r 1) = pq(r 1) = pqr pq r − − − elements of order r. Let G have nq Sylow q-subgroups. By Sylow’s Theorem, n 1 (mod q) and n pr. q ≡ q | Now p<q,sop 1 (mod q). Hence nq =1,r or pr.Ifnq = 1, then nq r. The Sylow q-subgroups≡ of G are cyclic of order q, so any distinct two intersect in the identity and hence they contain n (q 1) r(q 1) >qp q − − elements of order q. There are now more than pqr pq + pq = G elements in G, a contradiction. − | | Hence if n = 1, then n =1,soG has a unique, normal, Sylow q-subgroup Q. r q (b) Consider the quotient group G/Q. It has order pqr/q = pr.Let¯nr be the number of Sylow r-subgroups in G/Q. Then by Sylow’s Theorem, n¯ 1 (mod r) andn ¯ p. r ≡ r | 2 But p<r,sop 1 (mod r) and hencen ¯r = 1. Thus G/Q has a normal Sylow r-subgroup. ≡ The Correspondence Theorem implies that this normal subgroup is K/Q, where K is a normal subgroup of G containing Q. As K/Q = r and Q = q,we deduce that K = qr. | | | | | | (c) Apply Sylow’s Theorem to K.LetK have nr Sylow r-subgroups. Then n 1 (mod r) and n q. r ≡ r | Since q<r, we deduce that nr = 1. Thus K has a normal Sylow r-subgroup R. Now R is one of the Sylow r-subgroups of G. All nr Sylow r-subgroups of G are conjugate, so any other Sylow r-subgroup of G has the form Rx for some x G. ∈ Now R K G,so x x R K = K. This means that Rx is a subgroup of K of order r, that is, a Sylow r-subgroup of K. But K has a unique Sylow r-subgroup, so Rx = R. Hence G has a unique Sylow r-subgroup, contrary to the assumption we made in (a). (d) We now know our assumption made in (a) was incorrect. Thus nr = 1 and G has a normal Sylow r-subgroup, R. Consider G/R, of order pq.LetG/R have n˜q Sylow q-subgroups. Sylow’s Theorem gives n˜ 1 (mod q) andn ˜ p, q ≡ q | son ˜q =1(asp<q). Thus G/R has a normal Sylow q-subgroup, L/R, where R L G and L = qr. | | (e) Take G0 = G, G1 = L, G2 = R and G3 = 1 (with R and L as in (d)). Then G = G0 >G1 >G2 >G3 =1 with Gi Gi 1 for all i. (In fact, we have shown that this is a normal series: − G G for all i.) Certainly G = L = qr and G = R = r. The factors i | 1| | | | 2| | | G0/G1,G1/G2,G2/G3 have orders p, q and r, respectively. Consequently these are all simple, cyclic, groups. Thus this is a composition series for G. 3. Let p be a prime number and let G be a non-trivial p-group. Show that G has a chain of subgroups G = G >G >G > >G =1 0 1 2 ··· n such that G is a normal subgroup of G and G : G = pi for i =0,1,...,n. i | i | What are the composition factors of G? [Hint: Use the fact that Z(G) =1toproduceanelementx Z(G)oforderp.Considerthe ∈ quotient group G/K where K = x and apply induction.] 3 n Solution: Let G = p and induct on n.Ifn = 0, then G = G0 = 1 is the required composition| | series. Suppose then that n>1, and that the result holds for all finite p-groups of order smaller than G. Now G is a non-trivial p-group and hence Z(G) = 1.Letx be any non-identity element in Z(G). Then the order r 1 of x is a non-trivial power of p,say x = pr with r>0. Then xp − is an element in Z(G) of order p. Rename so that| x| is this element of Z(G) of order p. Now X = x is a subgroup of G of order p that is contained in the centre of G. 1 Hence X G.(Ifg G and y Z(G), then g− yg = y.) Consider the quotient ∈ ∈ n 1 group G/X:ap-group of order G /p = p − . Hence by induction G/X has a | | composition series G/X = G0/X > G1/X > >Gn 1/X = X1 = 1 ··· − { } where G/X : G /X = pi for i =0,1,...,n 1. Each G /X corresponds to a | i | − i subgroup Gi of G containing X, so we obtain a chain of subgroups G = G0 >G1 > >Gn 1 = X>1. ··· − Here Gi Gi 1 for i =1,2,...,n 1 by the Correpondence Theorem. Also, − G = G /X p, and − | i| | i |· G : G = G/X : G /X = pi | i| | i | for i =0,1,...,n 1. Defining G = 1 produces a series for G − n G = G >G > >G = 1 0 1 ··· n i n i where G : Gi = p for all i. Consequently, Gi = p − and Gi 1/Gi = p for − all i. Hence| the| factors in this series are simple| | groups, this| is a composition| series