Group theoryAbstract group theory

English summaryNotes by Jørn B. Olsson

Version 2007

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

1

CONTENT:

1. On normal and characteristic subgroups. The Frattini argument2. On products of subgroups

3. Hall subgroups and complements4. Semidirect products.

5. Subgroups and Verlagerung/Transfer6. Focal subgroups, Grun’s theorems, Z-groups

7. On finite linear groups8. 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 Matematik3 AL. (Algebra 2 and Algebra 3). The notes about group theory in Algebra3 are written in English and are referred to as GT3 in the following.

Content of Chapter 1Definition 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 sub-

group of inner automorphisms of G.

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

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

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

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

∀α ∈ Aut(G) ∀m ∈M : α(m) ∈M,

then 〈M〉 char G.

Examples of subsets which are closed under automorphisms include (p isa 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 〈K(G)〉 =G′, G’s commutator group. (See GT3).

If G is abelian, then E(G) is exactly G’s torsion subgroup GT . In this casewe actually have E(G) = 〈E(G)〉.

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 givenin GT3, Section 1.18..

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

Proof: ...

4

The Frattini argument

If H, P ⊆ G are subgroups, then

NH(P ) = {h ∈ H | hPh−1 = 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 thesubset

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 ap–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 inG, satisfying NG(P ) ⊆ K. Then K = NG(K).

Proof: ...

5

2. On products of subgroupsG2 - 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 1.39and 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 groupG 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 Pof 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 | = mis called a p–(Sylow) complement in G. (M is a p′–Hall subgroup. See chapter3.) 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. Howeverif P ∈ Sylp(G), P � G, then G contains a p–Sylow complement. This is aspecial 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 theleast common multiple (lcm) of |H| and |K| is |G| then HK = G.

Proof: Analandous to (2B). �

(2E) Theorem: Let |G| = p1a1 · · · pr

ar , 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–Sylowcomplement Ki. Then |

⋂i∈I

Ki| =∏j /∈I

pjaj .

Proof: Induction by |I|. For |I| = 1 we use the definition. Assume thatthe Theorem is shown for subsets I ′ ⊆ I, I ′ 6= I. Let I = I∗ ∪ {k}, k /∈I∗. Let K∗ =

⋂i∈I∗

Ki. By the induction hypothesis |K∗| = pkak

∏j /∈I

pjaj .

As |Kk| =∏j 6=k

pjaj we get, that lcm (|K∗| , |Kk|) = |G|. By (2A) |G| =

|K∗| |Kk|/|⋂i∈I

Ki|. The result follows by an easy calculation. �

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 ∩Nis a p–subgroup of N , containing Q. As Q is a p–Sylow group in N , we getQ = 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 thereforep - |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–Sylowcomplement in G. Then N ∩K et p–Sylow complement in N , and KN/N isa p–Sylow complement in G/N .

Proof: ... �

We may also use the above results to extend the claims of (1C). The nextTheorem is about 〈Dp(G)〉 in (1C).

(2H) Theorem: Let G be finite, p a prime and D := 〈g ∈ G | p - |g|〉. ThenD is a characteristic subgroup of G and D is the smallest normal subgroupin G whose index in G is a power of p.

Proof: By (1C) D char G. In particular D � G. Let P ∈ Sylp(G). Weclaim DP = G. If q 6= p is a prime and q | |G|, then D contains a q-Sylowsubgroup 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 toshow D ⊆ N . It suffices to show that if g ∈ G, p - |g|, then g ∈ N . Considerthe coset g = 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 isone of four characteristic subgroups of a finite group G, associated to a givenprime p.

8

Let p be a prime. A p-group is a group of p-power order. A p′-group isa group of order prime til p. Correspondingly we define p- and p′-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 apower of p (p-subgroup).

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

• Op′(G) is the largest normal subgroup i G, whose order |Op′(G)| isprime til p (p′-subgroup).

