Group Theory Abstract Group Theory English Summary Notes by Jørn B

Home , Elementary abelian group, Holomorph (mathematics), Opposite group, Transfer (group theory)

Group theory Abstract group theory English summary Notes by Jørn B. Olsson Version 2007

b.. what wealth, what a grandeur of thought may spring from what slight beginnings. (H.F. Baker on the concept of groups)

1 CONTENT:

1. On normal and characteristic subgroups. The Frattini argument 2. On products of subgroups 3. Hall subgroups and complements 4. Semidirect products. 5. Subgroups and Verlagerung/Transfer 6. Focal subgroups, Gr¨un’stheorems, Z-groups 7. On ﬁnite linear groups 8. Frattini subgroup. Nilpotent groups. Fitting subgroup 9. Finite p-groups

2 1. On normal and characteristic subgroups. The Frattini argument G1 - 2007-version.

This continues the notes from the courses Matematik 2AL and Matematik 3 AL. (Algebra 2 and Algebra 3). The notes about group theory in Algebra 3 are written in English and are referred to as GT3 in the following. Content of Chapter 1 Deﬁnition of normal and characteristic subgroups (see also GT3, Section 1.12 for details) Aut(G) is the automorphism group of the group G and Auti(G) the subgroup of inner automorphisms of G.

(1A) Theorem: (Important properties of ¢ and char) This is Lemma 1.98 in GT3.

(1B) Remark: See GT3, Exercise 1.99.

If M is a subset of G, then hMi denotes the subgroup of G generated by M. See GT3, Remark 1.22 for an explicit desciption of the elements of hMi.

(1C): Examples and remarks on certain characteristic subgroups. If the subset M of G is “closed under automorphisms of G”, ie.

∀α ∈ Aut(G) ∀m ∈ M : α(m) ∈ M, then hMi char G. Examples of subsets which are closed under automorphisms include (p is a prime number)

•E(G) = {m ∈ G | |m| finite}

•Pp(G) = {m ∈ G | m is a p − element}

•Dp(G) = {m ∈ G | |m|finite, p - |m|} •K(G) = {m ∈ G | ∃ a, b ∈ G : m = [a, b]}.

3 Here [a, b] = aba−1b−1 is the commutator of a and b. Note, that hK(G)i = G0, G’s commutator group. (See GT3). If G is abelian, then E(G) is exactly G’s torsion subgroup GT . In this case we actually have E(G) = hE(G)i. If G is a ﬁnite group with a normal p–Sylow subgroup P , then

P = Pp(G).

(See GT3, Exercise 1.119). This also shows that if P ¢ G is a p–Sylow subgroup, then P char G.

We use some of the above remarks to prove the next theorem. A group G is called elementary abelian, if G is abelian and there is a prime p, s.t. xp = 1 for all x ∈ G. The deﬁnition of a solvable group is given in GT3, Section 1.18..

(1D) Theorem: Let N be a minimal normal subgroup in the ﬁnite solvable group G. Then N is elementary abelian. Proof: ...

4 The Frattini argument If H, P ⊆ G are subgroups, then

−1 NH (P ) = {h ∈ H | hP h = P } is called P ’s normalizer in H. This is a subgroup of H and P ¢ NG(P ). Moreover P ¢ G ⇔ NG(P ) = G. If H and K are subgroups of G, then their “product” HK = H · K is the subset HK = {g ∈ G | ∃h ∈ H, k ∈ K : g = hk}. (Se Chapter 2 and GT3, Section 1.6 for details.)

(1E) Theorem: (Frattini argument) Let G be ﬁnite, H ¢ G. Let P be a p–Sylow subgroup in H. Then

G = HNG(P ).

Proof: See eg. D Gorenstein: Finite groups, Theroem 1.3.7, p. 12

We may use the Frattini argument to prove (1F) Theorem: Let P be a p–Sylow subgroup in G. Let K be a subgroup in G, satisfying NG(P ) ⊆ K. Then K = NG(K). Proof: ...

5 2. On products of subgroups G2 - 2007-version

If H and K are subgroups of G, we deﬁne

HK = H · K = {g ∈ G|∃h ∈ H, k ∈ K : g = hk}.

For results on HK, see GT3, Section 1.5 and 1.6, especially Lemma 1.39 and Theorem 1.40. (2A) Theorem: Let H and K be subgroups of the ﬁnite group G.

(1) The subset HK satisﬁes

|HK| = |H| |K|/|H ∩ K|.

In particular |HK| | |H| |K|.

(2) We have |G : H ∩ K| ≤ |G : H| |G : K|.

(3) If HK is a subgroup of G, we have

|G : H ∩ K| | |G : H| |G : K|.

Proof: (1) See GT3, Lemma 1.39. (2) As HK ⊆ G we have |HK| ≤ |G| and then (1) shows, that

|H| |K|/|H ∩ K| ≤ |G|.

Divide this with |H||K| and multiply by |G| to get (2). (3) is proved analandously, using that we now know that |HK| | |G|, since HK is a subgroup. (2B) Theorem: If H and K are subgroups of subgroups of the ﬁnite group G and if (|G : H|, |G : K|) = 1, then G = HK. Proof: We show |G| | |HK|. As |HK| ≤ |G| we then get |HK| = |G|, ie. G = HK. Let p be a prime, p | |G|. Then p - |G : H| or p - |G : K|, by

6 assumption. We then get that either H or K contains en p–Sylow group P of G. We get |P | | |H| | |HK| or |P | | |K| | |HK| by (2A). In any case |P | | |HK|. As p was arbitrary we get |G| | |HK|, as desired. Deﬁnition: Let |G| = pam,(p, m) = 1. A subgroup M of G with |M| = m is called a p–(Sylow) complement in G.(M is a p0–Hall subgroup. See chapter 3.) The set of p–Sylow subgroups i G is denoted Sylp(G).

(2C) Corollary: If P ∈ Sylp(G) and if M is a p–Sylow complement in G, then G = PM,P ∩ M = {1}.

Proof: Apply (2B) with H = P , K = M to show G = PM. As (|P |, |M|) = 1 we get trivially P ∩ M = {1}. Remarks: 1. A group G need not contain a p–Sylow complement. However if P ∈ Sylp(G), P ¢ G, then G contains a p–Sylow complement. This is a special case of the Schur–Zassenhaus Theorem in Chapter 3. 2. Generally we call the subgroup K of G a complement to the subgroup M, if we have G = MK and M ∩ K = {1}.

(2D) Theorem: Let H and K be subgroups of the ﬁnite group G. If the least common multiple (lcm) of |H| and |K| is |G| then HK = G. Proof: Analandous to (2B).

a1 ar (2E) Theorem: Let |G| = p1 ··· pr , where pi 6= pj, i 6= j are primes. Let I ⊆ {1, . . . , r}, such that for all i ∈ I we have that G has a pi–Sylow T Q aj complement Ki. Then | Ki| = pj . i∈I j∈ /I Proof: Induction by |I|. For |I| = 1 we use the deﬁnition. Assume that the Theorem is shown for subsets I0 ⊆ I, I0 6= I. Let I = I∗ ∪ {k}, k∈ / ∗ ∗ T ∗ ak Q aj I . Let K = Ki. By the induction hypothesis |K | = pk pj . i∈I∗ j∈ /I Q aj ∗ As |Kk| = pj we get, that lcm (|K | , |Kk|) = |G|. By (2A) |G| = j6=k ∗ T |K | |Kk|/| Ki|. The result follows by an easy calculation. i∈I

Here is another application of (2A):

7 (2F) Theorem: Let G be ﬁnite and Q j N ¢ G. Let p be a prime. Then

Q ∈ Sylp(N) ⇔ ∃P ∈ Sylp(G): Q = P ∩ N.

Proof: ⇒ Let Q ∈ Sylp(N). Then in particular Q is a p–subgroup in G. Choose by Sylow’s 2. Theorem, P ∈ Sylp(H), such that Q ⊆ P . Now P ∩ N is a p–subgroup of N, containing Q. As Q is a p–Sylow group in N, we get Q = P ∩ N. ⇐ Let P ∈ Sylp(G), and put Q = P ∩ N. We have by (2A) . |N : Q| = |N : P ∩ N| = |N| |P ∩ N| = |NP |/|P | = |NP : P | which divides |G : P |. As P ∈ Sylp(G) we get p - |G : P |, and therefore p - |N : Q|. As Q is a p–subgroup in N, we get Q ∈ Sylp(N) (2G) Theorem: Let G be ﬁnite, N ¢ G. Assume that K is a p–Sylow complement in G. Then N ∩ K et p–Sylow complement in N, and KN/N is a p–Sylow complement in G/N. Proof: ... We may also use the above results to extend the claims of (1C). The next Theorem is about hDp(G)i in (1C). (2H) Theorem: Let G be ﬁnite, p a prime and D := hg ∈ G | p - |g|i. Then D is a characteristic subgroup of G and D is the smallest normal subgroup in G whose index in G is a power of p.

Proof: By (1C) D char G. In particular D ¢ G. Let P ∈ Sylp(G). We claim DP = G. If q 6= p is a prime and q | |G|, then D contains a q-Sylow subgroup Q of G, since all g ∈ Q satisfy p - |g|. Thus DP = G (eg. by (2D)). Then |G : D| = |P : P ∩ D| a power of p. Assume now that N ¢ G and that |G : N| is a power of p. We want to show D ⊆ N. It suﬃces to show that if g ∈ G, p - |g|, then g ∈ N. Consider the cosetg ¯ = gN as an element in the factor group G¯ = G/N. As |g¯| | |g| and p - |g| we get p - |g¯|. Thus |g¯| has on the one hand an order prime to p, and on the other hand a p-power order, since G¯ has p-power order. We get |g¯| = 1, ie. g ∈ N. The subgroup D in (2H) is denoted Op(G), as mentioned in (1C). It is one of four characteristic subgroups of a ﬁnite group G, associated to a given prime p.

8 Let p be a prime. A p-group is a group of p-power order. A p0-group is a group of order prime til p. Correspondingly we deﬁne p- and p0-elements. The following is only of interest when p | |G|. We deﬁne

• Op(G) is the smallest normal subgroup in G, whose index |G : Op(G)| is a power of p.

• Op(G) is the largest normal subgroup i G, whose order |Op(G)| is a power of p (p-subgroup).

• Op0 (G) is the smallest normal subgroup in G, whose index |G : Op0 (G)| is prime til p.

• Op0 (G) is the largest normal subgroup i G, whose order |Op0 (G)| is prime til p (p0-subgroup).

You of course need to convince yourself that these subgroups are well deﬁned. What is meant by smallest/largest? It could be by with respect to their order (number of elements) or with respect to the inclusion order. As we shall see for these subgroups it does not matter which order relation we choose. Let us mention, that if p - |G|, then the only p-element in |G| is the p p0 identity element 1. In that case O (G) = Op0 (G) = G and Op(G) = O (G) = {1}. How does this ﬁt with Theorem (2M) below? If H and K are subgroups of G and at least one of them is a normal subgroup in G, then HK is again a subgroup, as we have seen. We may then apply all of Theorem (2A). If H and K are both normal, then so is HK and H ∩ K is again normal. If for example |G : H| and |G : K| are both prime to p, then from Theorem (2A)(3) we get, that |G : H ∩ K| is prime to p. Thus also the intersection of all normal subgroups in G, of index prime to p is a normal subgroup with the same property. This subgroup is then Op0 (G). Correspondingly you may argue with the other subgroups. For example HK is a normal p0-subgroup of G, if both H and K are, since |HK| | |H| |K|, by (2A)(1). Thus the product of all such subgroups is normal p0-subgroup, ie. it is Op0 (G). We now have: (2I) Theorem: The subgroup Op(G) may be described in the following equivalent ways

9 p T (1) O (G) = {A¢G| |G:A| isa p−power} A (2) Op(G) = hg ∈ G | g p0 − elementi

p S (3) O (G) = hQ | Q ∈ {q prime 6= p} Sylq(G)i (4) Op(G) = hQ | Q p0 − subgroup of Gi

(2J) Theorem: The subgroup Op0 (G) may be described in the following equivalent ways (1) Op0 (G) = T A {A¢G | p - |G:A|} (2) Op0 (G) = hg ∈ G | g p − elementi

p0 (3) O (G) = hP |P ∈ Sylp(G)i (4) Op0 (G) = hP |P p − subgroup of Gi

(2K) Theorem: The subgroup Op(G) may be described in the following equivalent ways

(1) Op(G) = hA ¢ G |A er p − subgroupi (2) O (G) = T P p P ∈ Sylp(G)

(2L) Theorem: The subgroup Op0 (G) may be described in the following way

0 (1) Op0 (G) = hA ¢ G |A p − subgroupi

Proof of the Theorems: That the groups are described by the property (1) follows from Theorem (2A), as explained above in the cases Op0 (G) and Op0 (G). Proof for (2I): In Theorem (2H) it is shown that (1) and (2) are equivalent. S Let X = hQ | Q ∈ {q prime 6= p} Sylq(G)i. As the elements in each of the generating subgroups Q for X consists of p0-elements, (2) shows that X ⊆ Op(G). On the other hand X ¢ G, as the set of generating subgroups Q is closed under conjugation. As X contains q-Sylow groups of G for all primes q 6= p, no prime q 6= p can divide |G : X|. Thus |G : X| is a power of p and thus Op(G) ⊆ X. It is easily seen that the descriptions in (2) are (4) eqvivalent. (Why?)

