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 subgroup 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

(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

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

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

n 1 (mod r) and n q. r ≡ r |

Since q

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(aspG1 >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. Hence G1 contains the identity element and three elements from the following set of elements of order two: α2,β,αβ,α2β,α3β . { } Since α,α3 G , we see that there are only two possibilities: ∈ 1 G = 1,α2,β,α2β = α2,β 1 { } or G = 1,α2,αβ,α3β = α2,αβ . 1 { } Each of these is isomorphic to V = C C . Picking any element of order two 4 ∼ 2 × 2 in them determines G2. Hence we ﬁnd six more composition series: D > α2,β > α2 > 1 8 D > α2,β > β > 1 8 D > α2,β > α2β > 1 8 D > α2,αβ > α2 > 1 8 D > α2,αβ > αβ > 1 8 D > α2,αβ > α3β > 1 8

Thus there are seven composition series for D8.

5. How many diﬀerent composition series does the quaternion group Q8 have?

Solution: Recall that Q8 = 1, i, j, k . Since Q8 is a 2-group, Ques- tion 3 tells us that its composition{± factors± ± are± } all cyclic of order two and hence every composition series for Q8 has the form

Q8 = G0 >G1 >G2 >G3 = 1 where G = 4 and G = 2. Now Q has a unique element of order two, | 1| | 2| 8 namely 1. Hence G2 = 1 is immediately determined. The subgroup G1 is cyclic− of order 4 (it is not− isomorphic to C C as there are not enough 2 × 2 elements of order two). Thus G1 is determined by picking a generator. There are six elements of order 4 in Q8 and therefore three cyclic groups of order four. Hence the composition series for Q8 are: Q > i > 1 > 1 8 − Q > j > 1 > 1 8 − Q > k > 1 > 1 8 −

5 6. Let G be a group, N be a normal subgroup of G and suppose that

G = G >G >G > >G =1 0 1 2 ··· n is a composition series for G.DeﬁneN = N G for i =0,1,...,n. i ∩ i (a) Show that N is a normal subgroup of N for i =0,1,...,n 1. i+1 i − (b) Use the Second Isomorphism Theorem to show that

(Gi N)Gi+1 Ni /Ni+1 ∼= ∩ . Gi+1

[Hint: Note that N = G (G N).] i+1 i+1 ∩ i ∩ (c) Show that (G N)G is a normal subgroup of G containing G . i ∩ i+1 i i+1 Deduce that (G N)G /G is either equal to G /G or to the trivial group. [Re- i ∩ i+1 i+1 i i+1 member that Gi /Gi+1 is simple.]

(d) Deduce that N possesses a composition series. [Hint: Delete repeats in the series (Ni ).]

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 >G > >G = 1 0 1 ··· n be a composition series for G and let N be a normal subgroup of G. Deﬁne N = N G for each i. Then i ∩ i N = N N N N = 1 (3) 0 1 2 ··· n is a descending chain of subgroups. If x N G and g N G , then ∈ ∩ i+1 ∈ ∩ i 1 g− xg N as x, g N ∈ ∈ and 1 g− xg G as G G , ∈ i+1 i+1 i 1 so g− xg N Gi+1. Hence Ni+1 Ni for each i. Therefore (3) is a subnormal series for ∈N. We∩ show that it is a composition series. Now N G N /N = ∩ i i i+1 N G ∩ i+1 N G = ∩ i as G G (N G ) G i+1 i ∩ i ∩ i+1 (N G )G = ∩ i i+1 ∼ Gi+1 by the Second Isomorphism Theorem (as N G G and G G ). Since ∩ i i i+1 i N G, it follows that N Gi Gi, and since also Gi+1 Gi we deduce that (N G )G G . Therefore,∩ by the Correspondence Theorem, ∩ i i+1 i (N Gi)Gi+1 Gi ∩ . Gi+1 Gi+1

6 However, the latter group is simple, so either (N G )G G ∩ i i+1 = 1 or i . Gi+1 Gi+1

The above isomorphism implies that either Ni = Ni+1 or Ni/Ni+1(∼= Gi/Gi+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 >G >G > >G =1 0 1 2 ··· n

is a composition series for G.DeﬁneQi = Gi N/N for i =0,1,...,n.

