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.