School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet IV: Composition series and the Jordan Hölder Theorem (Solutions)

CMRD 2010 School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet IV: Composition series and the Jordan Hölder 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 0 /N > G 1 /N > >G r /N = {N1} = 1 be a composition series for G/N. By the Correspondence Theorem, each subgroup in this series has the form G i /N, where G i is a subgroup of G containing N. Furthermore, G i /N G i 1 /N for all i, so the Correspondence Theorem (applied to the correspondence between subgroups of G i 1 containing N and subgroups of G i 1 /N ) shows that G i G i 1. Therefore and G i G i 1 for all i. Let G = G 0 >G 1 > >G r = N, N = N 0 >N 1 > >N s = 1 be a composition series for N. Putting this together with the above chain of subgroups, we obtain Here G = G 0 >G 1 > >G r = N 0 >N 1 > >N s = 1. G i 1 /G i = G i 1 /N G i /N (by the Third Isomorphism Theorem) is a simple group since the G i /N form a composition series for G/N, while N j 1 /N j 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. Let G 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, where K 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? Remember that all 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 = G 0 >G 1 >G 2 >G 3 =1 where G 1 = qr and G 2 = r. Up to isomorphism, what are the composition factors of G? Solution: (a) Let G have n r Sylow r-subgroups. By Sylow s Theorem, n r 1 (mod r) and n r pq. Now p<q<r,sop, q 1 (mod r). Thus if n r = 1 then n r = pq. 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 elements of order r. n r (r 1) = pq(r 1) = pqr pq Let G have n q Sylow q-subgroups. By Sylow s Theorem, n q 1 (mod q) and n q pr. Now p<q,sop 1 (mod q). Hence n q =1,r or pr. If n q = 1, then n q 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 (q 1) r(q 1) >qp elements of order q. There are now more than pqr pq + pq = G elements in G, a contradiction. Hence if n r = 1, then n q =1,soG has a unique, normal, Sylow q-subgroup Q. (b) Consider the quotient group G/Q. It has order pqr/q = pr. Let n r be the number of Sylow r-subgroups in G/Q. Then by Sylow s Theorem, n r 1 (mod r) and n r p. 2

But p<r,sop 1 (mod r) and hence n 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 n r Sylow r-subgroups. Then n r 1 (mod r) and n r q. Since q<r, we deduce that n r = 1. Thus K has a normal Sylow r-subgroup R. Now R is one of the Sylow r-subgroups of G. All n r Sylow r-subgroups of G are conjugate, so any other Sylow r-subgroup of G has the form R x for some x G. Now R K G, so R x K x = K. This means that R x 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 R x = 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 n r = 1 and G has a normal Sylow r-subgroup, R. Consider G/R, of order pq. LetG/R have ñ q Sylow q-subgroups. Sylow s Theorem gives ñ q 1 (mod q) and ñ q p, so ñ q =1(asp<q). Thus G/R has a normal Sylow q-subgroup, L/R, where R L G and L = qr. (e) Take G 0 = G, G 1 = L, G 2 = R and G 3 = 1 (with R and L as in (d)). Then G = G 0 >G 1 >G 2 >G 3 =1 with G i G i 1 for all i. (In fact, we have shown that this is a normal series: G i G for all i.) Certainly G 1 = L = qr and G 2 = R = r. The factors G 0 /G 1,G 1 /G 2,G 2 /G 3 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 0 >G 1 >G 2 > >G n =1 such that G i is a normal subgroup of G and G : G i = p i for i =0,1,...,n. What are the composition factors of G? [Hint: Use the fact that Z(G) = 1toproduceanelementx Z(G) oforderp. quotient group G/K where K = x and apply induction.] Consider the 3