10 Proof of (2J): This is an exercise. Proof of (2K): Let X = T P. Assume that A ¢ G is a p-group. P ∈ Sylp(G) Let P ∈ Sylp(G). Then the subgroup AP also has p-power order by (2A). As P er en Sylow group we get AP = P, ie. A ⊆ P. Thus A ⊆ X. By (1) we see that Op(G) ⊆ X. Since the set of p-Sylow groups is closed under conjugation we have X ¢ G. Thus X ⊆ Op(G).

p p0 p p0 You may of course combine O ,O ,Op,Op0 . For example O (O (G)) is Op(H), where H = Op0 (G). This subgroup is also denoted Opp0 (G). You may then continue with Op0pp0 (G), etc. and get smaller subgroups of G. If we at some time reach the triviel subgroup {1}, we call G p-solvable. Analandously you may consider subgroups like Op0p(G), deﬁned as follows: p0 p0 p0 p p Op0p(G)/Op(G) = Op0 (G/Op(G)) Clearly O (O (G)) = O (G) and O (O (G)) = Op(G).

(2M) Theorem:

p p0 (1) We have Op0 (G) ⊆ O (G) and Op(G) ⊆ O (G).

p (2) Op0 (G) = O (G) ⇔ G has a normal p-complement.

p0 (3) Op(G) = O (G) ⇔ G has a normal p-Sylow group.

p Proof: (1) As p - |Op0 (G)| (2I)(4) shows that Op0 (G) ⊆ O (G). The other inclusion in (1) follows from (2J)(4). p (2) ⇒: Assume Op0 (G) = O (G) = K. We claim that K is a normal p- complement in G. Write |G| = pam, with p - m. It follows from (2I)(3), that p 0 m | |O (G)| = |K|. As K also equals Op0 (G), which is a p -group, we have m = |K|. Thus K is a normal p-complement. ⇐: If K is a normal p-complement in G, we easily get K = Op(G) = Op0 (G). a p0 (3) Write again |G| = p m. If Op(G) = O (G) = P, then P is a normal p-Sylow group in G : As |G : Op0 (G)| | m, we have pa | |Op0 (G)| = |P |. a Since P = Op(G) is a p-group, we get |P | = p . Conversely, if P is a normal p0 p-Sylow group in G, then P = Op(G) = O (G). The last part of this chapter is a preparation for later chapters. Deﬁnition: Let M be a subgroup in G. A subset X ⊆ G is called a (left) S˙ transversal to M in G if G = x∈X xM. Thus X contains exactly one element

11 from every coset of M in G. Remark that G = XM and |X| = |G : M|. If M has a complement in G, then this complement is also a transversal to M in G. We later need the following: (2N) Theorem: Let N ⊆ M ⊆ G be subgroups. If X is a transversal for M in G and Y a transversal for N in M, then XY is a transversal for N in G. S˙ S˙ Proof: We have M = y∈Y yN and G = x∈X xM, whence

[˙ [˙ [˙ [˙ G = xM = x( yN) = tN. x∈X x∈X y∈Y t∈XY

You may also consider right transversals to a subgroup of G. X ⊆ G is a S˙ right transversal to the subgroup M ⊆ G if G = x∈X Mx. In analandy with (2N) we have (2O) Theorem: Let N ⊆ M ⊆ G be subgroups. If X is a right transversal for M in G and Y a right transversal for N in M, then YX is a right transversal for N in G. Remarks: 1◦. X is a (left) transversal to M in G ⇔ X−1 is a right transversal to M in G, since we have (xM)−1 = Mx−1. 2◦. If M ¢G, then a right transversal X to M in G is also a left transversal and vice versa. If the subgroup M is not normal in G, then you may always ﬁnd a left transversal til M in G, which is not a right transversal to M in G. (Exercise).

12 3. Hall subgroups and complements G3 - 2007-version

We consider only finite groups in the Chapter. Let G be a finite group. Definition: A subgroup H of G is called a Hall subgroup, if it satisfies

(|H| , |G : H|) = 1 , ie. H’s order is relatively prime to its index in G. (Philip Hall 1904–1982). Clearly any p–Sylow group of a finite group is also a Hall subgroup. Also the trivial subgroups {1} and G of G are Hall subgroups. Let π be an arbitrary set of primes. If n is a natural number we write a1 ak n = p1 ··· pk , where p1, ··· , pk are different primes, and a1, ··· , ak ∈ N. We define Y ai nπ = pi , nπ0 = n/nπ .

{i|pi∈π}

(Thus if n = 60, π = {2, 3}, then nπ = 12, nπ0 = 5.) Remark, that ifπ ˜ is the 0 set of primes, which are not in π, then nπ = nπ˜. We deﬁne n∅ = 1. A π–Hall subgroup in the group G is a subgroup of order |G|π. For example the p–Sylow groups are also π–Hall subgroups, where π = {p}.A π0–Hall subgroup in G is a subgroup of order |G|π0 .

Eksemple: The alternating group A5 has no {3, 5}–Hall subgroup. Such a subgroup should have order 15. Every group of order 15 is cyclic. (GT3). But clearly A5 (or S5) does not contain an element of order 15. (Look at the cycle–structure of such an element). Remark: If M is a π–Hall subgroup of G and there exist also a π0–Hall subgroup N i G, then N is a complement to M in G. (See the previous Chapter). Generally Sylow’s Theorems may thus not be extended to statements about the existence of Hall subgroups. But a famous Theorem of Philip Hall shows that Sylow’s Theorems may be extended to arbitrary Hall subgroups, if we assume that G is solvable, and that they may only be extended in this way, when G is solvable. (3A) Theorem: (P. Hall) Let G be a solvable group. Assume that |G| = nm, where (n, m) = 1. Then:

13 (1) G contains a Hall subgroup of order m;

(2) Any two Hall subgroups of order m i G are conjugate in G;

(3) A subgroup of G, whose order divides m is contained in a Hall subgroup of order m.

Remark: There is also a statement on the number of Hall subgroups of G of order m (corresponding to Sylow’s 3. Theorem.) The statement is that this number is a product of factors ai, where for each ai there exists a prime divisor pi|m, st. ai ≡ 1 (mod pi) ,. We do not prove this although it is not difficult. Proof of (3A): See eg. M. Hall: Theory of groups, p. 141-143 Now we show that the statements of Theorem (3A) are only fulfilled solvable groups. The proof of this is based on the following Theorem, which may be proved using representation theory. There is also a purely group theoretic proof for the Theorem, which is quite complicated and we omit it. (3B) Theorem: (Burnside) Let G be a finite group of order paqb, where p and q are primes. Then G is solvable. Proof: Omitted! We need also the following result. (3C) Lemma: Assume that |G| = paqbm, where p and q are different primes, a, b ∈ N and (p, m) = (q, m) = 1. Assume in addition that G contains subgroups H, A and B, satisfying (1) |H| = paqb;

(2) A 6= G, |G : A| | pa;

(3) B 6= G, |G : B| | qb. Then G is not simple. Proof ... We can now show (3D) Theorem: Assume that the ﬁnite group G has a p-complement for all primes p with p | |G|. Then G is solvable.

14 Remark: A p–complement is a π0–Hall subgroup, where π = {p} and is also called a “p0–Hall subgroup” (like a p–Sylow group is a “p–Hall subgroup”). Thus (3D) has the following consequence. (3E) Corollary: Assume the the group G has en Hall subgroup of order m for enhver factorization |G| = mn, (m, n) = 1 of G’s order. Then G is solvable. Proof of (3D): See eg. M. Hall: Theory of groups, p. 144-145 By Hall’s Theorem a group G contains only π–Hall subgroups for all choices of the set of primes π, when G is solvable. But a non-solvable group may under certain conditions contain Hall subgroups for special choices of π. An example of this is Burnside’s Theorem on the existence of a normalt p–complement (shown in Chapter 6). Here we prove the Schur–Zassenhaus Theorem, stating that if G has a normal π–Hall subgroup, then this has complement (not necessarily normal) which is then a π0–Hall subgroup in G. First we show the Theorem under a stronger assumption. It should be mentioned that the method of proof for (3F) and the later Theorem (3I) er “cohomolandical” and the map t in the proofs is a “2- cocycle”. You may ﬁnd much more about cohomolandy theory for groups in many books, for example B. Huppert: Endliche Gruppen I. (3F) Theorem: (Schur) Let M be en normal abelian Hall subgroup i G. Then there exists a complement N til M i G. Proof: See eg. D Gorenstein: Finite groups p. 221-224 Next we show that the condition of M being abelian is not necessary in (3F), and get the Schur–Zassenhaus–Theorem. (Issai Schur 1875–1941, Hans Zassenhaus 1912–1991. Schur proved (3F) which ie really the hard part, and Zassenhaus then proved (3G) on the basis of (3F)). Theorem (3G): (Schur–Zassenhaus Theorem) Let M be a normal Hall subgroup in G of order m. Then there exists a complement N to M in G. Proof: See eg. D Gorenstein: Finite groups p. 221-224

(3H) Corollary: If P ∈ Sylp(G), then P has complement K in NG(P ), such that NG(P ) = PK.

(NG(P ) is a semidirect product of P and K. See the next Chapter). Proof: P is a normal Hall subgroup in NG(P ). Apply (3G).

15 Remark: In continuation of (3G) you my ask whether two complements to M in G are conjugate in G. This is the case, but we cannot prove it here. You may prove without much difficulty that if M ¢ G is a Hall subgroup, and if either M or G/M is solvable, then all complementer to M in G are conjugate in G. (See for example Theorem 6.2.1 i D. Gorensteins book Finite Groups). In 1963 Feit and Thompson proved that any finite group of odd order is solvable. But if M ¢ G is a Hall subgroup, then (|G : M|, |M|) = 1 and thus |G : M| or |M| must be odd. Thus according to Feit–Thompson, either G/M or M be must be solvable. Feit–Thompson’s proof is more than 250 pages long (Pacific Journal of Mathematics (1963)). There is a monandraph which treats an essential part of the proof (H. Bender–G. Glauberman: Local analysis for the odd–order theorem). The next quite remarkable theorem does not appear to have anything to do with Theorem (3F), but the proofs are quite similar. They have the existence of a normal abelian subgroup of G in common and both are contained in a more general Theorem of Gaschütz, see for example Hauptsatz 17.4 i Huppert: Endliche Gruppen I, p. 121. (3I) Theorem: (Gaschütz) Let M be a normal abelian p-subgroup in G. Let P be a p-Sylow group in G. The following statements are equivalent

(i) M has a complement in P

(ii) M has a complement in G.

Proof: See Kurzweil-Stellmacher: The theory of ﬁnite groups, p. 73-76.

16 4. Semidirect products - theory and examples G4 - 2007-version

This chapter extends GT3, Chapter 1.19 considerably. Let G be a group, and A a subgroup of then automorphism group Aut(G). We form a new group called G o A, (the semidirect product of G with/by A). The underlying set is G × A. The composition is deﬁned by

(g, ϕ)(g0, ψ) = (gϕ(g0), ϕψ) for g, g0 ∈ G, ϕ, ψ ∈ A. It is easily calculated that the associative law is fulﬁlled. The neutral/unit element is (1, 1), and the inverse element to (g, ϕ) is (ϕ−1(g−1), ϕ−1), since

(g, ϕ)(ϕ−1(g−1), ϕ−1) = (gϕ(ϕ−1(g−1)), ϕϕ−1) = (gg−1, ϕϕ−1) = (1, 1).

In particular G o Aut(G) is called the holomorph of G and denoted Hol(G). We are not going to discuss the holomorph of groups here. Assume that G ans H are groups, and that there exists a homomorphism α : H → Aut(G). We form a new group called GoαH (the semidirect product of G by H, relative to α). The underlying set is G×H, and the composition is deﬁned by (g, h)(g0, h0) = (gα(h)(g0), hh0). (You may here consider “conjugation of g0 by h”, ie. hg0h−1 in this group is being replaced by α(h)(g0). (For elements in an arbitrary group we know that ghg0h0 = g(hg0h−1)hh0.) In G oα H again (1, 1) is the neutral element, and the inverse element to (g, h) is (α(h−1)(g−1), h−1), since

(g, h)(α(h−1)(g−1), h−1) = (gα(h)(α(h−1)(g−1)), hh−1)

= (g(α(h)α(h−1))(g−1), hh−1) = (gα(1)(g−1), hh−1) = (gg−1, hh−1) = (1, 1). We used here that α is a homomorphism.

Special cases: (1) If H ⊆ Aut(G) and α is the embedding of H in Aut(G), them the two above constructions coincide. Therefore the ﬁrst is a special case of the second.

17 (2) If α : H → Aut(G) is deﬁned by α(h) = 1 (the identity), then G oα H = G × H is the usual direct product.

If α is clear from the context we often just write GoH instead of Goα H. You may want to realize a given group as a semidirect product: On the one hand we have: If X = G oα H is a semidirect product, then G∗ = {(g, 1) | g ∈ G},H0 = {(1, h) | h ∈ H} are subgroups of X. The maps

g 7→ g∗ = (g, 1) , h 7→ h0 = (1, h) are obviously isomorphisms between G and G∗ and between H and H0. Clearly X = G∗H0 and G∗ ∩ H0 = {(1, 1)}. Furthermore G∗ ¢ X, which is easily seen from the multiplication formula and the formula for inverse elements. If g∗ = (g, 1) ∈ G∗ and h0 = (1, h) ∈ H0, then h0g∗(h0)−1 = (1, h)(g, 1)(1, h−1) = (α(h)(g), h)(1, h−1) = (α(h)(g), 1) = (α(h)(g))∗

On the other hand we may consider the following situation: Assume that Y is a group with subgroups G and H, which satisfy

G ¢ Y,Y = GH.

Then Y may be “connected” with a semidirect product of G by H: For h ∈ H we let α(h) be the restriction of the inner automorphism κh of Y to G. We thus have α(h)(g) = hgh−1 for h ∈ H, g ∈ G. Then α is a homomorphism from H to Aut(G), and we may therefore form X = G oα H. How are X and Y related? Here is the answer:

(4A) Theorem: Let Y = GH and X = G oα H be as above. Then ρ(g, h) = gh deﬁnes a surjective group homomorphism from X to Y . We have

G ∩ H = {1} ⇔ ρ is an isomorphism.