for G, and the n composition factors for G are each isomorphic to Cp. 4. The dihedral group D8 has seven different composition series. Find all seven. Solution: Recall that D8 = α,β where α has order four, β has order two, 1 and αβ = βα− . Since D8 is a 2-group, the previous question tells us that its composition factors are all cyclic of order two and hence every composition series for D8 has the form D8 = G0 >G1 >G2 >G3 = 1 where G = 4 and G =2. | 1| | 2| One possibility for G1 is that it is cyclic of order four. All elements of D8 not in α have order two (see Tutorial Sheet I). Hence α is the only cyclic subgroup of order 4 and this has a unique subgroup of order two, namely α2 . Thus D > α > α2 > 1 8 is a composition series for D8 and this is the only one with G1 cyclic. 4 Consider now the other possibilities for G1. If it is not cyclic then it must contain no elements of order four.
Load more

Recommended publications
  • THE UPPER BOUND of COMPOSITION SERIES 1. Introduction. a Composition Series, That Is a Series of Subgroups Each Normal in the Pr
    THE UPPER BOUND OF COMPOSITION SERIES ABHIJIT BHATTACHARJEE Abstract. In this paper we prove that among all finite groups of or- α1 α2 αr der n2 N with n ≥ 2 where n = p1 p2 :::pr , the abelian group with elementary abelian sylow subgroups, has the highest number of composi- j Pr α ! Qr Qαi pi −1 ( i=1 i) tion series and it has i=1 j=1 Qr distinct composition pi−1 i=1 αi! series. We also prove that among all finite groups of order ≤ n, n 2 N α with n ≥ 4, the elementary abelian group of order 2 where α = [log2 n] has the highest number of composition series. 1. Introduction. A composition series, that is a series of subgroups each normal in the previous such that corresponding factor groups are simple. Any finite group has a composition series. The famous Jordan-Holder theo- rem proves that, the composition factors in a composition series are unique up to isomorphism. Sometimes a group of small order has a huge number of distinct composition series. For example an elementary abelian group of order 64 has 615195 distinct composition series. In [6] there is an algo- rithm in GAP to find the distinct composition series of any group of finite order.The aim of this paper is to provide an upper bound of the number of distinct composition series of any group of finite order. The approach of this is combinatorial and the method is elementary. All the groups considered in this paper are of finite order. 2. Some Basic Results On Composition Series.
    [Show full text]
  • Topics in Module Theory
    Chapter 7 Topics in Module Theory This chapter will be concerned with collecting a number of results and construc- tions concerning modules over (primarily) noncommutative rings that will be needed to study group representation theory in Chapter 8. 7.1 Simple and Semisimple Rings and Modules In this section we investigate the question of decomposing modules into \simpler" modules. (1.1) De¯nition. If R is a ring (not necessarily commutative) and M 6= h0i is a nonzero R-module, then we say that M is a simple or irreducible R- module if h0i and M are the only submodules of M. (1.2) Proposition. If an R-module M is simple, then it is cyclic. Proof. Let x be a nonzero element of M and let N = hxi be the cyclic submodule generated by x. Since M is simple and N 6= h0i, it follows that M = N. ut (1.3) Proposition. If R is a ring, then a cyclic R-module M = hmi is simple if and only if Ann(m) is a maximal left ideal. Proof. By Proposition 3.2.15, M =» R= Ann(m), so the correspondence the- orem (Theorem 3.2.7) shows that M has no submodules other than M and h0i if and only if R has no submodules (i.e., left ideals) containing Ann(m) other than R and Ann(m). But this is precisely the condition for Ann(m) to be a maximal left ideal. ut (1.4) Examples. (1) An abelian group A is a simple Z-module if and only if A is a cyclic group of prime order.