Solution: Let G = p n and induct on n. If n = 0, then G = G 0 = 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 of x is a non-trivial power of p, say x = p r with r>0. Then x pr 1 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. Hence X G. (Ifg G and y Z(G), then g 1 yg = y.) Consider the quotient group G/X: a p-group of order G /p = p n 1. Hence by induction G/X has a composition series G/X = G 0 /X > G 1 /X > >G n 1 /X = {X1} = 1 where G/X : G i /X = p i for i =0,1,...,n 1. Each G i /X corresponds to a subgroup G i of G containing X, so we obtain a chain of subgroups G = G 0 >G 1 > >G n 1 = X>1. Here G i G i 1 for i =1,2,...,n 1 by the Correpondence Theorem. Also, G i = G i /X p, and G : G i = G/X : G i /X = p i for i =0,1,...,n 1. Defining G n = 1 produces a series for G G = G 0 >G 1 > >G n = 1 where G : G i = p i for all i. Consequently, G i = p n i and G i 1 /G i = 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 C p. 4. The dihedral group D 8 has seven different composition series. Find all seven. Solution: Recall that D 8 = α, β where α has order four, β has order two, and αβ = βα 1. Since D 8 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 D 8 has the form D 8 = G 0 >G 1 >G 2 >G 3 = 1 where G 1 = 4 and G 2 =2. One possibility for G 1 is that it is cyclic of order four. All elements of D 8 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 α 2. Thus D 8 > α > α 2 > 1 is a composition series for D 8 and this is the only one with G 1 cyclic. 4

Consider now the other possibilities for G 1. If it is not cyclic then it must contain no elements of order four. Hence G 1 contains the identity element and three elements from the following set of elements of order two: {α 2,β,αβ,α 2 β,α 3 β}. Since α, α 3 G 1, we see that there are only two possibilities: or G 1 = {1,α 2,β,α 2 β} = α 2,β G 1 = {1,α 2,αβ,α 3 β} = α 2,αβ. Each of these is isomorphic to V 4 = C2 C 2. Picking any element of order two in them determines G 2. Hence we find six more composition series: D 8 > α 2,β > α 2 > 1 D 8 > α 2,β > β > 1 D 8 > α 2,β > α 2 β > 1 D 8 > α 2,αβ > α 2 > 1 D 8 > α 2,αβ > αβ > 1 D 8 > α 2,αβ > α 3 β > 1 Thus there are seven composition series for D 8. 5. How many different composition series does the quaternion group Q 8 have? Solution: Recall that Q 8 = {±1, ±i, ±j, ±k}. Since Q 8 is a 2-group, Question 3 tells us that its composition factors are all cyclic of order two and hence every composition series for Q 8 has the form Q 8 = G 0 >G 1 >G 2 >G 3 = 1 where G 1 = 4 and G 2 = 2. Now Q 8 has a unique element of order two, namely 1. Hence G 2 = 1 is immediately determined. The subgroup G 1 is cyclic of order 4 (it is not isomorphic to C 2 C 2 as there are not enough elements of order two). Thus G 1 is determined by picking a generator. There are six elements of order 4 in Q 8 and therefore three cyclic groups of order four. Hence the composition series for Q 8 are: Q 8 > i > 1 > 1 Q 8 > j > 1 > 1 Q 8 > k > 1 > 1 5

6. Let G be a group, N be a normal subgroup of G and suppose that G = G 0 >G 1 >G 2 > >G n =1 is a composition series for G. Define N i = N G i for i =0,1,...,n. (a) Show that N i+1 is a normal subgroup of N i for i =0,1,...,n 1. (b) Use the Second Isomorphism Theorem to show that [Hint: Note that N i+1 = G i+1 (G i N).] N i /N i+1 = (G i N)G i+1 G i+1. (c) Show that (G i N)G i+1 is a normal subgroup of G i containing G i+1. Deduce that (G i N)G i+1 /G i+1 is either equal to G i /G i+1 or to the trivial group. [Remember that G i /G i+1 is simple.] (d) Deduce that N possesses a composition series. [Hint: Delete repeats in the series (N i ).] Solution: (This is the proof of a theorem, namely that a normal subgroup of a group with a composition series also has a composition series, so we present the proof as a whole. The solution to each part can be extracted quite easily.) Let G = G 0 >G 1 > >G n = 1 be a composition series for G and let N be a normal subgroup of G. N i = N G i for each i. Then Define N = N 0 N 1 N 2 N n = 1 (3) is a descending chain of subgroups. If x N G i+1 and g N G i, then g 1 xg N as x, g N and g 1 xg G i+1 as G i+1 G i, so g 1 xg N G i+1. Hence N i+1 N i for each i. Therefore (3) is a subnormal series for N. We show that it is a composition series. Now N i /N i+1 = N G i N G i+1 = N G i (N G i ) G i+1 as G i+1 G i = (N G i )G i+1 G i+1 by the Second Isomorphism Theorem (as N G i G i and G i+1 G i ). Since N G, it follows that N G i G i, and since also G i+1 G i we deduce that (N G i )G i+1 G i. Therefore, by the Correspondence Theorem, (N G i )G i+1 G i+1 G i G i+1. 6

