Algebraic Structures II

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1 MAS 305 Algebraic Structures II Notes 9 Autumn 2006 Composition series soluble groups Definition A normal subgroup N of a group G is called a maximal normal subgroup of G if (a) N G; (b) whenever N M G then either M = N or M = G. By the Correspondence Theorem, if N G N G then every normal subgroup of G/N corresponds to a normal subgroup of G containing N. So a normal subgroup N is maximal if only if G/N is simple. Definition Given a group G, a composition series for G of length n is a sequence of subgroups G = B 0 > B 1 > > B n = 1 G } such that (a) B i B i 1 for i = 1,..., n, (b) B i 1 /B i is simple for i = 1,..., n. In particular, B 1 is a maximal normal subgroup of G B n 1 is simple. The (isomorphism classes of the) quotient groups B i /B i 1 are called composition factors of G. Example S 4 has the following composition series of length 4, where K is the Klein group (1), (12)(34), (13)(24), (14)(23)}. S 4 > A 4 > K > (12)(34) > 1} We know that A 4 S 4 ; the composition factor S 4 /A 4 = C2. We have seen that K A 4 ; A 4 /K = C 3. All subgroups of K are normal in K, because K is Abelian. Both K/ (12)(34) (12)(34) /1} are isomorphic to C 2. So the composition factors of S 4 are C 2 (three times) C 3 (once). 1

2 S 4 A 4 K (12)(34) = H 1} A 4 S 4 S 4 /A 4 = C2 K A 4 A 4 /K = C 3 H K K/H = C 2 1} H H/1} = C 2 Example If G is simple then its only composition series is G > 1}, of length 1. Example (Z,+) has no composition series. If H Z then H is cyclic of infinite order. If H = x then 2x is a subgroup of H with 0} 2x H, 2x H because H is Abelian. So H is not simple. If B 0 > B 1 > > B n is a composition series then B n 1 is simple, so there can be no composition series. Theorem Every finite group G has a composition series. Proof We use induction on G. If G = 1 then the composition series is just G = B 0 = 1}. Assume that G > 1 that the result is true for all groups of order less than G. Since G is finite, G has at least one maximal normal subgroup N. Then N < G, so by induction N has a composition series N = B 1 > B 2 > > B n = 1} with B i B i 1 B i 1 /B i simple for i = 2,..., n. Putting B 0 = G gives the composition series G = B 0 > B 1 > > B n = 1} for G, because B 1 B 0 B 0 /B 1 = G/N, which is simple. The next theorem shows that statements such as the composition factors of S 4 are C 2 (three times) C 3 do not depend on the choice of composition series. 2

3 Jordan-Hölder Theorem Suppose that the finite group G has two composition series G = B 0 > B 1 > > B n = 1} G = C 0 > C 1 > > C m = 1}. Then n = m the lists of composition factors for the two series are identical in the sense that if H G φ(h) = i 1 : B i 1 /B i = H} then φ(h) = ψ(h). ψ(h) = i 1 : C i 1 /C i = H} Proof We use induction on G. The result is true if G = 1. Assume that G > 1 that the result is true for all groups of order less than G. Then n m are both positive. Put Then φ 1 (H) = i 2 : B i 1 /B i = H} ψ 1 (H) = i 2 : C i 1 /C i = H}. φ1 (H) + 1 if H φ(h) = = G/B 1 φ 1 (H) otherwise ψ1 (H) + 1 if H ψ(h) = = G/C 1 ψ 1 (H) otherwise. First suppose that B 1 = C 1. Then B 1 has the following two composition series: of length n 1, B 1 > > B n = 1} B 1 = C 1 > > C m = 1} of length m 1. Now, B 1 < G, so by the inductive hypothesis the result is true for B 1, so n = m φ 1 (H) = ψ 1 (H) for all H. If H = G/B 1 = G/C 1 then φ(h) = φ 1 (H) + 1 = ψ 1 (H) + 1 = ψ(h); otherwise φ(h) = φ 1 (H) = ψ 1 (H) = ψ(h). Therefore the result is true for G. 3