    [Show full text]
  • Composition Series of Arbitrary Cardinality in Abelian Categories
    COMPOSITION SERIES OF ARBITRARY CARDINALITY IN ABELIAN CATEGORIES ERIC J. HANSON AND JOB D. ROCK Abstract. We extend the notion of a composition series in an abelian category to allow the mul- tiset of composition factors to have arbitrary cardinality. We then provide sufficient axioms for the existence of such composition series and the validity of “Jordan–Hölder–Schreier-like” theorems. We give several examples of objects which satisfy these axioms, including pointwise finite-dimensional persistence modules, Prüfer modules, and presheaves. Finally, we show that if an abelian category with a simple object has both arbitrary coproducts and arbitrary products, then it contains an ob- ject which both fails to satisfy our axioms and admits at least two composition series with distinct cardinalities. Contents 1. Introduction 1 2. Background 4 3. Subobject Chains 6 4. (Weakly) Jordan–Hölder–Schreier objects 12 5. Examples and Discussion 21 References 28 1. Introduction The Jordan–Hölder Theorem (sometimes called the Jordan–Hölder–Schreier Theorem) remains one of the foundational results in the theory of modules. More generally, abelian length categories (in which the Jordan–Hölder Theorem holds for every object) date back to Gabriel [G73] and remain an important object of study to this day. See e.g. [K14, KV18, LL21]. The importance of the Jordan–Hölder Theorem in the study of groups, modules, and abelian categories has also motivated a large volume work devoted to establishing when a “Jordan–Hölder- like theorem” will hold in different contexts. Some recent examples include exact categories [BHT21, E19+] and semimodular lattices [Ro19, P19+]. In both of these examples, the “composition series” arXiv:2106.01868v1 [math.CT] 3 Jun 2021 in question are assumed to be of finite length, as is the case for the classical Jordan-Hölder Theorem.
    [Show full text]
  • A Splitting Theorem for Linear Polycyclic Groups
    New York Journal of Mathematics New York J. Math. 15 (2009) 211–217. A splitting theorem for linear polycyclic groups Herbert Abels and Roger C. Alperin Abstract. We prove that an arbitrary polycyclic by finite subgroup of GL(n, Q) is up to conjugation virtually contained in a direct product of a triangular arithmetic group and a finitely generated diagonal group. Contents 1. Introduction 211 2. Restatement and proof 212 References 216 1. Introduction A linear algebraic group defined over a number field K is a subgroup G of GL(n, C),n ∈ N, which is also an affine algebraic set defined by polyno- mials with coefficients in K in the natural coordinates of GL(n, C). For a subring R of C put G(R)=GL(n, R) ∩ G.LetB(n, C)andT (n, C)bethe (Q-defined linear algebraic) subgroups of GL(n, C) of upper triangular or diagonal matrices in GL(n, C) respectively. Recall that a group Γ is called polycyclic if it has a composition series with cyclic factors. Let Q be the field of algebraic numbers in C.Every discrete solvable subgroup of GL(n, C) is polycyclic (see, e.g., [R]). 1.1. Let o denote the ring of integers in the number field K.IfH is a solvable K-defined algebraic group then H(o) is polycyclic (see [S]). Hence every subgroup of a group H(o)×Δ is polycyclic, if Δ is a finitely generated abelian group. Received November 15, 2007. Mathematics Subject Classification. 20H20, 20G20. Key words and phrases. Polycyclic group, arithmetic group, linear group.