However, the latter group is simple, so either (N G i )G i+1 G i+1 = 1 or G i G i+1. The above isomorphism implies that either N i = N i+1 or N i /N i+1 ( = G i /G i+1 ) is simple. It follows that the factors in (3) are either trivial or simple. Hence, if we delete any repetitions occurring in (3), we obtain a composition series for N. 7. Let G be a group, N be a normal subgroup of G and suppose that G = G 0 >G 1 >G 2 > >G n =1 is a composition series for G. Define Q i = G i N/N for i =0,1,...,n. (a) Show that Q i is a subgroup of G/N such that Q i+1 is a normal subgroup of Q i for i =0,1,..., n 1. (b) Use Dedekind s Modular Law to show that G i+1 N G i =(G i N)G i+1. Show that Q i /Q i+1 = G i /G i+1 (G i N)G i+1 /G i+1. [Hint: Use the Third Isomorphism Theorem, the Second Isomorphism Theorem and note that G i N = G i (G i+1 N)sinceG i+1 G i.] (c) Show that (G i N)G i+1 is a normal subgroup of G i containing G i+1. Deduce that (G i N)G i+1 /G i+1 is either equal to G i /G i+1 or to the trivial group. [Remember that G i /G i+1 is simple.] Hence show that the quotient on the right hand side in (b) is either trivial or isomorphic to G i /G i+1. (d) Deduce that G/N possesses a composition series. Solution: Let (As with the previous question, we now produce a complete proof.) G = G 0 >G 1 > >G n = 1 be a composition series for G. Since N G, we obtain a collection of subgroups G = G 0 N G 1 N G n N = N and the Correspondence Theorem gives a chain of subgroups of G/N: G/N = G 0 N/N G 1 N/N G n N/N = 1. (4) Define Q i = G i N/N. Ifx G i+1 and g G i, then g 1 xg G i+1,so (Ng) 1 (Nx)(Ng) G i+1 N/N = Q i+1 and hence Q i+1 Q i. Now G i+1 G i, so Dedekind s Modular Law gives G i+1 N G i =(G i N)G i+1. 7

Then Q i /Q i+1 = G in/n G i+1 N/N G i N = G i+1 N = (G ig i+1 )N G i+1 N = = G i G i G i+1 N by the Third Isomorphism Theorem as G i+1 G i by the Second Isomorphism Theorem (G i G i N, G i+1 N G i N) G i (G i N)G i+1 by Dedekind s Modular Law = G i /G i+1 (G i N)G i+1 /G i+1 by the Third Isomorphism Theorem. Now G i /G i+1 is a simple group and we are factoring it by some normal subgroup. Consequently, the quotient is either trivial or is isomorphic to G i /G i+1. Thus Q i /Q i+1 = 1 or Gi /G i+1. Upon deleting repetitions from the series (4), we obtain a series whose factors are isomorphic to some of the G i /G i+1. Thus G/N has a composition series. 8. The purpose of this question is to prove the Jordan Hölder Theorem. Let G be a group and suppose that and G = G 0 >G 1 > >G n =1 G = H 0 >H 1 > >H m =1 are two composition series for G. Proceed by induction on n. (a) If n =0,showthatG =1andthattheJordan HölderTheoremholds. (b) Suppose n 1. Suppose that G 1 = H 1. Observe that the theorem holds for this case. (c) Now suppose that G 1 = H 1. Use the fact that G 0 /G 1 and H 0 /H 1 are simple to show that G 1 H 1 = G. [Hint: Observe that it is a normal subgroup containing both G 1 and H 1.] (d) Define D = G 1 H 1. Show that D is a normal subgroup of G such that G 0 /G 1 = H1 /D and H 0 /H 1 = G1 /D. (e) Use Question 6 to see that D possesses a composition series D = D 2 >D 3 > >D r =1. (f) Observe that we now have four composition series for G: G = G 0 >G 1 >G 2 > >G n =1 G = G 0 >G 1 >D= D 2 > >D r =1 G = H 0 >H 1 >D= D 2 > >D r =1 G = H 0 >H 1 >H 2 > >H m =1. Apply the case of part (b) (twice) and the isomorphisms of part (d) to complete the induction step of the proof. 8