18 Proof: Let g, g0 ∈ G, h, h0 ∈ H. We have

ρ((g, h)(g0, h0)) = ρ(gα(h)(g0), hh0) (multiplication in X)

= ρ(g(hg0h−1), hh0) (definition of α(h)) = g(hg0h−1)hh0 (definition of ρ) = ghg0h0 (cancel h−1h) = ρ(g, h)ρ(g0, h0) (definition of ρ) Thus ρ is a homomorphism, and since Y = GH it is clear that ρ is surjective. If G ∩ H = {1} we see that

ρ(g, h) = 1 ⇔ gh = 1 ⇒ g = h−1 ∈ G ∩ H = {1} ⇒ g = h = 1 , so that ρ is injective in this case. If G ∩ H 6= {1} then ρ(x, x−1) = 1 when x 6= 1, x ∈ G ∩ H, so that ρ is not injective. The above Theorem shows that when G ∩ H = {1}, then Y yields an “inner” characterization of et semidirect product. An example of a group Y realized as a semidirect product, is

Y = Sn,G = An,H = h(1, 2)i n ≥ 2.

Another example we have seen is the result of (3G). If G has a normal Hall subgroup M, then there exists a complement L to M in G. Thus G is a semidirect product of M with L. Let us consider various other examples. (4B) Example: Dihedral groups. If G = hgi is a cyclic group, then the map ι : g 7→ g−1 is an automorphism of G. In this situation G o hιi is a dihedral group. If |G| = n ≤ ∞, we denote G o hιi by Dn. We then have that |Dn| = 2n. The group Dn is generated by two elements (g, 1) and (1, ι). Let us remark that |(1, ι)| = |(g, ι)| = 2 since (g, ι)2 = (gι(g), 1) = (gg−1, 1) = 1. As Dn is generated by (g, ι) and (1, ι), we see that a dihedral group is generated by 2 elements of order 2 (so-called “involutions”). On the other hand a group generated by 2 involutions is isomorphic to a dihedral group. This is seen as follows. Assume that D = hx, yi, where x2 = y2 = 1, x 6= y. We then have that x−1 = x and y−1 = y. Put g = xy.

19 Then G := hgi is cyclic and D = hg, yi. Furthermore ygy−1 = yxyy−1 = yx = y−1x−1 = (xy)−1 = g−1. Thus conjugation by y correponds to the map ι above. When n ∈ N, n ≥ 3, then Dn may be realized as a subgroup of Sn, if we consider the subgroups

∗ Dn = h(1, 2, ··· , n), τ = (1, n)(2, n − 1) · · ·i

∗ ∗ of Sn. If G = h(1, 2, ··· , n)i, H = hτi then Dn = GH and G ¢ Dn, G ∩ H = {1}, since

τ(1, 2, ··· , n)τ −1 = (n, n − 1, ··· , 2, 1) = (1, 2, ··· , n)−1.

∼ ∗ ∗ ∼ It is the clear that Dn = Dn (use (4A)). We also have that D3 = S3, as they ∗ have the same order. Let us remark that for n = 4 we have |D4| = 8, so that ∗ D4 is a 2–Sylow group in S4, (and also in S5. Why?). The definition may be extended to the case where G is an abelian group. The map ι : g 7→ g−1,G → G, is still an automorphism of G, so you may form a “di-abelian” group G o hιi of order 2|G|. In the following we do not distinguish between the abstract group Dn and the concrete permutation ∗ group Dn. (4C) Example: Permutation matrices and monomial groups. If R is a commutative ring with unit element 1, n ∈ N, then the set of invertible n × n–matrices with coefficients from R form a group called n GL(n, R) (= {A ∈ Rn | det A invertible in R}). When π ∈ Sn we define a n matrix P (π) ∈ Rn by

P (π) = [aij] where aij = δiπ(j)

(δ is “Kronecker delta”). Clearly P (π) has exactly one element 6= 0 in each column and in each row. Repeated application of the column rule for deter- minants show that det P (π) = ± 1, so that P (π) ∈ GL(n, R). If π, ρ ∈ Sn we have P (π)P (ρ) = P (πρ): Let P (π)P (ρ) = [cij]; then X cij = δiπ(k)δkρ(j) 6= 0 ⇔ k There exists a k such that k = ρ(j) and π(k) = i ⇔ πρ(j) = i,

20 ie. cij = δiπρ(j). This means that P is a homomorphism from Sn to GL(n, R). Clearly P is injective so we may consider Sn as a subgroup of GL(n, R). A matrix on the form P (π), π ∈ Sn is called a (n × n−) permutation matrix. We have:

det P (π) = sign(π), (π0s sign)

t −1 P (π) = P (π ) for all π ∈ Sn. Both claims are proved by expressing π as a product of transpositions (ie. permutations on the form (i, j)): The permutation matrix P ((i, j)) is ob- tained from the unit matrix En by interchanging the i’th and the j’te column. Thus det P (τ) = −1, when τ is a transposition. Clearly we also have t P (τ) = P (τ) , when τ is a transposition. If π = τ1τ2 ··· τk, where all τi are transpositions then

n Y k det(P (π)) = det(P (τi)) = (−1) = sign(π) i=1 and t t t t P (π) = [P (τ1) ··· P (τk)] = P (τk) ··· P (τi) −1 = P (τk) ··· P (τ1) = P (τk ··· τ1) = P (π ). A permutation matrix consists of zeros except exactly one 1 in each row and each column. You may now replace the numbers 1 in a permutation matrix P (π), π ∈ Sn by n elements from a given group G. If g1, g2, . . . , gn ∈ G, π ∈ Sn we put P (g1, . . . , gn; π) = (δiπ(j)gi). This is of course no longer an element in GL(n, R), but a “formal” matrix with elements from the set G ∪ {0} (when we deﬁne that δiig = g, δijg = 0 for i 6= j). If then G is a group and A a subgroup of Sn we put

Mon(G, A) = {P (g1, ··· , gn; π) | gi ∈ G π ∈ A}, a set of “G–monomial” matrices. If in addition we deﬁne that 0 + g = g + 0 = g for g ∈ G, then G’s composition and the usual rule for matrix multiplication induces a composition on Mon(G, A). If we “multiply” the matrices

P (g1, ··· , gn; π) and P (h1, ··· , hn; ρ)

21 and use the rules mentioned above we get a matrix [cij], where X cij = δiπ(k)giδkρ(j)hk = δiπρ(j)gihρ(j) = δiπρ(j)gihπ−1(i) k so that

P (g1, ··· , gn; π)P (h1, ··· , hn; ρ) = P (g1hπ−1(1), ··· , gnhπ−1(n); πρ).

Using this “matrix multiplication” Mon(G, A) becomes a group with P (1, ··· , 1; (1)) as neutral element and where

P (g1, ··· , gn; π) has −1 −1 −1 P (gπ(1), ··· , gπ(n); π ) as inverse element. We call Mon(G, A) a (G–)monomial group. Now Mon(G, A) is an “inner” semidirect product of the normal subgroup

∗ G = {P (g1, ··· , gn; (1)) | gi ∈ G, i = 1, 2, ··· , n},

(which is isomorphic to G × · · · × G) with the subgroup A∗ = {P (1, ··· , 1; π) | | {z } n π ∈ A} (which is isomorphic to A.)

(4D) Example: Let us look at the monomial group Mon(G, A) (as a semidirect product) from the“outside”. ∗ n If A ⊆ Sn and G = G × · · · × G (= G ), we may deﬁne a homomorphism | {z } n α : A → Aut(G∗) by

α(π)(g1, ··· , gn) = (gπ−1(1), ··· , gπ−1(n)).

It may look odd that the map β : A → Aut(G∗) given by

β(π)(g1, ··· , gn) = (gπ(1), ··· , gπ(n)) is not a homomorphism. The connection between α and β is that α(π) = β(π−1), so that if one of the maps is a homomorphism, then the other is an antihomomorphism. That α is a homomorphis is seen as follows:

22 Assume that π, ρ ∈ A. Let

α(ρ)(g1, ··· , gn) = (h1, ··· , hn)

α(π)(h1, ··· , hn) = (k1, ··· , kn).

By deﬁnition hi = gρ−1(i) and ki = hπ−1(i) for i = 1, ··· , n. We then get that ki = hπ−1(i) = gρ−1(π−1(i)) = g(πρ)−1(i). Thus

α(π) ◦ α(ρ)(g1, ··· , gn) = (k1, ··· , kn)

= (g(πρ)−1(1), ··· , g(πρ)−1(n)) = α(πρ)(g1, ··· , gn), ie. α(π)◦α(ρ) = α(πρ). Only when A is abelian, β will be a homomorphism. We may now deﬁne an isomorphism ϕ

ϕ ∗ ∼ G oα A = Mon(G, A)

∗ by ϕ(g1, ··· , gn; π) = P (g1, ··· , gn; π). The multiplication in G o A is

(g1, ··· , gn; π)(h1, ··· , hn; ρ) = (g1hπ−1(1), ··· , gnhπ−1(n); πρ).

It is necessary but not very “nice” that you have to apply π−1 on the indices of the hi’s. This may be avoided by “interchanging ” G and A, as we do in the next example.

(4E) Example: (Wreath product, Kransprodukt). As in (4D) A ⊆ Sn and G∗ = G × · · · × G . We deﬁne a composition on A × G∗ by | {z } n

(π; g1, ··· , gn)(ρ; h1, ··· , hn) = (πρ; gρ(1)h1, ··· , gρ(n)hn).

∗ Then A×G is a group with ((1); 1, ··· , 1) as a neutral element, and (π; g1, ··· , gn) −1 −1 has (π ; h1, ··· , hn) as inverse element, where hi = gπ−1(i). This group is called the wreath productet of G by A, and denoted G o A. Now G o A may also by realized by “monomial” matrices, and

G o A ' Mon(G, A).

If (π; g1, ··· , gn) ∈ G o A we put

∗ P (π; g1, ··· , gn) = (δiπ(j)gj).

23 The only diﬀerence between this and P (g1, ··· , gn; π) is that “gi” is replaced ∗ ∗ by “gj”. If we “multiply” P (π; g1, ··· , gn) and P (ρ; h1, ··· , hn) we get ∗ ∗ ∗ P (πρ; gρ(1)h1, ··· , gρ(n)hn) so that P –matrices form a group Mon (G, A), which is isomorphic to G o A. We have now using G and A constructed four groups, two “matrix groups” ∗ ∗ Mon(G, A) and Mon (G, A), and two “abstract” groups G oα A (i (4D)) and G o A here. We have shown that

∗ ∗ G oα A ' Mon(G, A) and at G o A ' Mon (G, A). To complete the picture we want to show that

∗ G oα A ' G o A. First a Remark.

(4F) Remark: The“opposite” group. If G ia an arbitrary group we may form its opposite group Gop, as follows. The underlying set is G’s elements, and the composition ◦ i Gop, is given by

g ◦ h = hg

(where we have used the composition in G on the right hand side!) It is clear that Gop is a group with the same neutral element and same inverse elements. Moreover the map ι : g 7→ g−1 is an isomorphism between G and Gop, since

ι(gh) = (gh)−1 = h−1g−1 = ι(h)ι(g) = ι(g) ◦ ι(h).

∗ (4G) Theorem: Let G o A and G oα A be as before. By

−1 −1 −1 ψ :(π; g1, ··· , gn) 7→ (g1 , ··· , gn ; π )

∗ op is deﬁned an isomorphism between G o A and (G oα A) . Thus we also have

∗ G o A ' G oα A.

Proof: We have

−1 −1 −1 −1 −1 −1 ψ(π; g1, ··· , gn) ◦ ψ(ρ; h1, ··· , hn) = (h1 , ··· , hn ; ρ )(g1 , ··· , gn , π )

24 −1 −1 −1 −1 −1 −1 = (h1 gρ(1), ··· , hn gρ(n); ρ π ) = ψ(πρ; gρ(1)h1, ··· , gρ(n)hn) =

ψ((π; g1, ··· , gn)(ρ; h1, ··· , hn)). It is easy to ﬁnd wreath products as subgroups in symmetric groups:

(4H) Remark: If G ⊆ Sm and A ⊆ Sn, then G o A is (isomorphic to) a subgroup of Smn.

Proof: Let (π; g1, ··· , gn) ∈ G, and consider

∗ P (π; g1, ··· , gn) = [δiπ(j)gj].

∗ If replace gj by P (gj), then P (π; P (g1), ··· ,P (gn)) becomes a mn × mn– permutation matrix.

(4I) Example: (of (4H)) Let m = 3, n = 2, π = (1, 2, 3) ∈ S3, g1 = (1, 2), ∗ g2 = (1), g3 = (1, 2) ∈ S2 such that P (π; P (g1), P (g2), P (g3) should be a 6 × 6-permutation matrix. Let us calculate this and the correponding permutation. First we consider the permutation matrix P (π)

 0 0 1  P (π) =  1 0 0  0 1 0 by (4C). In this matrix we replace the 1 in the j’th column by the 2×2-matrix P (gj) j = 1, 2, 3 and the 0’s by 2 × 2-zero matrices

 0 0 0 0 0 1   0 0 0 0 1 0    ∗  0 1 0 0 0 0  P (π; P (g1),P (g2),P (g3)) =    1 0 0 0 0 0     0 0 1 0 0 0  0 0 0 1 0 0

This is a 6 × 6-permutation matrix. The corresponding permutation ρ is calculated by considering the position of the 1 in the various columns. We get 1 2 3 4 5 6 ρ = 4 3 5 6 2 1

25 (Here for example ρ(1) = 4 because the 1 in the ﬁrst column is in row 4). Thus ρ = (1, 4, 6)(2, 3, 5).