    [Show full text]
  • Classifying Categories the Jordan-Hölder and Krull-Schmidt-Remak Theorems for Abelian Categories
    U.U.D.M. Project Report 2018:5 Classifying Categories The Jordan-Hölder and Krull-Schmidt-Remak Theorems for Abelian Categories Daniel Ahlsén Examensarbete i matematik, 30 hp Handledare: Volodymyr Mazorchuk Examinator: Denis Gaidashev Juni 2018 Department of Mathematics Uppsala University Classifying Categories The Jordan-Holder¨ and Krull-Schmidt-Remak theorems for abelian categories Daniel Ahlsen´ Uppsala University June 2018 Abstract The Jordan-Holder¨ and Krull-Schmidt-Remak theorems classify finite groups, either as direct sums of indecomposables or by composition series. This thesis defines abelian categories and extends the aforementioned theorems to this context. 1 Contents 1 Introduction3 2 Preliminaries5 2.1 Basic Category Theory . .5 2.2 Subobjects and Quotients . .9 3 Abelian Categories 13 3.1 Additive Categories . 13 3.2 Abelian Categories . 20 4 Structure Theory of Abelian Categories 32 4.1 Exact Sequences . 32 4.2 The Subobject Lattice . 41 5 Classification Theorems 54 5.1 The Jordan-Holder¨ Theorem . 54 5.2 The Krull-Schmidt-Remak Theorem . 60 2 1 Introduction Category theory was developed by Eilenberg and Mac Lane in the 1942-1945, as a part of their research into algebraic topology. One of their aims was to give an axiomatic account of relationships between collections of mathematical structures. This led to the definition of categories, functors and natural transformations, the concepts that unify all category theory, Categories soon found use in module theory, group theory and many other disciplines. Nowadays, categories are used in most of mathematics, and has even been proposed as an alternative to axiomatic set theory as a foundation of mathematics.[Law66] Due to their general nature, little can be said of an arbitrary category.
    [Show full text]
  • Group Theory
    Chapter 1 Group Theory Most lectures on group theory actually start with the definition of what is a group. It may be worth though spending a few lines to mention how mathe- maticians came up with such a concept. Around 1770, Lagrange initiated the study of permutations in connection with the study of the solution of equations. He was interested in understanding solutions of polynomials in several variables, and got this idea to study the be- haviour of polynomials when their roots are permuted. This led to what we now call Lagrange’s Theorem, though it was stated as [5] If a function f(x1,...,xn) of n variables is acted on by all n! possible permutations of the variables and these permuted functions take on only r values, then r is a divisior of n!. It is Galois (1811-1832) who is considered by many as the founder of group theory. He was the first to use the term “group” in a technical sense, though to him it meant a collection of permutations closed under multiplication. Galois theory will be discussed much later in these notes. Galois was also motivated by the solvability of polynomial equations of degree n. From 1815 to 1844, Cauchy started to look at permutations as an autonomous subject, and introduced the concept of permutations generated by certain elements, as well as several nota- tions still used today, such as the cyclic notation for permutations, the product of permutations, or the identity permutation. He proved what we call today Cauchy’s Theorem, namely that if p is prime divisor of the cardinality of the group, then there exists a subgroup of cardinality p.
    [Show full text]
  • Composition Series and the Hölder Program
    Composition Series and the Holder¨ Program Kevin James Kevin James Composition Series and the H¨olderProgram Remark The fourth isomorphism theorem illustrates how an understanding of the normal subgroups of a given group can help us to understand the group itself better. The proof of the proposition below which is a special case of Cauchy's theorem illustrates how such information can be used inductively to prove theorems. Proposition If G is a finite Abelian group and p is a prime dividing jGj, then G contains an element of order p. Kevin James Composition Series and the H¨olderProgram Definition A group G is called simple if jGj > 1 and the only normal subgroups of G are f1G g and G. Note 1 If G is Abelian and simple then G = Zp. 2 There are both infinite and finite non-abelian groups which are simple. The smallest non-Abelian simple group is A5 which has order 60. Kevin James Composition Series and the H¨olderProgram Definition In a group G a sequence of subgroups 1 = N0 ≤ N1 ≤ · · · ≤ Nk−1 ≤ Nk = G is called a composition series for G if Ni E Ni+1 and Ni+1=Ni is simple for 0 ≤ i ≤ k − 1. If the above sequence is a composition series the quotient groups Ni+1=Ni are called composition factors of G. Theorem (Jordan-Holder)¨ Let 1 6= G be a finite group. Then, 1 G has a composition series 2 The (Jordan H¨older)composition factors in any composition series of G are unique up to isomorphism and rearrangement.
