The Jordan Hölder Theorem

Transcription

1 Chapter 4 The Jordan Hölder Theorem Definition 4.1 Let G be a group. A subnormal series of G is a finite chain of subgroups G = G 0 >G 1 >G 2 > >G n = 1 such that G i+1 is a normal subgroup of G i for i =0,1,...,n 1. The collection of quotient groups G 0 /G 1,G 1 /G 2,...,G n 1 /G n are the factors of the series, and the length of the series is n. Note that we do not require each subgroup in the subnormal series to be normal in the whole group, only that it is normal in the previous subgroup in the chain. A normal series is a series where G i is a normal subgroup of G for all i. Note also that the length n is also the number of factors occurring. We will be interested in three different types of subnormal series in this course, and for all three we will require special properties of the factors. The first case is where the factors are all required to be simple groups. Definition 4.2 A composition series for a group G is a finite chain of subgroups G = G 0 >G 1 > >G n = 1 such that, for i =0,1,...,n 1, G i+1 is a normal subgroup of G i and the quotient group G i /G i+1 is simple. The quotient groups G 0 /G 1,G 1 /G 2,...,G n 1 /G n are called the composition factors of G. Example 4.3 Let G = S 4,andconsiderthefollowingchainofsubgroups: S 4 >A 4 >V 4 > (1 2)(3 4) > 1. (4.1) 44

2 We know that A 4 S 4 and V 4 A 4. Since V 4 is an abelian group, (1 2)(3 4) V 4. Certainly 1 (1 2)(3 4). Hence (4.1) is a series of subgroups, each normal in the previous one. We can calculate the order of each subgroup, and hence calculate the order of the quotient groups: S 4 /A 4 =2 A 4 /V 4 =3 V 4 / (1 2)(3 4) =2 (1 2)(3 4) =2. Thus the quotients are all of prime order. We now make use of the factthat agroupg of prime order p is both cyclic and simple (see Example 3.6), to see that the factors for the series (4.1) are cyclic simple groups. Thus (4.1) is composition series for S 4,withcompositionfactors C 2,C 3,C 2,C 2. Example 4.4 The infinite cyclic group G has no composition series. Proof: Let G = x, where x =, andsuppose G = G 0 >G 1 >G 2 > >G n = 1 is a composition series for G. Then G n 1 is a non-trivial subgroup of G, so G n 1 = x k for some positive integer k (see Tutorial Sheet I). Hence G n 1 is an infinite cyclic group, so is not simple. This contradicts ourseries being a composition series. Suppose that there are subgroups G N>M 1 with M N. (TheideahereisthatM and N will be successive terms in a series which we are testing to see whether or not it is a composition series.) The Correspondence Theorem tells us that subgroups of N/M correspond to subgroups of N which contain M. Furthermore,underthiscorrespondence, normal subgroups of N/M correspond to normal subgroups of N which contain M. Weconcludethat N/M is simple if and only if the only normal subgroups of N containing M are N and M themselves. Proposition 4.5 Every finite group possesses a composition series. 45

3 Proof: Let G be a finite group. We know at least one subnormal series, namely G 1. Let G = G 0 >G 1 > >G n = 1 (4.2) be a longest possible subnormal series of G. ThiscertainlyexistssinceG has only finitely many subgroups, so only finitely many subnormal series. We claim that (4.2) is a composition series. Suppose that it is not. Then one of the factors, say G j /G j+1,isnot simple. This quotient then possesses a non-trivial proper normal subgroup and this corresponds to a subgroup N of G with Then G j+1 <N G j. G = G 0 >G 1 > >G j >N>G j+1 > >G n = 1 is a subnormal series in G (note that G j+1 N since G j+1 G j )whichis longer than (4.2). This contradicts our assumption that (4.2) is the longest such series. Hence (4.2) is indeed a composition series for G. On the other hand, we have an example of an infinite group (namely the infinite cyclic group) which does not possess a composition series. There are infinite groups that possess composition series, but infinite simple groups (which necessarily occur as some of the composition factors) are much less well understood than finite simple groups. The important thing about composition series is that the composition factors occurring are essentially unique. This is the content of the following important theorem. Theorem 4.6 (Jordan Hölder Theorem) Let G be a group and let and G = G 0 >G 1 >G 2 > >G n = 1 G = H 0 >H 1 >H 2 > >H m = 1 be composition series for G. Thenn = m and there is a one-one correspondence between the two sets of composition factors and {G 0 /G 1,G 1 /G 2,...,G n 1 /G n } {H 0 /H 1,H 1 /H 2,...,H m 1 /H m } such that corresponding factors are isomorphic. Proof: See Tutorial Sheet IV. 46