Another way of calculating ρ is as follows: Consider g1 as a permutation of {1, 2}, g2 as a permutation of {3, 4} and g3 as a permutation of {5, 6}. Then g1g2g3 = (1, 2)(3)(4)(5, 6) = (1, 2)(5, 6). Next you consider π = (1, 2, 3) as a permutation ∆(π) of 6 which permutes the sets {1, 2}, {3, 4}, {5, 6} by 1 → 3 → 5 → 1 and 2 → 4 → 6 → 2, ie.

∆(π) = (1, 3, 5)(2, 4, 6).

For the product ∆(π)g1g2g3 we then get

(1, 3, 5)(2, 4, 6)(1, 2)(5, 6) = (1, 4, 6)(2, 3, 5), the same permutation ρ as before! (Writing ∆ in front of π, ie. ∆(π) means intuitively that we are dealing with a “duplicartion” of π. If ρ = (1, 3) ∈ S3 then ∆(ρ) = (1, 5)(2, 6).)

n (4J) Remark: If G is ﬁnite, and A ⊆ Sn, then |G o A| = |G| |A|.

(4K) Exaeple: (Sylow groups in the symmetric groups Spa ). Let p be a prime. If n ∈ N we let νp(n) be the largest non negative ν (n) 2 number such that p p | n. For example ν3(72) = 2 as 72 = 3 · 8. Let a us consider νp(|Spa |), a ≥ 1. We have νp(|(Spa |) = νp(p !). It is clear that pa−1 among the numbers 1, ··· , pa are divisible by p, namely p, 2p, ··· , pa−1p. Moreover pa−2 are divisible by p2 (if a ≥ 2), namely p2, 2p2, ··· , pa−2p2. By continuing with p3 etc. we get

a a−1 a−2 a νp(p !) = p + p + ··· + 1 = (p − 1)/(p − 1).

For a = 1, a p–Sylow group in Sp is cyclic, generated by for example p+1 (1, 2, ··· , p), ie. ' Zp. By (4J) |Zp o Zp| = p . Thus Zp o Zp has the same order as a p–Sylow group in Sp2 . This is generalized: Let us induc- tively deﬁne the group Xa by X1 = Zp, Xa = Xa−1 o Zp. Using induction over a we see that Xa is isomorphic to a subgroup of Spa (use (4H)) and a that νp|Xa| = νp(|Spa |) = νp(p !) (use (4J)), such that the p–Sylow group of Spa is an iterated wreath product of a cyklic groups of order p. Let us

26 consider a concrete realization of the p–Sylow group i Sp2 . This group has an elementary abelian subgroup of order pp, namely

h(1, 2, ··· , p)i × h(p + 1, ··· , 2p)i × · · · × h(··· , p2)i.

This corresponds to the subgroup G × · · · × G in the generel case. The group A becomes in this example the group generated by h(1, p + 1, 2p + 1, ··· , (p − 1)p + 1)(2, p + 2, ··· , (p − 1)p + 2) ··· (p, 2p, ··· , p2)i

In the case p = 2 the 2–Sylow group of S4 is generated by (1, 2) and (1, 3)(2, 4). The group “G×G” becomes h(1, 2)i×h(3, 4)i and A = h(1, 3)(2, 4)i. If we go to S8 its 2–Sylow group is generated by

(1, 2), (1, 3)(2, 4) and (1, 5)(2, 6)(3, 7)(4, 8).

Think about this! How does it look in S16? Notice that each of these elements is a “duplication” of the previous one in the same way as in (4I)!

(4L) Remark: You can show that the p–Sylow group in Sn, n arbitrary, may be described as follows: Write n “p–adically”, ie.

k n = a0 + a1, p + ··· + akp , where 0 ≤ ai ≤ p − 1.

Then in Sn the p–Sylow group is isomorphic to

a1 a2 ak X1 × X2 × · · · × Xk ,

ai where Xi = Xi × · · · × Xi and Xi is as in the previous example. | {z } ai

We remind about the deﬁnition of centralizer (GT3). If a ∈ G, then a’s centralizer in G is the subgroup

−1 CG(a) = {g ∈ G |gag = a}.

Correpondingly CG(X) is deﬁned for a subset X of G.

(4M) Example: The wreath product also plays a rˆolein the description of centralizers of elements in symmetric groups. We consider only the case of

27 “homocyclic” elements, ie. elements which is a product of cycles of the same length. Let k, ` ∈ N and assume that κ ∈ Sk` is a product of k disjoint cycles of the same length `. We explain that ∼ CSk` (κ) = Z` o Sk, where Z` ´ısa cyclic group of order `. We assume that

κ = (a11, a12, ··· , a1`)(a21, a22, ··· , a2`) ··· (ak1, ak2, ··· , ak`), where the aij’s are diﬀerent numbers between 1 and k`. When ϕ ∈ Sk`

−1 ϕκϕ = (ϕ(a11), ϕ(a12), ··· , ϕ(a1`)) ··· (ϕ(ak1), ϕ(ak2), ··· , ϕ(ak`)).

Therefore ϕ ∈ C(κ) = CSk` (κ) if and only if the sets of cycles

{(a11, a12, ··· , a1`), ··· , (ak1, ak2, ··· , ak`)} and

{(ϕ(a11), ϕ(a12), ··· , ϕ(a1`)), ··· , (ϕ(ak1), ϕ(ak2), ··· , ϕ(ak`))} are identical. This means that if ϕ ∈ C(κ) and we know ϕ(ai1), then also ϕ(ai2), ··· , ϕ(ai`) are determined since the order in cycles must be respected: 0 0 If ϕ(ai1) = ai0j0 where 1 ≤ i ≤ k and 1 ≤ j ≤ `, then also

(∗) ϕ(aij) = ai0(j0+j−1) for 1 ≤ j ≤ `, where the second index is calculated modulo `. An element ϕ ∈ C(κ) is thus completely determined by

ϕ(a11), ϕ(a21), ··· , ϕ(ak1).

As a11, a21, ··· , ak1 are all in different cycles i κ, then also ϕ(a11), ϕ(a21), ··· , ϕ(ak1) must be in different cycles. On the other hand every choice of ϕ(a11), ϕ(a21), ··· , ϕ(a`1) in different cycles yields an element in C(κ) using (∗) above. Given π ∈ Sk we may in particular define its “duplication element” ∆(π) ∈ C(κ) by (∗∗) ∆(π)ai1 = aπ(i)1 1 ≤ i ≤ k.

28 (By (∗) we also have ∆(π)aij = aπ(i)j for all i, j). Let now ϕ ∈ C(κ) be determined by

(∗ ∗ ∗) ϕ(ai1) = aπ(i)ti for 1 ≤ i ≤ k.

Here 1 ≤ ti ≤ ` for all i. As ϕ(a11), ··· , ϕ(ak1) are diﬀerent cycles, π is a −1 permutation of 1, ··· , k, ie. π ∈ Sk. By deﬁnition of ∆(π ) we then get from (∗∗) and (∗ ∗ ∗) −1 ∆(π )ϕ(ai1) = aiti . −1 −1 It is clear that ∆(π ) = ∆(π) , since ∆ is a homomorphism Sk → C(κ). Let us put ψ = ∆(π−1)ϕ = ∆(π)−1ϕ. We then have

ψ(ai1) = aiti for 1 ≤ i ≤ k.

Let us denote the cycles in κ by z1, ··· , zk, ie.

zi = (ai1, ai2, ··· , ai`).

As disjoint cycles are permutable it is clear that zi ∈ C(κ) for all i. Thus also the element ψ0 deﬁned by

0 t1−1 t2−1 tk−1 ψ = z1 z2 ··· zk ∈ C(κ). Now for 1 ≤ i ≤ k

0 ti−1 ψ (ai1) = zi (ai1)u (why?)

= aiti (why??) We conclude that ψ = ψ0 and get that

t1−1 tk−1 ϕ = ∆(π) · z1 ··· zk . If we deﬁne a map α : Zk o Sk → C(κ) by s1 sk α(π; s1, ··· , sk) = ∆(π) · z1 ··· zk , where the si’s are calculated modulo ` = |zi|, then α is surjective by the above. It is easy to see that α is a homomorphism, and that the kernel of α is triviel. Thus α is an isomorphism.

In conclusion we can say that if you want to consider GoA, where G ⊆ Sm ∗ and A ⊆ Sn as a subgroup of Smn, you let the n copies of G in G = G×· · ·×G (n gange) operate on pairwise disjoint subsets of {1, ··· , mn}, where each of these subsets have m elements. By “duplication” the elements of A are blown up to permute the n disjoint subsets, on which the G’s operate.

29 On subgroups in a direct product

We describe a method so that you may in principle determine all subgroups of the direct product of two groups. This parametrization of the subgroups is usually not found in textbooks on group theory although it is fairly simple. Let G and H be groups and X = G × H the direct product of G and H. Consider the following set of 5-tuples:

U(G, H) = {(G1,G2,H1,H2, ϕ)} where G2 ¢ G1 ⊆ G are subgroups in G, H2 ¢ H1 ⊆ H subgroups in H, and ϕ is a group isomorphism ϕ : G1/G2 → H1/H2. If T = (G1,G2,H1,H2, ϕ) ∈ U(G, H) we put

UT = {(g, h) ∈ G × H | g ∈ G1, h ∈ H1 and ϕ(gG2) = hH2}.

Here we consider gG2 ( hH2) as an element in G1/G2 (H1/H2).

(4N) Theorem: The map T → UT is a bijection between the sets U(G, H) and the set of subgroups of G × H.

Proof: We ﬁrst remark that if T ∈ U(G, H), then UT is a subgroup of G×H: −1 −1 If (g, h), (g1, h1) ∈ UT we have ϕ(gG2) = hH2 and ϕ(g1 G2) = h1 H2, since ϕ is a homomorphism. Thus we get that

−1 −1 −1 −1 ϕ(gg1 G2) = ϕ(gG2)ϕ(g1 G2) = hH2h1 H2 = hh1 H2

−1 −1 −1 ie. (gg1 , hh1 ) = (g, h)(g1, h1) ∈ UT . It is also clear that if T,T 0 ∈ U(G, H) and T 6= T 0 (ie. if at least one of 0 the ﬁve “coordinates” in T and T is diﬀerent ), then UT 6= UT 0 . Let now U be a subgroup of G × H. We show that there exists T ∈ U(G, H), such that U = UT . Put

G1 = {g ∈ G | There exists h ∈ H, s˚a g, h) ∈ U}

G2 = {g ∈ G | (g, 1) ∈ U}

H1 = {h ∈ H | There exists g ∈ G, s˚a(g, h) ∈ U}

H2 = {h ∈ H | (1, h) ∈ U}.

30 It is clear that G1,G2 are subgroups of G, and H1,H2 are subgroups of H and G2 ⊆ G1, H2 ⊆ H1. Assume now that h ∈ H1, x ∈ H2. Choose g ∈ G, such that (g, h) ∈ U. We also have (1, x) ∈ U so that

(g, h)(1, x)(g, h)−1 = (1, hxh−1) ∈ U,

−1 ie. hxh ∈ H2. Thus H2 ¢ H1 (and analogously G2 ¢ G1). Assume that g ∈ G, and that h, h1 ∈ H both satisfy (g, h) ∈ U,(g, h1) ∈ U. By definition of H1 we get h, h1 ∈ H1. Furthermore g ∈ G1. We have −1 −1 −1 (g, h) (g, h1) = (1, h h1) ∈ U so that h h1 ∈ H2. Thus hH2 = h1H2. We see that ψ : g 7→ hH defines a map from G1 → H1/H2. It is clear that this map is a homomorphism. If x ∈ ker(ψ), there exists a h2 ∈ H2, such that (x, h2) ∈ U. As (1, h2) ∈ U (because h2 ∈ H2) we get (x, 1) ∈ U, ie. x ∈ G2. It is also easily seen that ψ is surjective. By the first isomorphism Theorem for groups ψ induces an isomorphism ϕ : G1/G2 → H1/H2 and we then get that U = UT , where

T = (G1,G2,H1,H2, ϕ) ∈ U(G, H)

(4O) Remark: Let (G1,G2,H1,H2, ϕ) = T ∈ U(G, H) as above. We than have |UT | = |G1||H2| = |G2||H1|, if the groups are ﬁnite. (Why?)

(4P) Remark: A special class of subgroups of G × H are those on the forme G1 × H1, G1 subgroup in G, H1 subgroup in H. The corresponding T ∈ U(G, H) is then (G1,G1,H1,H1, 1).

A generalization of (4N) to a direct product of three or more subgroups is much more complicated than you might expect!

31 5. Subgroups and Verlagerung G5 - 2007-version

Some parts of this Chapter are also discussed in GT3, Chapter 1.16. (5A) Remark: Let us consider a wreath product G o A as in (4E). The map

(π; g1, ··· , gn) 7→ π is a en homomorphism from G o A onto A with G∗ as a kernel. But generally none of the following maps are homomorphisms.

(π; g1, ··· , gn) 7→ gi (G o A → G)

∗ (π; g1, ··· , gn) 7→ (g1, ··· , gn)(G o A → G )

(π; g1, ··· , gn) 7→ g1g2 ··· gn (G o A → G). The ﬁrst two maps may only be homomorphisms when A = {(1)} which is uninteresting. However the third map may be a homomorphism, when G is abelian. This may be generalized as follows: Assume that N ¢ G, and that G/N is abelian. Consider the map

pG/N : G o A → G/N deﬁned by pG/N (π; g1, ··· , gn) = g1g2 ··· gnN.

Then pG/N is a surjective homomorphism. We remark that

g1 ··· gnN = (g1N) ··· (gnN) , and as G/N is abelian, we get for each ρ ∈ Sn that

g1N ··· gnN = gρ(1)N ··· gρ(n)N = gρ(1) ··· gρ(n)N.