Solution: Let 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. We prove the Jordan Hölder Theorem by induction on n (the length of some composition series for G). If n = 0, then G = 1, which forces m = 0 and the required correspondence between the composition factors vacuously exists (neither series has any factors!). Suppose then that n 1 and that the theorem holds for all groups possessing shorter composition series. Consider two cases: Case 1: G 1 = H 1. Then G 1 >G 2 > >G n = 1 G 1 = H 1 >H 2 > >H m = 1 are composition series for G 1 (since in both cases the factors are simple) and the first has length n 1. By induction, n 1=m 1 and there is a one-one correspondence between and {G 1 /G 2,G 2 /G 3,...,G n 1 /G n } {H 1 /H 2,H 2 /H 3,...,H m 1 /H m } such that corresponding factors are isomorphic. Since G 0 /G 1 = H 0 /H 1, this immediately extends to a one-one correspondence between 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. Case 2: G 1 = H 1. Now G 1 G and H 1 G, sog 1 H 1 G. Now G 1 G 1 H 1 G 0 = G and G 0 /G 1 is simple, so G 1 H 1 = G or G 1. If G 1 H 1 = G 1, then H 1 G 1 H 1 = G 1 and H 1 G 1 G = H 0. Since H 0 /H 1 is simple, this implies that H 1 = G 1, contrary to assumption. Therefore G 1 H 1 = G. Let D = G 1 H 1 G. By the Second Isomorphism Theorem G 1 /(G 1 H 1 ) = G 1 H 1 /H 1 = G/H 1 = H 0 /H 1 H 1 /(G 1 H 1 ) = G 1 H 1 /G 1 = G/G 1 = G 0 /G 1. 9

Thus G 1 /D = H 0 /H 1 and H 1 /D = G 0 /G 1. (5) Now let D = D 2 >D 3 > >D r = 1 be a composition series for D. (This exists since D G, by Question 6.) We now have four series with simple factors, i.e., composition series, for G: G = G 0 >G 1 >G 2 > >G n = 1 (6) G = H 0 >H 1 >H 2 > >H m = 1 (7) G = G 0 >G 1 >D= D 2 >D 3 > >D r = 1 (8) G = H 0 >H 1 >D= D 2 >D 3 > >D r = 1 (9) ((5) implies that the second factors in (8) and (9) are both simple groups.) Apply Case 1 to (6) and (8). We deduce that r = n and that the composition factors occurring in the two series are in one-one correspondence such that corresponding factors are isomorphic. Now compare (8) and (9). The only composition factors which are not identical are the first two in both cases. However, (5) says that the first factor for (8) is isomorphic to the second for (9) and vice versa. Hence we obtain the required correspondence between the factors of (8) and (9) (so G 0 /G 1 corresponds to H 1 /D and G 1 /D to H 0 /H 1 ). Finally apply Case 1 to (9) and (7). (We know the former has length n(= r), so it is valid to apply Case 1.) We deduce that r = m and the composition factors are in the required one-one correspondence. Putting this all together, we deduce that n = r = m and, upon composing the sequence of three correspondences, that there is a one-one correspondence between the composition factors of (6) and (7) such that corresponding factors are isomorphic. This completes the inductive step. 9. Let G be a simple group of order 60. (a) Show that G has no proper subgroup of index less than 5. Show that if G has a subgroup of index 5, then G = A 5. [Hint for both parts: If H is a subgroup of index k, actonthesetofcosetstoproducea permutation representation. What do we know about the kernel?] (b) Let S and T be distinct Sylow 2-subgroups of G. Show that if x S T then C G (x) 12. Deduce that S T =1. [Hint: WhyareS and T abelian?] (c) Deduce that G has at most five Sylow 2-subgroups and hence that indeed G = A 5. [Hint: The number of Sylow 2-subgroups equals the index of a normaliser.] Thus there is a unique simple group of order 60 up to isomorphism. 10