    [Show full text]
  • Section VII.35. Series of Groups
    VII.35. Series of Groups 1 Section VII.35. Series of Groups Note. In this section we introduce a way to consider a type of factorization of a group into simple factor groups. This idea is similar to factoring a natural number into prime factors. The big result of this section is the Jordan-H¨older Theorem which gives an indication of why it is important to classify finite simple groups (first encountered in Section III.15). Definition 35.1. A subnormal (or subinvariant) series of a group G is a finite sequence H0, H1,...,Hn of subgroups of G such that Hi < Hi+1 and Hi is a normal subgroup of Hi+1 (i.e., Hi / Hi+1) with H0 = {e} and Hn = G. A normal (or invariant) series of group G is a finite sequence H0, H1,...,Hn of normal subgroups of G (i.e., Hi / G) such that Hi < Hi+1, H0 = {e}, and H0 = G. Note. If Hi is normal in G, then gHi = Hig for all g ∈ G and so Hi is normal in Hi+1 ≤ G. So every normal series of G is also a subnormal series of G. Note. If G is an abelian group, then all subgroups of G are normal, so in this case there is no distinction between a subnormal series and a normal series. Example. A normal series of Z is: {0} < 16Z < 8Z < 4Z < Z. VII.35. Series of Groups 2 Example 35.3. The dihedral group on 4 elements D4 has the subnormal series: {ρ0} < {ρ0, µ1} < {ρ0, ρ2, µ1, µ2} <D4.
    [Show full text]
  • Reflexive Unitary Subsemigroups of Left Simple Semigroups
    REFLEXIVE UNITARY SUBSEMIGROUPS OF LEFT SIMPLE SEMIGROUPS A. Nagy Introduction Ideal series of semigroups [1] play an important role in the examination of semigroups which have proper two-sided ideals. But the corresponding theorems cannot be used when left simple (or right simple or simple) semigroups are con- sidered. So it is a natural idea that we try to use the group-theoretical methods (instead of the ring-theoretical ones) for the examination of these semigroups. The purpose of this paper is to find a suitable type of subsemigroups of left simple semigroups which makes possible for us to generalize some notions (the notion of the normal series and the composition series of groups) and some results concerning the groups to the left simple semigroups. We note that the subsemigroups we are looking for are the reflexive unitary subsemigroups of left simple semigroups. For notations and notions not defined here, we refer to [1] and [2]. Reflexive unitary subsemigroups As it is known [1], a subset H of a semigroup S is said to be reflexive in S if ab ∈ H implies ba ∈ H for all a,b ∈ S. We say that a subset U of a semigroup S is left [right] unitary in S (see [1]) if ab,a ∈ U implies b ∈ U [ab,b ∈ U implies a ∈ U] for all a,b ∈ S. A subset of S which is both left and right unitary in S is said to be unitary in S. We note that if A ⊆ B are subsemigroups of a semigroup S and B is unitary in S, then A is unitary in B if and only if A is unitary in S.