Therefore pG/N ((π; g1, ··· , gn)(ρ; h1, ··· , hn))

= pG/N (πρ; gρ(1)h1, ··· , gρ(n)hn)

= gρ(1)h1 ··· gρ(n)hnN

32 = (g1 ··· gnN)(h1 ··· hnN) (see above)

= pG/N (π; g1, ··· , gn)pG/N (ρ; h1, ··· , hn) .

(5B) Notation: Let H be a subgroup in the group G of ﬁnite index |G : H| = n. Let T = {g1, ··· , gn} be a transversal til H i G (See Chapter 2), • [ (∗) G = giH.

gi∈T Let x ∈ G. By (∗) there exists for 1 ≤ i ≤ n a uniquely determined T T integer αx (i) ∈ {1, ··· , n}, and a uniquely determined element hxi ∈ H such that T xg = g T h . i αx (i) xi 0 Let us try to replace T by another transversal T = {g1eh1, ··· , gnehn} T T 0 T T 0 where ehi ∈ H, and for x ∈ G compare αx with αx , and hxi with hxi . We have (5C) Lemma: In the above notation we have

T T 0 αx = αx .

(Thus we put, independently of the choice of T ,

T αx = αx).

Moreover hT 0 = h−1 hT h . xi eαx(i) xiei

Proof: If xgi ∈ gjH we also have that xgiehi ∈ gjH. Thus, by deﬁnition,

T T 0 αx (i) = j = αx (i) ,

T T 0 T ie. αx = αx . As xgi = gαx(i)hxi, we get

T x(giehi) = gαx(i)hxiehi

= (g h )(h−1 hT h ) , αx(i)eαx(i) eαx(i) xiei

33 and thus hT 0 = h−1 hT h . xi eαx(i) xiei (5D) Theorem Let notation be as in (5B) and (5C). Then the map

αG/H : x → αx is a homomorphism G → Sn (called the permutation repræsentationen of G on the cosets of H) with kernel \ K = gHg−1 . g∈G

Proof: Let x ∈ G. We ﬁrst show that αx ∈ Sn. It is clear that αx : {1, ··· , n} → {1, ··· , n} so we need only show that αx is injective: If αx(i) = 0 j = αx(i ) then xgi ∈ gjH and xgi0 ∈ gjH, so that

−1 −1 gi gi0 = (xgi) (xgi0 ) ∈ H.

0 This means giH = gi0 H, ie. i = i .

Let now x, y ∈ G. For 1 ≤ i ≤ n we have ygi ∈ gαy(i)H, so that x(ygi) ∈ xgαy(i)H = gαx(αy(i))H. From this we get αxy = αx ◦ αy. Let x ∈ G. Then

αx = (1) ⇔ xgi ∈ giH for 1 ≤ i ≤ n

−1 ⇔ x ∈ giHgi for 1 ≤ i ≤ n n \ −1 \ −1 ⇔ x ∈ giHgi = gHg . i=1 g∈G (5E) Corollary: If the group G has a subgroup H of index |G : H| = n, then there exists a normal subgroup N of G, satisfying

(i) |G : N| | n!

(ii) N ⊆ H.

34 Proof: As N we may use ker α, where α is as in (5D), since α induces an injective homomorphism from G/ ker α into Sn. Thsu |G/ ker α| = |G/N| | |Sn| = n!

(5F) Example: Let n ≥ 5. Then the alternating group An has no subgroup of index k, when 1 < k < n. Otherwise the simple group An would contain a normal subgroup of index ≤ k ! diﬀerent from An, and this is not possible. In particular we see that An−1 must be a maximal subgroup in An (of index n). (5H) Example: If G has a subgroup H of index 3, then G has also a normal subgroup of index 2 or 3. Either H ¢ G or G has a normal subgroup N with H ⊆ N, |G : N| = 3! = 6. In that case G/N ' S3 which has A3 as a normal subgroup of index 2. Thus G also has a normal subgroup of index 2.

(5I) Remark: If we choose H = {1} in (5D), and G is ﬁnite then αG/H = αG is a monomorphism G → S|G|, which is called the regular representation of G.

(5J) Theorem: Let the notation be as in (5B) and (5C). By

T T VT (x) = (αx; hx1, ··· , hxn) is deﬁned a monomorphism VT : G → H o Sn. T Proof: Let x, y ∈ G. As ygi = gαy(i)hyi we get

T T T (xy)gi = x(ygi) = (xgαy(i))hyi = (gαx(αy(i))hxαy(i))hyi such that T T T T VT (xy) = (αxαy; hxαy(1)hy1, ··· , hxαy(n)hyn) T T T T = (αx; hα1, ··· , hxn)(αy; hy1, ··· , hyn) = VT (x)VT (y) .

If x ∈ Ker(VT ) then xgi = gi1 for all i, ie. x = 1. (5K) Theorem: Notation as above. Assume in addition that K ¢ H such that H/K is abelian. Then

V er = pH/K ◦ VT deﬁnes a homomorphism G → H/K which is independent of the choice of T . It is called Verlagerung from G to H/K.

35 Proof: VT : G → H o Sn and pH/K : H o Sn → H/K are homomorphisms, whence V er is also a homomorphism. If T 0 is another transversal it must be shown that pH/K ◦ VT = pH/K ◦ VT 0 . This follws easily from (5C). We use the German term “Verlagerung” in these notes. In English textbooks it is often called “transfer”.

36 6. Focal subgroups, Gr¨un’sTheorems, Z–groups G6 - 2007-version

First a couple of general remarks. As usual G0 = [G, G] denotes the commutator subgroup of the group G. (6A) Lemma: Let H be a subgroup of G. We have G0 ⊆ H ⇔ H ¢ G and G/H abelsk .

Proof: Exercise or Chapter 1.18 in GT3

(6B) Example: Let us consider the dihedral group D20 of order 40 (see (4B)). We write

20 2 −1 −1 D20 = hx, y | x = 1, y = 1, yxy = x i .

0 2 0 It is easy to see that D40’s commutator group is D40 = hx i, |D40| = 10. The 0 commutator factor group D40/D40 is Klein’s four group Z2 × Z2. By (6A) the list of normal subgroups N/D20, with D20/N abelian is as follows:

2 2 2 D20, hxi, hx , yi, hx , xyi, hx i . In this list the 1., 3. and 4. group are dihedral groups and the other are cyclic. (Why?)

In this Chapter G is a ﬁnite group. A p–focal subgroup in G is a p–Sylow group i G0. By (2F) the p–focal 0 subgroups in G are exactly P ∩ G , P ∈ Sylp(G). In the above Example (6B) the 2-focal subgroup has order 2 and the 5-focal subgroup has order 5. The p–focal subgroups are important because they may under certain circumstances be described “locally” using certain “proper” subgroups in G. This is particularly important in the study of non-abelian simple groups. In such a group G0 = G, and the p–focal subgroup are exactly the p-Sylow groups in G. Generally we have 0 (6C) Theorem: Let P ∈ Sylp(G). Then P/P ∩ G is an abelian p–group, and there exists a normal subgroup K ¢ G, such that G/K ' P/P ∩ G0 .

37 Proof: The factor group G/G0 is abelian and contains therefore a p–complement (= p0–Hall subgroup) K/G0. As G0 ⊆ K we have K ¢ G, and G/K is abelian by (6A). As |G : K| is a power of p we have G = KP, eg. by (2B). We get G/K ' P/P ∩ K. It is clear that P ∩ G0 ⊆ P ∩ K. As |K : G0| is prime to p then a p–Sylow group for G0 is also a p–Sylow group for K. Therefore (2F) shows that P ∩ G0 = P ∩ K. Thus G/K = KP/K ∼= P/P ∩ K = P/P ∩ G0, as desired.

If x, y ∈ G and H ⊆ G is a subgroup we write x ∼H y, if there exists a h ∈ H so that hxh−1 = y, and we call x and y conjugate i H. It is clear that ∼H is an equivalence relation on the elements of G. When x, y ∈ G then [x, y] = xyx−1y−1 is x and y’s commutator. We have G0 = h[x, y] | x, y ∈ Gi. The next Theorem is very useful.

(6D) Theorem: (Generators for focal subgroups) Let P ∈ Sylp(G). Then

0 −1 P ∩ G = hxy | x, y ∈ P , x ∼G yi .

Proof D. Gorenstein: Finite groups, 7.3.3-7.3.4

Deﬁnition: Let K ⊆ L ⊆ G be subgroups. We say that L controls fusion in K, if we have:

∀ a, b ∈ K : a ∼G b ⇒ a ∼L b .

Here is a simple example of control of fusion:

(6E) Lemma: If P ∈ Sylp(G) is abelian, then NG(P ) controls fusion in P .

Proof Assume that a, b ∈ P, g ∈ G and gag−1 = b. We then have that −1 gCG(a)g = CG(b). As P is abelian, we have P ∈ Sylp(CG(a)) and P ∈ Sylp(CG(b)). By the above we also have

−1 −1 gP g ∈ Sylp(gCG(a)g ) = Sylp(CG(b)).

−1 If we apply Sylows Theorem on the Sylow groups gP g and P in CG(b), we −1 −1 see that there exists c ∈ CG(b), such that gP g = cP c . If we then put −1 −1 n = c g we see that n ∈ NG(P ) and nan = b.

38 (6F) Theorem: If P ∈ Sylp(G), and if L ⊇ P controls fusion in P , then P ∩ G0 = P ∩ L0 .

Proof: By assumption

∀ x, y ∈ P : x ∼G y ⇔ x ∼L y . Thus by (6D) P ∩ G0 and P ∩ L0 have the same generators so that

P ∩ G0 = P ∩ L0 .

0 0 (6G) Corollary: If P ∈ Sylp(G) is abelian then P ∩ G = P ∩ NG(P ) . Proof: Use (6E) and (6F).

(6H) Example: Let us check what (6D) means for a 2–Sylow group in S4 (and in S5). By (4B)

P = h(1, 2, 3, 4), (1, 4)(2, 3)i ∈ Syl2(S4) . P is a dihedral group and contains the following elements

P = {(1), (1, 2, 3, 4), (1, 4, 3, 2), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3), (1, 3), (2, 4)} .

Of these elements only the following are conjugate in S4 (and in S5!)

(i) (1, 2, 3, 4) and (1, 4, 2, 3) = (1, 2, 3, 4)−1 (ii) (1, 2)(3, 4) and (1, 3)(2, 4) and (1, 4)(2, 3) (iii) (1, 3) and (2, 4).

∗ 0 Let P = P ∩(S4) . From (i) and (iii) we get only (1, 3)(2, 4) as generator in P ∗. From (ii) we get (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3), since every element in (ii) is a product of the two other. Thus P ∗ = {(1), (1, 3)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}, Klein’s four group. In S5 we get the same result! This example also shows that (6G) is wrong when P is not abelian: In S4 there are three 2–Sylow groups, so that |S4 : NS4 (P )| = 3, ie. NS4 (P ) = P . Therefore

0 0 ∗ 0 P ∩ NS4 (P ) = P = {(1), (1, 3)(2, 4)}= 6 P = P ∩ S4 .

39 Corollary (6G) may be generalized as follows:

(6I) Theorem: (Gr¨un’sTheorem) Let P ∈ Sylp(G). We have

0 0 −1 0 P ∩ G = hP ∩ NG(P ) ,P ∩ g P g | g ∈ Gi .

Remarks: If P is abelian then P 0 = {1} and thus P ∩ g−1P 0g = 1 for all g ∈ G. Therefore (6I) implies (6G). We again have to consider Verlagerung into an abelian factor group of P , and need to choose coset representatives for P in G in a special way. You start by dividing G into double cosets modulo P so we need to add some facts about double cosets. This is done generally here. If H, K are subgroups in G, g ∈ G we call the subset

HgK = {x ∈ G | ∃h ∈ H , k ∈ K : x = hgk} a double (H,K)–coset. The double cosets are equivalence classes for the following equivalence relation H ≡K on G:

gH ≡K g1 ⇔ ∃h ∈ H , k ∈ K : g = hg1k .

Thus the (H,K) double cosets divide G into disjoint subsets. Contrary to a result on cosets we do not have that |HgK| | |G|. In fact

(6J) Theorem: The double coset HgK is a union of |H : H ∩ gKg−1| right cosets to K (and of |K : K ∩ g−1Hg| left cosets to H). In particular

|HgK| = |H : H ∩ gKg−1 | |K| (= |K : K ∩ g−1Hg| |H|) .

Proof. Easy direct calculation. You may also prove the result by applying −1 (2A) to the subgroups H and gKg of G. ∗ When u ∈ P , P ∈ Sylp(G), g ∈ G we give a description of some coset representatives in P g∗P which are useful in the calculation of of V er(u). ∗ (6K) Lemma: Let u ∈ P , P ∈ Sylp(G), g ∈ G. There exists g ∈ G, 1 = a0, a1, ··· , ar ∈ P , t0, t1, ··· , tr ≥ 0, such that the following is fulﬁlled: (1) P g∗P = P gP .

r pti −1 S S j (2) P gP = u aigP . i=0 j=0

40 −1 −1 pti (3) For all i we have g ai u aig ∈ P (4) For all i we have g−1upti g ∈ P .

r (5) P pti = |P : P ∩ gP g−1|. i=0

Proof: Huppert: Endliche Gruppen I, IV.3.4, p. 423.

This is used to prove (6I). Proof for (6I): Huppert: Endliche Gruppen I, IV.3.4, p. 423.

Let P ∈ Sylp(G). Z(P ) denotes as usual P ’s center. The group G is called p–normal if:

∀g ∈ G : gZ(P )g−1 ⊆ P ⇒ gZ(P )g−1 = Z(P ).

(For example G is p–normal, if P is abelian.)

The next result may be seen as a generalization of (6G) in another direction.