(a) Show that Qi is a subgroup of G/N such that Qi+1 is a normal subgroup of Qi for i =0,1, ..., n 1. − (b) Use Dedekind’s Modular Law to show that G N G =(G N)G .Showthat i+1 ∩ i i ∩ i+1

Gi /Gi+1 Qi /Qi+1 = . ∼ (G N)G /G i ∩ i+1 i+1 [Hint: Use the Third Isomorphism Theorem, the Second Isomorphism Theorem and note that Gi N = Gi (Gi+1N)sinceGi+1 Gi .] (c) Show that (G N)G is a normal subgroup of G containing G . i ∩ i+1 i i+1 Deduce that (G N)G /G is either equal to G /G or to the trivial group. [Re- i ∩ i+1 i+1 i i+1 member that Gi /Gi+1 is simple.] Hence show that the quotient on the right hand side in (b) is either trivial or isomorphic to Gi /Gi+1. (d) Deduce that G/N possesses a composition series.

Solution: (As with the previous question, we now produce a complete proof.) Let G = G >G > >G = 1 0 1 ··· n be a composition series for G. Since N G, we obtain a collection of subgroups

G = G N G N G N = N 0 1 ··· n and the Correspondence Theorem gives a chain of subgroups of G/N:

G/N = G N/N G N/N G N/N = 1. (4) 0 1 ··· n 1 Deﬁne Q = G N/N.Ifx G and g G , then g− xg G ,so i i ∈ i+1 ∈ i ∈ i+1 1 (Ng)− (Nx)(Ng) G N/N = Q ∈ i+1 i+1

and hence Qi+1 Qi.

Now Gi+1 Gi, so Dedekind’s Modular Law gives G N G =(G N)G . i+1 ∩ i i ∩ i+1 7 Then

GiN/N Qi/Qi+1 = Gi+1N/N G N = i by the Third Isomorphism Theorem ∼ Gi+1N (GiGi+1)N = as Gi+1 Gi Gi+1N by the Second Isomorphism Theo- G = i rem ∼ Gi Gi+1N ∩ (Gi GiN, Gi+1N GiN) G = i by Dedekind’s Modular Law (Gi N)Gi+1 ∩ G /G = i i+1 by the Third Isomorphism Theorem. ∼ (G N)G /G i ∩ i+1 i+1 Now Gi/Gi+1 is a simple group and we are factoring it by some normal subgroup. Consequently, the quotient is either trivial or is isomorphic to Gi/Gi+1. Thus

Qi/Qi+1 ∼= 1 or Gi/Gi+1. Upon deleting repetitions from the series (4), we obtain a series whose factors are isomorphic to some of the Gi/Gi+1. Thus G/N has a composition series.

8. The purpose of this question is to prove the Jordan–H¨older Theorem. Let G be a group and suppose that

G = G >G > >G =1 0 1 ··· n and G = H >H > >H =1 0 1 ··· m are two composition series for G.Proceedbyinductiononn.

(a) If n =0,showthatG =1andthattheJordan–H¨olderTheoremholds.

(b) Suppose n 1. Suppose that G1 = H1.Observethatthetheoremholdsforthiscase. (c) Now suppose that G = H .UsethefactthatG /G and H /H are simple to show that 1 1 0 1 0 1 G1H1 = G.[Hint:ObservethatitisanormalsubgroupcontainingbothG1 and H1.] (d) Deﬁne D = G H .ShowthatD is a normal subgroup of G such that 1 ∩ 1

G0/G1 ∼= H1/D and H0/H1 ∼= G1/D.

(e) Use Question 6 to see that D possesses a composition series

D = D >D > >D =1. 2 3 ··· r (f) Observe that we now have four composition series for G:

G = G >G >G > >G =1 0 1 2 ··· n G = G >G >D= D > >D =1 0 1 2 ··· r G = H >H >D= D > >D =1 0 1 2 ··· r G = H >H >H > >H =1. 0 1 2 ··· m 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 >G >G > >G = 1 0 1 2 ··· n G = H >H >H > >H = 1 0 1 2 ··· m be composition series for G. We prove the Jordan–H¨older 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: G1 = H1. Then G >G > >G = 1 1 2 ··· n G = H >H > >H = 1 1 1 2 ··· m are composition series for G1 (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− − −