Solution: (a) Let H be a subgroup of G of index r where 2 r 5. Act by right multiplication on the cosets of H: (Hx,g) Hxg. This determines a homomorphism ρ: G S r. The kernel of ρ is a normal subgroup of G contained in H, sokerρ = 1. Hence G = im ρ S r,so 60 = G = im ρ r!. Thus r {2, 3, 4}; that is, G has no proper subgroup of index less than 5. If r = 5, then G is isomorphic to a subgroup of S 5 of index 2 and hence im ρ = A 5 and G = A 5 (by Tutorial Sheet III, Quesion 1). (b) Suppose x S T. Since S and T have order four, they are, in particular, abelian. Thus x commutes with every element of S and with every element of T. This gives S, T C G (x) and ST C G (x). Note ST = S T / S T 4 4/2 = 8. Thus C G (x) has at least 8 elements. On the other hand, C G (x) is a subgroup of G so has order dividing 60. Furthermore, S is a subgroup of C G (x), so 4 divides the order of C G (x). This implies that C G (x) 12 (since 8 60). Therefore G :C G (x) 5. Part (a) implies that C G (x) =G or G :C G (x) =5. IfS T = 1, then there exists a non-identity element x S T.IfC G (x) =G, then x Z(G) and x is a non-trivial proper normal subgroup of G, contradicting the simplicity of G. Thus if 1 = x S T, then G :C G (x) = 5. Hence G = A 5 by part (a). The Sylow 2-subgroups of A 5 have order 4 and thus V 4 = {1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} is one such Sylow 2-subgroup. Also, V 4 A 4,soN A5 (V 4 ) A 4. Consequently the number of conjugates of V 4 in A 5 is A 5 :N A5 (V 4 ) A 5 : A 4 =5. We can make five Sylow 2-subgroups by changing the fixed point (V 4 corresponds to fixing the point 5). Hence A 5 has exactly five Sylow 2-subgroups and a calculation shows that any two Sylow 2-subgroups of A 5 intersect in the identity. (For example, the only element of V 4 which fixes 5 and another point is the identity.) 11

Thus if S T = 1, we obtain a contradiction. We have shown that either C G (x) =G which produces a non-trivial central subgroup, or G = A 5 where Sylow 2-subgroups actually intersect in the identity. Hence S T = 1. (c) We now use part (b) to count the elements in G. Let n 2 be the number of Sylow 2-subgroups of G. By Sylow s Theorem, n 2 15, so n 2 =3,5or15. (Remember that G is simple, so n 2 = 1.) If n 2 = 3, then the normaliser of a Sylow 2-subgroup would have index 3, which is impossible by part (a). Thus n 2 = 5 or 15. If n 2 = 15, then by part (b), any two distinct Sylow 2-subgroups intersect in the identity, so they contain a total of 15 3=45 non-identity elements. Let n 5 be the number of Sylow 5-subgroups. Sylow s Theorem gives n 5 1 (mod 5) and n 5 12, so n 5 = 6. Hence the Sylow 5-subgroups (any pair of which intersect trivially) account for 6 4=24 non-identity elements. We now have too many elements: the Sylow 3- and 5-subgroups account for 45 + 24 = 69 elements. Thus n 2 =15. We deduce that n 2 = 5 and if S is a Sylow 2-subgroup, then G :N G (S) = n 2 =5. Now part (a) gives G = A 5, as required. 12