(6L) Theorem: (Gr¨un’s2. Theorem) Let G be p–normal. Let P ∈ Sylp(G) 0 0 and H = NG(Z(P )). Then P ∩ G = P ∩ H . Proof: Huppert: Endliche Gruppen I, IV.3.7, p. 425.

The next Theorem is one of the most useful applications of Verlagerung.

(6M) Theorem: (Burnside) Let P ∈ Sylp(G). If P ⊆ Z(NG(P )), then G has a normal p–complement K. Proof: We show that P ∩G0 = {1}, and then the existence of K follows from (6C). Put N = NG(P ). We have that P ⊆ Z(N), and as P ⊆ N = NG(P ) we see that P is abelian. Thus by (6G)

(∗∗) P ∩ G0 = P ∩ N 0 .

But P ∩N 0 may be calculated by (6D). As P ⊆ Z(N) we get for x, y ∈ P that 0 0 x ∼N y ⇔ x = y. By (6D) P ∩ N = {1}, so that (∗∗) yields P ∩ G = {1}, as desired. In the following we need

41 (6N) Lemma: If L is a subgroup of G, then CG(L) ¢ NG(L), and the faktor group NG(L)/CG(L) is isomorphic to a subgroup of Aut(L).

Proof: If x ∈ NG(L) −1 αx : ` 7→ x`x defines an automorphism of L. The map x → αx is a homomorphism from NG(L) into Aut(L) with CG(L) as kernel. (6P) Remark: If L is cyclic then Aut(L) is abelian. If L is an elementary abelian p–group of order pn then L may be considered as a vector space over the field Zp with p elements, dimL = n. The homomorphismes from the group L to itself “correspond” then to linear maps of the vector space L. It follows that Aut(L) ' GL(n, Zp), the set of invertible n × n–matrices with coefficients from Zp.

(6Q) Theorem: Let p be smallest prime dividing |G|. If P ∈ Sylp(G) is cyclic, then G has a normal p–complement. In particular a group with a cyclic 2–Sylow group always has a normal 2–complement. Proof: By (6N) NG(P )/CG(P ) is isomorphic to a subgroup of Aut(P ). But Aut(P ) is an abelian group of order pn−1(p − 1), if |P | = pn. (See eg. GT3, n Chapter 1.13) As p | |CG(P )| (because P is abelian) we get p - |NG(P ): CG(P )|. Now p is the smallest prime divisor of |G|, so that ((p − 1), |G|) = ((p − 1), |NG(P )|) = 1. We get that |NG(P ): CG(P )| = 1, ie. NG(P ) = CG(P ). Thus P ⊆ Z(NG(P )). Apply (6M). (6R) Theorem: A simple group of order 60 is isomorphic to the alternating group A5.

Proof. Let G be simple, |G| = 60. Choose P ∈ Syl2(G), so that |P | = 4. Then |G : NG(P )| =: n2(G) is the number of 2–Sylow groups in G. We have n2(G)|15 = |G : P |. By (5E) we have n2(G) ≥ 5, so that n2(G) ∈ {5, 15}. 60 If n2(G) = 15, then P = NG(P ) as |NG(P )| = 15 = 4. By (6M) this is impossible. Thus n2(G) = |G : NG(P )| = 5, and the claim follows from (5E). (6S) Lemma: If G/Z(G) is cyclic, then G is abelian. (See also GT3). Proof: Exercise

You are reminded that G(i) is the i’th commutator group in G. G(0) = G; G(1) = [G, G]; G(i) = [G(i−1),G(i−1)] .

42 (GT3, Chapter 1.18) We have G(i)charG for all i, by (1A)(2).

(6T) Theorem: Assume that i ≥ 1 and that the factor groups G(i)/G(i+1) and G(i+1)/G(i+2) are both cyclic. Then G(i+1) = G(i+2). Proof: The assumptions are independent of what G(0), ··· ,G(i−1) are, or what G(j) is for j ≥ i + 2. We may therefore assume that i = 1, and that G(i+2) = G(3) = {1}. We then know that G(2)/G(3) = G(2) = hai is cyclic. As (2) 00 G = G = hai ¢ G, we have NG(hai) = G. Let X = CG(a). By (6N) G/Z is isomorphic to a subgroup of Aut(hai). Also Aut(hai) is abelian ((6P)). Thus G/X is abelian. This means that G0 ⊆ X. We then have

00 0 G = hai ⊆ G ⊆ X = CG(a) .

0 00 0 As G ⊆ CG(a), we get G = hai ⊆ Z(G ). Thus, by an isomorphism theorem, . . G0/Z(G0) ' G0 hai (Z(G0)/hai) .

As G0/hai = G0/G00 is cyclic, we get that G0/Z(G0) is cyclic. By (6S) G0 is (2) 0 0 (2) (3) abelian, so that G = (G ) = {1}. Thus G = G (= {1}), as desired. We can now give a description of the ﬁnite groups, where all Sylow groups are cyclic (for all primes). Such a group is called a Z-group.(Z is an abreviation for Zassenhaus. Hans Zassenhaus 1912–1991. (See also Chapter 3)). (6U) Theorem: A Z–group is solvable. Proof: Let G be a Z–group. The proof is by induction by |G|. Let p be the smallest prime divisor in |G|. By (6Q) G has a normalt p–complement K. Thus G/K is a (cyclic) p–group, ie. solvable, and K is solvable by the induction hypothesis (See also Theorem (2F), which shows that K is a Z–group). Thus G is solvable. (6V) Lemma: An abelian Z–group is cyclic. Proof: Exercise

(6W) Theorem: A Z–group G is a semidirect product

G = A o B,

43 where A and B are cyclic. Proof: G is solvable by (6U). Consider the commutator series

G = G(0) ⊃ G(1) ⊃ · · · ⊃ G(k) = {1} .

The factor groups G(i)/G(i+1) are abelian Z–groups, and thus cyclic by (6V). Then (6T) shows that G(2) = {1}! We now have that G/G0 and G0 are cyclic. Let |G0| = m, |G : G0| = n, so that |G| = mn. Let G0 = hai and choose b ∈ G with G/G0 = hbG0i. We have G = ha, bi. As hai ¢ G, we get b−1ab = ar, where hari = hai (so that (r, m) = 1). Furthermore

a−1b−1ab = ar−1 . If X = har−1i then G/X = haX, bXi. As [a, b] ∈ X, G/X is abelian, so that G0 ⊆ X. Hence G0 = X = har−1i. Since |a| = |ar−1|, we have (r − 1, m) = 1. As G/G0 = hbG0i has order n, we have bn ∈ G0, say

bn = as .

Then bn = b−1bnb = b−1asb = (b−1ab)s = ars, ie. as = ars. Therefore |a| = m | (r − 1)s. As (m, r − 1) = 1 we get m|s, and as am = 1 we get as = 1. Thus bn = as = 1 and |b| = n. We can now show (m, n) = 1: If m1 n1 p|m, p|n, p a prime and m = m1p, n = n1p, then a and b are elements of order p, and as hai ∩ hbi = {1} we see that ham1 , bn1 i is an abelian group of order p2, which is not cyclic. This is impossible as G’s p–Sylow groups are cyclic. Thus (m, n) = 1. We have shown that G is a semidirect product of A = G0 = hai and B = hbi. Let us remark that as (m, n) = 1, A and B are Hall subgroups in G. In the case where all Sylow groups have prime order there is a stronger statement than (6W):

(6X) Theorem: Assume that |G| = p1p2 ··· pr, where p1 < p2 < . . . < pr are primes. Then G = A o B, where there exists an s, 1 ≤ s ≤ r such that

|B| = p1 ··· ps , |A| = ps+1 ··· pr .

44 7. Groups of a given ﬁnite order G7- 2007-version

This chapter is essentially diﬀerent from the other chapters. It is inspirered by the pages 203-215 in the book D.S.Dummitt, R.S. Foote: Abstract algebra, Prentice-Hall 1991. We illustrate methods to at study ﬁnite groups of a given order. You may for instance want to show that such a group cannot be simple. We use Sylow’s Theorems and results from the previous 2 Chapters. First we make a number of Remarks, which might be useful for this type of problems and then we illustrate the methods in a series of examples. Remarks:

(I) If P ∈ Sylp(G), then the number np(G) := |Sylp(G)| of p–Sylow groups of G satisﬁes

np(G) = |G : NG(P )| and np(G) ≡ 1(mod p).

If G er simple and p | |G|, then np(G) > 1, unless G is cyclic of order p. (This follows from Sylow’s Theorem.)

(II) If P ∈ Sylp(G) has order p, then G has precisely (p − 1)np(G) elements of order p.

(We know that as |P | = p, none of the np(G) p-Sylow groups can have elements 6= 1 in common. Each Sylow group contais (p − 1) elements 6= 1. This counting argument does not work, if |P | > p, unless you can show ∗ ∗ the 2 arbitrary diﬀerent subgroups P,P ∈ Sylp(G) satisfy P ∩ P = 1. If this is the case we say that the Sylow groups are TI, ie. “Trivial Intersection.” If the p-Sylow groups of G are TI then G contains (|P |− 1)np(G) elements 6= 1 of p-power order. ) (III) If you want to show that the Sylow groups in G are TI (eg. to count p-elements, see (II)), you may do this indirectly by considering a Sylow group intersection R = P ∩ P ∗ 6= 1 of maximal order where P and

45 P ∗ are diﬀerent. As p-groups are nilpotent (see Chapters 8-9) the subgroup R of P (or P ∗) cannot be equal to its own normalizer in P (or ∗ P .) You then know that NG(R) must contain at least (p + 1) p-Sylow groups, since the p-subgroups NP (R) and NP ∗ (R) of NG(R) cannot be contained in the same p-Sylow group T of NG(R), due to the maximality ∗ of R. If they were and T ⊆ U, U ∈ Sylp(G), then P = U = P , as ∗ |P ∩ U| ≥ |NP (R)| > |R| and |P ∩ U| ≥ |NP ∗ (R)| > |R|. Thus there is a chance that NG(R) may be a very large subgroup of G. See also Remark (XI).

(IV) If G is simple and p is the largest prime dividing |G|, then G has no subgroup U of index k, 1 < k < p. (If it did then G would have a normal subgroup NN ⊆ U ⊆ G, with G/N isomorphic to a subgroup of Sk. (Corollary (5E)) It is clear that N = 1. As p | |G|, but p - k!, because k < p, we would get a contradiction.)

(V) We may generalize (IV) as follows: If G is a group, we deﬁne s(G) := min{t ∈ N | |G| | t!}. Then we have: If G is simple then G has no subgroup of index k, when 1 < k < s(G).

(If it did then G would be isomorphic to a subgroup of Sk, as in (IV). In particular then |G| | k!, contradicting that k < s(G).)

(VI) Remark (V) may also be improved a little: If G is simple of order ≥ 3 and has a subgroup of index k > 1, then every subgroup of G is isomorphic to a subgroup of the alternating Ak.

(We know that G is isomorphic to a subgroup H of Sk. But actually H must be a subgroup of Ak : If H * Ak then H ∩ Ak has index 2 in H. But as H is simple (isomorphic to G) we have a contradiction.)

(VII) If p is an odd prime and P ∈ Sylp(Ak) then 1 |N (P )| = |N (P )|. Ak 2 Sk

(This follows easily from the Frattini argument in Chapter 1.)

(VIII) If p is an odd prime, k = p eller k = p + 1 and P ∈ Sylp(Ak) then 1 |NAk (P )| = 2 p(p − 1).

46 (It is not diﬃcult to see that in these cases for k we have |NSk (P )| = p(p − 1). (Exercise.) Then use (VII).)

(IX) Assume that p < q are primes, such that p - (q − 1). (We then know that every group of order pq is cyclic). Assume that P ∈ Sylp(G) and Q ∈ Sylq(G) are both cyclic of order p and q respectively. Then

p | |NG(Q)| ⇔ q | |NG(P )|.

∗ (If p | |NG(Q)|, then NG(Q) contains a subgroup P of order p. Then P ∗Q is a subgroup of order pq in G, which then is cyclic. Therefore Q ∗ ∗ ∗ and P centralizes each other so that Q ⊆ CG(P ) ⊆ NG(P ). As P ∗ ∗ and P are conjugate in G, then also NG(P ) and NG(P ) are conjugate. We get q | |NG(P )|. The other direction is proved analogously.) (X) You may also ’play primes against each other’ in other situations than the one in (IX). Assume that P is a p-subgroup and at the prime q is a divisor in |NG(P )|. You may then look at the number of q- Sylow groups in NG(P ) applying for example (I). If it turns out that nq(NG(P )) = 1, then NG(P ) ⊆ NG(Q), where Q ∈ Sylq(NG(P )). There my be other subgroups in NG(Q), eg. if Q 6∈ Sylq(G), which may make this subgroup very large. Then you may apply for example (IV)-(VI).

2 ∗ (XI) Assume that np(G) 6≡ 1(mod p ). Then there exist p-Sylow groups P,P in G, such that R = P ∩ P ∗ has index p in P and in P ∗. In particular then NG(R) contains at least (p + 1) p-Sylow groups. ∗ (This is seen as follows: Choose P ∈ Sylp(G). Consider for P ∈ ∗x Sylp(G) the set {P |x ∈ P }, ie. the set of P -conjugate subgroups to P ∗. If x ∈ P and P ∗x = P ∗, then < P ∗, x >= P ∗ < x > is a p- subgroup in G. As P ∗ is a Sylow group, we have x ∈ P ∗, ie. x ∈ P ∩P ∗. Therefore the set {P ∗x|x ∈ P } contains exactly |P : P ∩ P ∗| Sylow 2 groups. The assumption np(G) 6≡ 1(mod p ) shows then that there exists a p-Sylow group P ∗ 6= P, such that p2 - |P : P ∩ P ∗|. In that case we put R = P ∩ P ∗. As p-groups are nilpotent (see Chapters 8-9) the subgroup R of P (respectivelt P ∗) not be its own normalizer in P ∗ ∗ ( respectively P .) Therefore P,P ⊆ NG(R).)