4 Hence, once we have determined one composition series for a (say, finite) group, then we have uniquely determined composition factors which can be thought of as the ways of breaking our original group down into simple groups. This is analogous to a prime factorisation for groups. To recognise when we have a composition series, we need to be able to recognise simple groups. We have already observed (Example 3.6) that cyclic groups of prime order are simple. It is not hard to show that these are all the abelian simple groups. The following was proved in MT4003. Theorem 4.7 Let n 5. ThenthealternatinggroupA n is simple. It is worth pointing out that much, much more is known. A mammoth effort by hundreds of mathematicians from the 1950s to the 1980s succeeded in classifying the finite simple groups. The complete proof runs to tens of thousands of pages of extremely complicated mathematics. More work is still being done to check, clarify and simplify the proof. Nevertheless, it is generally accepted that this Classification is correct, though when relying upon it a mathematician would normally state that he or she is doing so. Theorem 4.8 (Classification of Finite Simple Groups) Let G be a finite simple group. Then G is one of the following: (i) acyclicgroupofprimeorder; (ii) an alternating group A n where n 5; (iii) one of sixteen infinite families of groups of Lie type; (iv) one of twenty-six sporadic simple groups. The groups of Lie type are essentially matrix-like groups which preserve geometric structures on vector spaces over finite fields. For example, the first (and most easily described) family is PSL n (q) = SL n(q) Z(SL n (q)). To make it, first construct the group GL n (q) ofinvertiblen n matrices with entries from the finite field F of order q. Next, take the subgroup of matrices of determinant 1: SL n (q) ={ A GL n (q) det A =1}, the special linear group. Now,factorbythecentreofSL n (q), which consists of all scalar matrices of determinant 1: Z(SL n (q)) = { λi λ F, λ n =1}. 47

5 The result is PSL n (q). It can be shown that PSL n (q) issimpleifandonlyif either n =2andq 4, or n 3. The construction of the other 15 families of groups of Lie type is similar but harder. The twenty-six sporadic groups are: Name Symbolic name Order Mathieu M Mathieu M Janko J Mathieu M Janko J Mathieu M Higman Sims HS Janko J Mathieu M McLaughlin McL Held He Rudvalis Ru Suzuki Suz O Nan O N Conway Co Conway Co Fischer Fi Harada Norton HN Lyons Ly Thompson Th Fischer Fi Conway Co Janko J Fischer Fi Baby Monster B Monster M see below M = Unsurprisingly, we omit the proof of Theorem 4.8. We finish this chapter with a few examples. Example 4.9 Let G be a finite abelian group of order n. Write n = p r 1 1 pr prs s where p 1, p 2,...,p s are the distinct prime factors of n. If G = G 0 >G 1 >G 2 > >G m = 1 48

6 is a composition series, then the composition factors G 0 /G 1,G 1 /G 2,...,G m 1 /G m are abelian simple groups. They are therefore cyclic of prime order. Now G = G 0 /G 1 G 1 /G 2... G m 1 /G m. This must be the prime factorisation of G = n, andhencethecomposition factors of G are C p1,c p1,...,c p1,c p2,c p2,...,c p2,...,c ps,c ps,...,c ps. }{{}}{{}}{{} r 1 times r 2 times r s times Although the Jordan Hölder Theorem tells us that the composition factors are essentially uniquely determined, the composition series need not be unique. For example, if G = x is cyclic of order 30, then there are several different composition series; e.g., G = x > x 2 > x 6 > 1 where the composition factors are C 2, C 3 and C 5,and G = x > x 3 > x 15 > 1 where the composition factors are C 3, C 5 and C 2 ;etc. Later in the course we shall characterise the finite groups whose composition factors are cyclic as being the soluble groups. Our final example has a unique composition series: Example 4.10 Let n 5andconsiderthesymmetricgroupS n of degree n. We already know the following series: S n >A n > 1 (4.3) which has factors C 2 and A n. Both of these are simple groups, so (4.3) is acompositionseriesfors n. It can be shown that S n has precisely three normal subgroups (namely those occurring in the above series) and hence (4.3) is the only composition series for S n. The Jordan Hölder Theorem again raises the question of how we put the composition factors back together. We have a unique decomposition, but how complicated is the reverse process? The answer turns out to be rather difficult, but in the next chapter we shall meet some ways of creating new groups and this will give some ways of putting the composition factors back together. 49