4 Secondly, suppose that B 1 C 1, put D = B 1 C 1. Because B 1 C 1 are both normal subgroups of G, so is B 1 C 1. If B 1 C 1 = B 1 then B 1 > C 1, but this cannot be true, because C 1 is a maximal normal subgroup of G. Hence B 1 B 1 C 1 G B 1 C 1 B 1, so B 1 C 1 = G. By the Third Isomorphism Theorem, G/B 1 = B 1 C 1 /B 1 = C1 /B 1 C 1 = C 1 /D. Similarly, G/C 1 = B1 /D. Let D = D 0 > D 1 > > D t = 1} be a composition series for D, put θ(h) = i 1 : D i 1 /D i = H}. B 2 G B 1 C 1 D C 2 D 1 B n 1 D t 1 C m 1 1 G } Now C 1 /D is simple, so C 1 > D > D 1 > > D t = 1} is a composition series for C 1. So is C 1 > C 2 > > C m = 1}. But C 1 < G, so by inductive hypothesis t + 1 = m 1 θ(h) + 1 if H = C1 /D ψ 1 (H) = θ(h) otherwise. 4

5 Applying the similar argument to B 1 gives t + 1 = n 1 θ(h) + 1 if H = B1 /D φ 1 (H) = θ(h) otherwise. Hence n = m. Moreover, since G/B 1 = C1 /D G/C 1 = B1 /D, either (a) G/B 1 θ(h) + 2 if H = G/B1 = G/C1 φ(h) = ψ(h) = θ(h) otherwise, or (b) G/B 1 θ(h) + 1 if H = G/B1 or H = G/C1 φ(h) = ψ(h) = = G/C 1 θ(h) otherwise. Definition A finite group is soluble if all its composition factors are cyclic of prime order. Example S 4 is soluble. Example S 5 is not soluble, because its only composition series is S 5 > A 5 > 1}. We have already shown that if G = p n for some prime p then G has subgroups 1 G } = G 0 < G 1 < < G n = G with G i G G i = p i for i = 0,..., n. So G i+1 /G i = p so G i+1 /G i = Cp for i = 0,..., n 1. Thus every finite p-group is soluble. A composition series in which every subgroup is normal in the whole group is called a chief series. A finite group is supersoluble if it has a chief series all of whose composition factors are cyclic of prime order. So all finite p-groups are supersoluble. Theorem If H is a normal subgroup of a finite group G, if H G/H are both soluble then G is soluble. Proof Let H = H 0 > H 1 > > H r = 1} be a composition series for H. Let G/H = K 0 > K 1 > > K s = H} be a composition series for G/H. By the Correspondence Theorem, there are subgroups G 0,..., G s of G containing H such that G i /H = K i for i = 0,..., s G i G i 1 for i = 1,..., s. By the Second Isomorphism Theorem, Then K i 1 /K i = (G i 1 /H)/(G i /H) = G i 1 /G i. G = G 0 G 1 G s = H = H 0 H 1 H r = 1} is a composition series for G in which every composition factor is cyclic of prime order. 5

6 This proof also shows that if H G H has a composition series of length r G/H has a composition series of length s then G has a composition series of length r + s. In other words, (the composition length of G) = (the composition length of H) + (the composition length of G/H). Corollary All finite Abelian groups are soluble. Proof Use induction on G. If G = 1 then G is soluble. Assume that G is Abelian, that G > 1 that all Abelian groups of order less than G are soluble. By Cauchy s Theorem, G contains a subgroup H of prime order. Thus H is soluble. Since G is Abelian, H G G/H is Abelian. But H > 1 so G/H < G, so G/H is soluble, by inductive hypothesis. By the preceding theorem, G is soluble. Theorem Let G be a finite group. Then G is soluble if only if there is a sequence of subgroups G = B 0 > B 1 > > B n = 1} such that (a) B i B i 1 for i = 1,..., n (b) B i 1 /B i is Abelian for i = 1,..., n. Proof If G is soluble then any composition series satisfies (a) (b) with each B i 1 /B i cyclic of prime order, hence Abelian. Conversely, use induction on the order of G. The result is true if G = 1. Now assume that G > 1 that the result is true for all groups of smaller order. Suppose that G has a such a sequence. Then B i 1 /B i is Abelian for i = 2,..., n, so B 1 satisfies the conditions. Also, B 1 < G. By inductive hypothesis, B 1 is soluble, Moreover, B 1 G G/B 1 is Abelian, hence soluble, by the preceding corollary. Hence G is soluble, by the preceding theorem. 6