And here are the Examples:

47 Example 1: A group G of order 132 = 22 · 3 · 11 is not simple. Proof: If you want to make the assumption that G is simple lead to a contradiction, you may here ’count elements’. It is usually good to start with the largest prime. Remark that if G is simple, then np(G) 6= 1 for p = 2, 3, 11. As n11(G) | 12, we egt n11(G) = 12, so G has 12 · 10 = 120 elements of order 11 by (I). As n3(G) ≥ 4, G has at least 4 · 2 = 8 element´s of order 3 by (I). In total you have at least 128 elements of order 11 and 3 in G. Only 4 elements remains. Among these 4 elements must be all elements in all 2-Sylow groups in G. We get n2(G) = 1, a contradiction . Example 2: A group G of order p2 · q, where p and q are primes, is not simple: Either a p-Sylow group or a q-Sylow group is normal in G.

Proof: Assume that np(G) > 1, ie. that a p-Sylow group P is not normal. 2 We get np(G) = q, so that |NG(P )| = p . This shows that NG(P ) = P, and as P is abelian (as any group of prime square order is ), we get P ⊆ Z(NG(P )). Theorem (6M) shows that G har et normalt p-complement, which then must have order q. Alternatively you may by elementary number theoretic considerations show that vise, at np(G) > 1 ⇒ nq(G) = 1. Example 3: A group G of order 3393 = 32 · 13 · 29 is not simple. Proof: This and the next 2 examples use ’subgroups of small index’. If G is simple, G cannot contain a subgroup U of index 1 < k < 29, by (IV). But n3(G) = 13, as all other divisors 6= 1 of 13 · 31 are not congruent to 1 (mod 3). Thus the 3-Sylow-normalizer must have index 13, by (I), contradiction. The idea in this example may often be used. Example 4: A group G of order 396 = 22 · 32 · 11 is not simple. Proof: Assume G simple. As 12 is the only divisor > 1 of 22 · 32, which is congruent to 1 modulo 11, we have n11(G) = 12.G and its subgroups are thus isomorphic to subgroups of A12, by (IV). If P ∈ Syl11(G), we have |NG(P )| = 33 = 3 · 11. But A12 (or even S12) has no subgroup of order 33. Such a subgroup had to be cyclic, as any group of order 33 er cyclic. An element of order 33 in a symmetric group needs at least 11+3=14 elements to operate on. Here we could alternatively use (VIII). Example 5: A group G of order 224 = 25 · 7 is not simple.

48 Proof: Assume G simple. We get n2(G) = 7, so that G is isomorphic to a 5 subgroup of A7, by (VI). But 2 - |A7|. Example 6: A group G of order 4095 = 32 · 5 · 7 · 13 is not simple.

Proof: Assume G simple. By (I) we have n7(G) = 15 and n13(G) = 105 = 3·5·7. If P ∈ Syl7(G) and Q ∈ Syl13(G) we get |NG(P )| = 4095/15 = 3·7·13 and |NG(Q)| = 4095/105 = 3 · 13. This contradicts (IX). Example 7: A group G of order 351 = 34 · 13 is not simple.

Proof: Assume G simple. Let P ∈ Syl3(G). We have that n3(G) = 13 by 2 ∗ (I), so that n3(G) 6≡ 1 (mod 3 ). By (XI) there exists P ∈ Syl3(G), so ∗ 3 that R = P ∩ P has order 3 . Let X = NG(R). Then X contains at least 2 3-Sylow groups, namely P and P ∗. Thus |P | is a proper divisor of |X|, since otherwise P = X. We get X = G, so that R / G, a contradiction. Example 8: A group G of order 2205 = 32 · 5 · 72 is not simple. 2 Proof: Assume G simple. We get n7(G) = 15 6≡ 1 (mod 7 ). Choose by ∗ ∗ (XI) P,P ∈ Syl7(G), so that R = P ∩ P has order 7. Thus we have for X = NG(R) that n7(X) > 1. As n7(X) | 45 = |G : P |, we get n7(X) = n7(G) = 15. Thus X contains all 7-Sylow groups from G. As G is generated by these we get X = G, a contradiction.

49 8. The Frattini subgroup. Nilpotent groups. The Fitting subgroup G8 - 2007-version

Let G be a group. The subgroup M ⊆ G, M 6= G, is called maximal in G, if there is no subgroup K satisfying M ⊂ K ⊂ G. Max(G) denotes the set of maximal subgroup in G. If G 6= {1} is a ﬁnite group, then Max(G) 6= ∅, and we put \ Φ(G) = M. M∈Max(G) We call Φ(G) The Frattini(sub)group in G. If M ∈ Max(G) and α ∈ Aut(G), then also α(M) ∈ Max(G). Thus Φ(G) char G and in particular Φ(G) ¢ G. If G = {1} we put Φ(G) = {1}. It turns out that Φ(G) has a connection with “non-generators” for G: If X ⊆ G, then as in Chapter 1 hXi is the subgroup of G generated by X, ie. the smallest subgroup i G, containing X. If hXi = G, we call X a generating set for G. Consider elements g ∈ G satisfying:

∀ X ⊆ G : hX, gi = G ⇒ hXi = G.

Such elements are called non-generators, becuse they are superﬂuous in all generating sets for G. Let I(G) = {g ∈ G | g is non − generator}. (8A) Theorem: Let G be a ﬁnit group. Then

Φ(G) = I(G) .

Remark: This Theorem also holds when G is inﬁnite. Proof: ...

(8B) Theorem: Let G be a ﬁnite group. Let P ∈ Sylp(Φ(G)). Then P ¢G. Proof: We apply the Frattini argument (1E) on Φ(G) ¢ G and get

G = Φ(G) · NG(P ) .

50 Using (8A) we see

G = hΦ(G),NG(P )i = hI(G),NG(P )i = hNG(P )i = NG(P ) , ie. P ¢ G. Remark: This shows that when G is ﬁnite then all Sylow groups in the group Φ(G) are normal in G and therefore also in Φ(G). This actually means that Φ(G) is an (inner) direct product of its Sylow groups (ie. a ﬁnite nilpotent group, see Theorem (8J)).

We give in the following a description of nilpotent groups. Here the group G is not necessarily ﬁnite. When H and K are subsets of G, then

[H,K] = h[h, k] | h ∈ H , k ∈ Ki is the subgroup of G generated by all commutators of elements from H and K. As usual Z(G) is the center of the group G. It is easily seen that

H char G , K char G ⇒ [H,K] char G because the automorphisms of G then “mix” the generators for [H,K], see Chapter 1. (8C) Lemma: Let K ¢ G and K ⊆ H ⊆ G. Then

[H,G] ⊆ K ⇔ H/K ⊆ Z(G/K) .

Proof: We have

[H,G] ⊆ K ⇔ ∀h ∈ H ∀g ∈ G :[h, g] ∈ K

⇔ ∀h ∈ H/K , ∀g ∈ G/K :[h, g] = 1 ⇔ H/K ⊆ Z(G/K) . (8D) Lemma: Let ϕ : G → H be a surjective homomorphism. Then

ϕ(Z(G)) ⊆ Z(H) .

51 Proof: Let a ∈ Z(G), h ∈ H. We show that [ϕ(a), h] = 1. As ϕ is surjektive there exists a g ∈ G such that ϕ(g) = h. Then [a, g] = 1, as a ∈ Z(G) and thus 1 = ϕ([a, g]) = [ϕ(a), ϕ(g)] = [ϕ(a), h] . In connection with the definitione of nilpotent group central series are important. Definition: We definere two series of subgroups of an arbitrary group G: The lower central series for G

G = L1(G) ⊇ L2(G) ⊇ · · · ⊇ Lc(G) ⊇ · · · is deﬁned by

L1(G) = G,Li(G) = [Li−1(G),G] for i ≥ 2 . the upper central series for G

{1} = Z0(G) ⊆ Z1(G) ⊆ · · · ⊆ Zc(G) ⊆ · · · is deﬁned by

Z0(G) = {1} ,Zi(G)/Zi−1(G) = Z G/Zi−1(G) for i ≥ 2 .

(In particular Z1(G) = Z(G).)

0 (8E) Examples: (1) Let G = S4. We have Z(G) = {1}, and thus Z (G) = 1 2 0 Z (G) = Z (G) = ··· = {1}. We have that G = A4. Furthermore 0 [G ,G] = [A4,S4] = A4, since for example [(1, 2, 3), (1, 3, 2, 4)] = (1, 2, 4) and [(1, 2, 3), (2, 3, 4)] = (1, 4)(2, 3) generate A4. Therefore

L1(G) = S4 ,L2(G) = L3(G) = ··· = A4 .

(2) Let G = D8, a dihedral group of order 16. G = hx, y | x8 = y2 = 1 , y−1xy = x−1i .

4 ∼ Here we have Z(G) = hx i and that G/Z(G) = D4. The group D4 has also et center of order 2, so we get

Z(G/hx4i) = hx2i/hx4i .

52 Now G/hx2i is a four group. We get

Z0(G) = {1} ,Z1(G) = hx4i ,Z2(G) = hx2i

Zi(G) = G for i ≥ 3. Furthermore G0 = hx2i (as we have seen earlier). Now [G0,G] = h[x2, y]i = hx4i and [hx4i,G] = 1, da x4 ∈ Z(G). We get

2 4 L1(G) = G,L2(G) = hx i ,L3(G) = hx i

Li(G) = {1} for i ≥ 4 .

The ﬁrst example shows that den lower (respectively upper) central series need not ﬁnish in {1} (respectively G). In the second example we had 2 Z (G) = G and L3(G) = {1}. The connection is explained here: (8F) Theorem: For any group G and any m ≥ 0 we have

m Z (G) = G ⇔ Lm+1(G) = {1} .

Furthermore in this case we have

m−i (∗) Li+1(G) j Z (G) for alle i

Proof: ...

Deﬁnition: A group G is called nilpotent if there exists an m ≥ 0 such that Zm(G) = G. The smallest integer m with Zm(G) = G is called the nilpotency class of G. We then say that G is nilpotent of class m. Remark: G = {1} is nilpotent of class 0. A group G 6= {1} is nilpotent of class 1, if and only if it is abelian. A non-abelian group G is nilpotent of klasse 2 if and only if G0 ⊆ Z(G). (Why?) (8F) Theorem: Assume that G is nilpotent.

(1) Any subgroup of G is nilpotent.

(2) If N ¢ G, then G/N is nilpotent.

53 Proof: (1) Let U ⊆ G be subgroup. From the deﬁnition we get easily

Li(U) ⊆ Li(G) for all i ≥ 1

(use induction on i). As G is nilpotent, we get Lm(G) = 1 for some m ≥ 1. Then also Lm(U) = {1}. (2) It is easily seen from the deﬁnition that

Li(G/N) = Li(G)N/N for all i ≥ 1 .

Thus if Lm(G) = 1, then also Lm(G/N) = 1, and thus G/N is nilpotent.

(8G) Remark: If G = S3, N = A3, then both G/N and N are abelian (and thus nilpotent), but G is not nilpotent. (Ln(G) = A3 for n ≥ 2.) Thus the group S3 is solvable, but not nilpotent.

(8H) Theorem: Assume that G is nilpotent, and that H ⊂ G is a subgroup 6= G. Then H 6= NG(H).

Proof: Assume that Lm(G) = {1}. As L1(G) = G and H 6= G, there exists an i, 1 ≤ i ≤ m − 1 such that Li+1(G) ⊆ H, but Li(G) * H. Then

[Li(G),H] ⊆ [Li(G),G] = Li+1(G) ⊆ H.

But the inclusion [Li(G),H] ⊆ H means that Li(G) ⊆ NG(H). (Why?) As Li(G) * H, we get NG(H) 6= H. (8I) Theorem: If G and H are nilpotent, then also G × H is nilpotent. Proof: Show by induction that

Li(G × H) ⊆ Li(G) × Li(H) for i ≥ 1.

Remark: A ﬁnite p-group is nilpotent. This follows from the fact that any ﬁnite p-group 6= {1} has a center 6= {1}. (See Theorem (9D) below or Chapter 1.11 in GT3).

(8J) Theorem: For a ﬁnite group G the following conditions are equviva- lent: (1) G is nilpotent.

54 (2) For all subgroups H 6= G we have NG(H) 6= H. (3) For alle primes p G has a normal p-Sylow group. (4) G er et direct product of its Sylow groups. Proof:(1) ⇒ (2). Use (8H). (2) ⇒ (3). Let P ∈ Sylp(G). Put H = NG(P ). By (1F) we have H = NG(H), so that from the assumption (2) we get H = G, ie. P ¢ G. (3) ⇒ (4). If p 6= q are primes and P ∈ Sylp(G), Q ∈ Sylq(G), then [P,Q] ⊆ P ∩ Q = {1}, as P ¢ G, Q ¢ G. Thus the elements in P and Q are permutable. As every element may in a unique way be written as a product of permutable elements of prime power order (see Remark below), we get (4). (4) ⇒ (1) follows from the Remark above and (8I). Remark: Assume that g ∈ G, |g| = mn where (m, n) = 1. Write 1 = km+`n, where k, ` ∈ Z. Then g = gkm·g`n, where gkm has order n and g`n has order m. If g = xy, |x| = n, |y| = m and xy = yx, then gkm = xkmykm = xkm, since ykm = (ym)k = 1. Thus gkmx−1 = xkm−1 = x−`n = (xn)−` = 1, as xn = 1. We get gkmx−1 = 1 or x = gkm. Analogously we see that y = g`n. (See also Chapter 1.15 in GT3.)