G1/G2,G2/G3,...,Gn 1/Gn { − } and H1/H2,H2/H3,...,Hm 1/Hm { − } such that corresponding factors are isomorphic. Since G0/G1 = H0/H1, this immediately extends to a one-one correspondence between

G0/G1,G1/G2,...,Gn 1/Gn { − } and H0/H1,H1/H2,...,Hm 1/Hm { − } such that corresponding factors are isomorphic. Case 2: G = H . 1 1 Now G1 G and H1 G,soG1H1 G. Now G1 G1H1 G0 = G and G0/G1 is simple, so G1H1 = G or G1.

If G1H1 =G1, then H1 G1H1 = G1 and H1 G1 G = H0. Since H0/H1 is simple, this implies that H1 = G1, contrary to assumption. Therefore

G1H1 = G. Let D = G H G. By the Second Isomorphism Theorem 1 ∩ 1 G1/(G1 H1) = G1H1/H1 = G/H1 = H0/H1 ∩ ∼ H /(G H ) = G H /G = G/G = G /G . 1 1 ∩ 1 ∼ 1 1 1 1 0 1

9 Thus G1/D ∼= H0/H1 and H1/D ∼= G0/G1. (5) Now let D = D >D > >D = 1 2 3 ··· r 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 >G >G > >G = 1 (6) 0 1 2 ··· n G = H >H >H > >H = 1 (7) 0 1 2 ··· m G = G >G >D= D >D > >D = 1 (8) 0 1 2 3 ··· r G = H >H >D= D >D > >D = 1 (9) 0 1 2 3 ··· r ((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 G0/G1 corresponds to H1/D and G1/D to H0/H1). 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 ∼= A5. [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.Showthatifx S T then C (x) 12. ∈ ∩ | G | 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 ∼= A5.[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 . The kernel of ρ is a normal → r subgroup of G contained in H,sokerρ = 1. Hence G ∼= im ρ Sr,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 S5 of index 2 and hence im ρ = A5 and G ∼= A5 (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 CG(x) and ST C (x). ⊆ G Note ST = S T / S T 4 4/2 = 8. Thus C (x) has at least 8 elements. On | | | |·| | | ∩ | · G the other hand, CG(x) is a subgroup of G so has order dividing 60. Furthermore, S is a subgroup of CG(x), so 4 divides the order of CG(x). This implies that

C (x) 12 | G |

(since 8 60). Therefore G :CG(x) 5. | | Part (a) implies that C (x)=G or G :C (x) =5.IfS T = 1, then there G | G | ∩ exists a non-identity element x S T .IfCG(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 (x) = 5. Hence G = A by part (a). The ∈ ∩ | G | ∼ 5 Sylow 2-subgroups of A5 have order 4 and thus

V = 1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3) 4 { }

is one such Sylow 2-subgroup. Also, V4 A4,soNA5 (V4) A4. Consequently the number of conjugates of V4 in A5 is A :N (V ) A : A =5. | 5 A5 4 | | 5 4|

We can make five Sylow 2-subgroups by changing the fixed point (V4 corresponds to fixing the point 5). Hence A5 has exactly five Sylow 2-subgroups and a calcu- lation shows that any two Sylow 2-subgroups of A5 intersect in the identity. (For example, the only element of V4 which fixes 5 and another point is the identity.)

11 Thus if S T = 1, we obtain a contradiction. We have shown that either ∩ CG(x)=G which produces a non-trivial central subgroup, or G ∼= A5 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.Letn2 be the number of Sylow 2-subgroups of G. By Sylow’s Theorem, n 15, so n =3,5or15. 2 | 2 (Remember that G is simple, so n2 = 1.) If n2 = 3, then the normaliser of a Sylow 2-subgroup would have index 3, which is impossible by part (a).

Thus n2 =5or15.Ifn2 = 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 n5 be the number of Sylow 5-subgroups. Sylow’s Theorem gives n 1 (mod 5) and n 12, 5 ≡ 5 | so n5 = 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 =15. 2 We deduce that n2 = 5 and if S is a Sylow 2-subgroup, then

G :N (S) = n =5. | G | 2 Now part (a) gives G ∼= A5, as required.