    [Show full text]
  • Chains and Lengths of Modules
    Chains and Lengths of Modules A mathematical essay by Wayne Aitken∗ Fall 2019y In this essay begins with the descending and ascending chain conditions for modules, and more generally for partially ordered sets. If a module satisfies both of these chain conditions then it has an invariant called the length. This essay will consider some properties of this length. The first version of this essay was written to provide background material for another essay focused on the Krull{Akizuki theorem and on Akizuki's other theorem that every Artinian ring is Noetherian.1 The current essay considers Noetherian and Artinian modules, the follow-up essay considers Noetherian and Artinian rings. I have attempted to give full and clear statements of the definitions and results, and give indications of any proof that is not straightforward. However, my philos- ophy is that, at this level of mathematics, straightforward proofs are best worked out by the reader. So some of the proofs may be quite terse or missing altogether. Whenever a proof is not given, this signals to the reader that they should work out the proof, and that the proof is straightforward. Often supplied proofs are sketches, but I have attempted to be detailed enough that the reader can supply the details without too much trouble. Even when a proof is provided, I encourage the reader to attempt a proof first before looking at the provided proof. Often the reader's proof will make more sense because it reflects their own viewpoint, and may even be more elegant. 1 Required background This document is written for readers with some basic familiarity with abstract algebra including some basic facts about rings (at least commutative rings), ideals, modules over such rings, quotients of modules, and module homomorphisms and isomorphisms.
    [Show full text]
  • Chapter I: Groups 1 Semigroups and Monoids
    Chapter I: Groups 1 Semigroups and Monoids 1.1 Definition Let S be a set. (a) A binary operation on S is a map b : S × S ! S. Usually, b(x; y) is abbreviated by xy, x · y, x ∗ y, x • y, x ◦ y, x + y, etc. (b) Let (x; y) 7! x ∗ y be a binary operation on S. (i) ∗ is called associative, if (x ∗ y) ∗ z = x ∗ (y ∗ z) for all x; y; z 2 S. (ii) ∗ is called commutative, if x ∗ y = y ∗ x for all x; y 2 S. (iii) An element e 2 S is called a left (resp. right) identity, if e ∗ x = x (resp. x ∗ e = x) for all x 2 S. It is called an identity element if it is a left and right identity. (c) The set S together with a binary operation ∗ is called a semigroup if ∗ is associative. A semigroup (S; ∗) is called a monoid if it has an identity element. 1.2 Examples (a) Addition (resp. multiplication) on N0 = f0; 1; 2;:::g is a binary operation which is associative and commutative. The element 0 (resp. 1) is an identity element. Hence (N0; +) and (N0; ·) are commutative monoids. N := f1; 2;:::g together with addition is a commutative semigroup, but not a monoid. (N; ·) is a commutative monoid. (b) Let X be a set and denote by P(X) the set of its subsets (its power set). Then, (P(X); [) and (P(X); \) are commutative monoids with respective identities ; and X. (c) x∗y := (x+y)=2 defines a binary operation on Q which is commutative but not associative.
    [Show full text]
  • A Some Results from Group Theory
    A Some Results from Group Theory A.1 Solvable Groups Definition A.1.1. A composition series for a group G is a sequence of sub- groups G = G0 ⊇ G1 ⊇ ...Gk ={1} with Gi a normal subgroup of Gi−1 for each i. A group G is solvable if it has a composition series with each quotient Gi /Gi−1 abelian. Example A.1.2. (1) Every abelian group G is solvable (as we may choose G0 = G and G1 ={1}). (2) If G is the group of Lemma 4.6.1, G = σ, τ | σ p = 1, τ p−1 = 1, τστ−1 = σ r where r is a primitive root mod p, then G is solvable: G = G0 ⊃ G1 ⊃ G2 ={1} with G1 the cyclic subgroup generated by σ . Then G1 is a normal subgroup of G with G/G1 cyclic of order p − 1andG1 cyclic of order p. (3) If G is the group of Lemma 4.6.3, G = σ, τ | σ 4 = 1, τ 2 = 1, −1 3 τστ = σ , then G is solvable: G = G0 ⊃ G1 ⊃ G2 ={1} with G1 the cyclic subgroup generated by σ . Then G1 is a normal subgroup of G with G/G1 cyclic of order 2 and G1 cyclic of order 4. (4) If G is a p-group, then G is solvable. This is Lemma A.2.4 below. (5) If G = Sn, the symmetric group on {1,...,n}, then G is solvable for n ≤ 4. S1 is trivial. S2 is abelian. S3 has the composition series S3 ⊃ A3 ⊃{1}.
    [Show full text]