We return to the Frattini group Φ(G) of a group. From (8B) and (8J) we get (8K) Theorem: Let G be finite. Then Φ(G) is a nilpotent and characteristic subgroup of G. Finally we consider the Fitting subgroup of a finite group G. In the rest of this Chapter we thus consider only finite groups. (8L) Lemma: Assume that H and K are normal nilpotent subgroups of the group G. Then also HK is a normal nilpotent subgroup of G.

Proof: It is clear that HK ¢ G. Let p be a prime, P1 ∈ Sylp(H), P2 ∈ Sylp(K). By (8J) er P1 ¢H, so that by one of the Remarks in (1C) P1charH. We get P1charH ¢ HK, so that P1 ¢ HK. Analogously P2 ¢ HK so that P1P2 ¢ HK. It is easily seen that P1P2 ∈ Sylp(HK). Thus in HK all Sylow groups are normal, so that HK is nilpotent by (8J). (8M) Theorem: Let F (G) = hH | H ¢ G , H nilpotenti .

55 Then F (G) is a characteristic nilpotent subgroup of G, containing all nilpotent normal subgroups of G. F (G) is called the Fitting group of G. Proof: If H ¢G is nilpotent and α ∈ Aut(G), then α(H)¢G and α(H) ' H is nilpotent. Therefore F (G) char G, by (1C). That F (G) is nilpotent is easily seen from (8L). The next Theorem may look rather technical, but it is useful in various connections: T −1 (8N) Theorem: Let M ∈ Max(G). Put L = x∈G xMx , so that L is the largest normal subgroup of G contained in M. Put G = G/L, and let F = F (G). If F 6= 1, then the following holds:

(1) F is a minimal normal subgroup in G.

(2) F is an elementary abelian p-group.

(3) CG(F ) = F . (4) F ∩ M = 1 (where M = M/L).

(5) |G : M| = pn for a suitable n.

Proof: We assume that F 6= 1, such that F is a normal nilpotent subgroup 6= 1 in G. Therefore also Z(F ) 6= {1}. Let p | |Z(F )| and put P = {x ∈ Z(F ) | xp = 1}. We have P char Z(F ) as the elements in P are permuted by automorphisms of Z(F ). P is abelian as P ⊆ Z(F ), and therefore it is an elementary abelian p-group. We get {1}= 6 P char Z(F ) char F = F (G) char G. As P ¢ G, P M is a subgroup of G containing M ∈ Max(G). 0 But P * M: If P j M, then P s inverse image in G (called P ) is a normal subgroup in G, contained in M. As L is the largest normal subgroup in G contained in M we get P ⊆ L, and thus P = P/L = {1}, a contradiction. As P * M and M is maximal in G we get G = P M. Then we get (5) using (2A).

Put C := CG(P ). As P ⊆ Z(F ) we get F ⊆ C. As P ¢ G we get C ¢ G so that C ∩ M ¢ M. As P centralizes C, and therefore also C ∩ M, we have

P ⊆ NG(C∩M). We get C∩M ¢G as G = P M. In inverse image of C∩M in G is a normal subgroup of G, contined in M. As before we get C ∩ M = {1}. But P ⊆ F ⊆ C, and as P M = G, we get P M = F M = C M = G.

56 As P ∩ M ⊆ F ∩ M j C ∩ M = {1} we get from (2A), that |P | = |F | = |C| = |G : M| ! Thus P = F = C. As P is elementary abelian and F = C we get (1) and (2). As C = F we get (3) and (4). (8O) Corollary: If G is a solvable group and M ∈ Max(M), then |G : M| is a prime power. T −1 Proof: Let L = x∈G xMx be as in (8N). As G is solvable, G = G/L is also solvable. Therefore F (G) 6= {1}, since for example a minimal normal subgroup in G is abelian and therefore nilpotent (see (1D)). Thus (1)–(5) in (8N) holds in this situation. By (5) |G : M| is a prime power. The technical Theorem (8N) plays also a rôlein the following interesting result describing connections between the Frattini group and the Fitting group in a finite group: (8P) Theorem: Let G be finite. Put Φ = Φ(G), F = F (G). We have (1) [F,F ] ⊆ Φ ⊆ F . (2) F/Φ = F (G/Φ).

Proof: As Φ¢G and Φ is nilpotent ((8K)) we get that Φ ⊆ F . To show that T −1 [F,F ] ⊆ Φ we choose M ∈ Max(G), and put L = x∈G xMx as in (8N). Since xMx−1 ∈ Max(G) for all x ∈ G, we must have Φ ⊆ L by deﬁnition of Φ. Put G = G/L and F (G) = K/L, where then K ¢ G is the invere image of F (G) in G. We have F ⊆ K: By an isomorphism Theorem F/F ∩L ∼= F L/L. There- fore F L/L is a normal nilpotent subgroup in G, as F/F ∩ L is nilpotent by (8F) (2). We get that F L/L ⊆ K/L, ie. F ⊆ K. By (8N) K/L is abelian (also if F (G) = 1, ie. K = L!) As F L/L ⊆ K/L F L/L is abelian; therefore [F,F ] ⊆ [F L, F L] j L, (see (6A)). This holds for all intersections L of conjugate maximal subgroups. Let M1,M2, ··· ,Mt be a set of representatives for the G-conjugacy classes of T −1 maximal subgroups. Put Li = xMix . Then Φ = L1 ∩ L2 ∩ · · · ∩ Lt. By x∈G the above [F,F ] ⊆ Li for all i. We get [F,F ] ⊆ Φ. Thus (1) is proved. We put Ge = G/Φ and F (Ge) = H/Φ, where H is the inverse image of F (Ge) in G. Then F ⊆ H because F/Φ (by (8F)(2)) is a nilpotent normal subgroup in G/Φ = Ge, ie. F/Φ ⊆ H/Φ. To show the inclusion H ⊆ F we show that H is nilpotent. Let P ∈ Sylp(H). Then P Φ/Φ ∈ Sylp(Ge), and

57 therefore P Φ/Φ¢G/Φ by (8J), ie. P Φ¢G. As Φ ⊆ H and P ∈ Sylp(H), we get P ∈ Sylp(P Φ). By the Frattini argument (1E) we get G = P ΦNG(P ) = ΦPNG(P ) = ΦNG(P ) = NG(P ), as Φ consists of “non-generators”, (8A). Therefore P ¢ G, and in particular P ¢ H. We get that H is nilpotent by (8F), as desired. Our last result in this chapter is about solvable groups: (8Q) Theorem: Let G be ﬁnite and solvable. Then

CG(F (G)) ⊆ F (G) .

Proof: Put F := F (G), C := CG(F ). Assume that C * F . We seek a contradiction. We have that Z(F ) = C ∩ F ⊂ C. Choose a normal series for G, including the groups C and C ∩ F with (elementary) abelian factors (see (1D)),

{1} = Gt ⊂ Gt−1 ⊂ · · · ⊂ Gs+1 = C ∩ F ⊂ · · · ⊂ Gr = C ⊆ · · · ⊆ G1 = G.

As F 6= 1 is nilpotent we have Gs+1 = C ∩ F = Z(F ) 6= {1}. Therefore Gs 6= Gs+1. (Remark that s + 1 > 1 since otherwisw Gs+1 = G, a contradiction). As Gs/Gs+1 is abelian we have [Gs,Gs] ⊆ Gs+1 = Z(F ). Since furthermore Gs ⊆ Gr = C = CG(F ) we get

[[Gs,Gs],Gs] j [Z(F ),CG(F )] = {1} .

Thus Gs is nilpotent, ie. Gs ⊆ F = F (G). We get Gs ⊆ C ∩ F (as Gs ⊆ Gr = C), a contradiction.

58 9. Finite p–groups G9 - 2007-version

In this chapter we consider (better late that never) ﬁnite groups of p- power order, ie. p-groups. (9A) Theorem: Let P be a p-group, {1}= 6 N ¢ P . Then N ∩ Z(P ) 6= {1}. In particular Z(P ) 6= {1}. Proof: As N ¢ P,N is a union of conjugacy classes in P . Let {1} = K1,K2, ··· ,Kr be the P –conjugacy classes contained in N. If xi ∈ Ki, then |Ki| = |P : CP (xi)|. We know that

(∗) |N| = |K1| + |K2| + ··· + |Kr| = 1 + |K2| + ··· + |Kr| .

If we for all 2 ≤ i ≤ r have CP (xi) 6= P , then p| |Ki| = |P : CP (xi)| for all i, 2 ≤ i ≤ r. This means by (∗), that p | |N| − 1. As p | |N|, because N 6= 1, this is a contradiction. Thus there is an i 6= 1 with CP (xi) = P . Then xi ∈ N ∩ Z(P ) and xi 6= 1.

(9B) Corollary: If N ¢ P and |N| = p, then N ⊆ Z(P ). (9C) Corollary: (1) If |P | = p2, then P is abelian. (2) If P is non–abelian then p2 | |P : Z(P )|. Proof: (1) By (9A) Z(P ) 6= 1, so that |P/Z(P )| ≤ p. Thus P/Z(P ) ic cyclic, so that P = Z(P ) by (6S). (2) follows also from (6S). We have that |P : Z(P )| 6= 1, and if |P : Z(P )| = p then P/Z(P ) is cyclic, contradiction. (9D) Theorem: A p-group is nilpotent. Proof: Let P 6= 1 be a p-group. By (9A) Z1(P ) = Z(P ) 6= {1}. If Z1(P ) 6= P then P/Z1(P ) 6= {1}, a p-group. By (9A) Z(P/Z1(P )) = Z2(P )/Z1(P ) 6= 1 , so that |Z2(P )| > |Z1(P )|. If Z2(P ) = P, then P is nilpotent. Otherwise 3 2 m |Z (P )| > |Z (P )|, etc. Finally we get Z (P ) = P for a suitable m ≥ 1. (9E) Corollary: Let U 6= P be a subgroup in then p-group P . Then

U ⊂ NP (U) .

59 Proof: Use (9D) and (8H). (9F) Theorem: Let |U| = pa, U ⊂ P . (1) There exists a subgroup V in P satisfying U ¢ V , |V | = pa+1.

(2) If U ¢ P, then also V may bechosen as normal i P .

Proof: Let Q = NP (U), so that U ¢Q, U 6= Q (by (9E)). Choose an element x of order p in Z(Q/U). (x exists by (9A), as Q/U 6= {1}). The inverse image V of hxi in Q has the properties U ⊆ V , |V : U| = |hxi| = |x| = p, and V ¢ Q as hxi ¢ Q = Q/U. a+1 Thus |V | = |V : U| |U| = p|U| = p and U ¢ V , as V ⊆ NP (U) = Q. This shows (1). As furthermore V ¢ Q, we see that if U ¢ P, then Q = P and then V ¢ P showing (2). (9G) Corollary: If |P | = pn then P contains a en (normal) subgroup N a med |N| = p for 1 ≤ a ≤ n. (9H) Corollary: Let M ⊆ P , |P | = pn. We have

M maximal subgroup in P

m |M| = pn−1 . If M is maximal in P , then M ¢ P . Proof: ⇑ is triviel. ⇓ follows easily from (9F) (1). The last statement follows from the ﬁrst and (9E). Let us now consider the Frattini group Φ(P ) of G (see Chapter 8). Max(P ) denotes again the set of maximal subgroups in P . (9I) Theorem: The factor group P/Φ(P ) is an elementary abelian p–group. ∼ Proof: If M ∈ Max(P ), then P/M = Zp by (9H). In particular (i) P 0 ⊆ M and (ii) For all x ∈ P we have xp ∈ M (since x = xM has order 1 or p in G/M).

60 As (i) and (ii) hold for all M ∈ Max(P ) we get from the deﬁnition of Φ(P ) = T M that M∈Max(P ) (iii) P 0 ⊆ Φ(P )

(iv) For all x ∈ P xp ∈ Φ(P ).

Then (iii) shows that P/Φ(P ) is abelian (see (6A)), and (iv) shows that for p all x ∈ P/Φ(P ) we have x = 1. Thus P/Φ(P ) is elementary abelian.

(9J) Theorem: Let x1, ··· , xk ∈ P . In the above notation we have

hx1, ··· , xki = P

hx1, ··· , xki = P, where xi is the image of xi in the Frattini factor group P = P/Φ(P ). Proof: ⇓ er trivial. ⇑: Assume hx1, ··· , xki = P . Put

U := hx1, ··· , xk, Φ(P )i ⊆ P.

We claim that U = P . If x ∈ P then by assumption and (9I) we may write

a1 ak x = x1 ··· xk where 0 ≤ ai ≤ p − 1 , ie. a a x = x1 1 ··· xk k

a1 ak −1 or u := x(x1 ··· xk ) ∈ Φ(P ). But then

a1 ak x = x1 ··· xk u ∈ hx1, ··· , xk, Φ(P )i = U.

We have shown P = hx1, ··· , xk, Φ(P )i. From (8A) we then get P = hx1, ··· , xki, as desired. (9K) Theorem: (Burnside) Assume that |P : Φ(P )| = pr.

(i) Each generating set for P contains at least r elements.

61 (ii) A generating set for P with s elements may be “contracted” to a generating set for P with r elements (by deleting some of the s elements).

(iii) In particular there exists a generating set for P with r elements.

Proof: We need (9J) and some linear algebra! The “contraction” theorem for ﬁnite-dimensional vector spaces states that a set of spanning vectors for the vector space may be reduced to a basis by removing some of the vectors. Now P/Φ(P ) is an elementary abelian group of order pr, and may thus be considered as a vector space of dimension r over the ﬁeld Zp med p ele- menter. (See (6P)). If hx1, ··· , xsi = P , then hx1, ··· , xsi = P and therefore {x1, ··· , xs} is a generating set of vectors for the vector space P . Therefore it contains at least r = dim P elements. This shows (i). (ii) follows from the contraction theorem for vector spaces and (9J), and then (iii) is trivial.