You of course need to convince yourself that these subgroups are welldefined. What is meant by smallest/largest? It could be by with respect totheir order (number of elements) or with respect to the inclusion order. Aswe shall see for these subgroups it does not matter which order relation wechoose.

Let us mention, that if p - |G|, then the only p-element in |G| is theidentity element 1. In that case Op(G) = Op′(G) = G and Op(G) = Op′(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 normalsubgroup in G, then HK is again a subgroup, as we have seen. We may thenapply all of Theorem (2A). If H and K are both normal, then so is HK andH ∩K is again normal. If for example |G : H| and |G : K| are both primeto 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 pis a normal subgroup with the same property. This subgroup is then Op′(G).Correspondingly you may argue with the other subgroups. For example HKis a normal p′-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 p′-subgroup, ie.it is Op′(G).

We now have:

(2I) Theorem: The subgroup Op(G) may be described in the following equiv-alent ways

9

(1) Op(G) =⋂

{A�G| |G:A| isa p−power}A

(2) Op(G) = 〈g ∈ G | g p′ − element〉

(3) Op(G) = 〈Q | Q ∈⋃

{q prime 6= p} Sylq(G)〉

(4) Op(G) = 〈Q | Q p′ − subgroup of G〉

(2J) Theorem: The subgroup Op′(G) may be described in the followingequivalent ways

(1) Op′(G) =⋂

{A�G | p - |G:A|}A

(2) Op′(G) = 〈g ∈ G | g p− element〉

(3) Op′(G) = 〈P |P ∈ Sylp(G)〉

(4) Op′(G) = 〈P |P p− subgroup of G〉

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

(1) Op(G) = 〈A�G |A er p− subgroup〉

(2) Op(G) =⋂

P ∈ Sylp(G) P

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

(1) Op′(G) = 〈A�G |A p′ − subgroup〉

Proof of the Theorems: That the groups are described by the property(1) follows from Theorem (2A), as explained above in the cases Op′(G) andOp′(G).

Proof for (2I): In Theorem (2H) it is shown that (1) and (2) are equivalent.Let X = 〈Q | Q ∈

⋃{q prime 6= p} Sylq(G)〉. As the elements in each of the

generating subgroups Q for X consists of p′-elements, (2) shows that X ⊆Op(G). On the other hand X � G, as the set of generating subgroups Q isclosed under conjugation. As X contains q-Sylow groups of G for all primesq 6= p, no prime q 6= p can divide |G : X|. Thus |G : X| is a power of pand 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 =⋂

P ∈ Sylp(G) P. Assume that A � G is a p-group.

Let P ∈ Sylp(G). Then the subgroup AP also has p-power order by (2A). AsP er en Sylow group we get AP = P, ie. A ⊆ P. Thus A ⊆ X. By (1) we seethat Op(G) ⊆ X. Since the set of p-Sylow groups is closed under conjugationwe have X �G. Thus X ⊆ Op(G). �

You may of course combine Op, Op′ , Op, Op′ . For example Op(Op′(G)) isOp(H), where H = Op′(G). This subgroup is also denoted Opp′(G). Youmay then continue with Op′pp′(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 Op′p(G), defined as follows:Op′p(G)/Op(G) = Op′(G/Op(G)) ClearlyOp′(Op′(G)) = Op′(G) andOp(Op(G)) =Op(G).

(2M) Theorem:

(1) We have Op′(G) ⊆ Op(G) and Op(G) ⊆ Op′(G).

(2) Op′(G) = Op(G) ⇔ G has a normal p-complement.

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

Proof: (1) As p - |Op′(G)| (2I)(4) shows that Op′(G) ⊆ Op(G). The otherinclusion in (1) follows from (2J)(4).

(2) ⇒: Assume Op′(G) = Op(G) = K. We claim that K is a normal p-complement in G. Write |G| = pam, with p - m. It follows from (2I)(3), thatm | |Op(G)| = |K|. As K also equals Op′(G), which is a p′-group, we havem = |K|. Thus K is a normal p-complement.

⇐: If K is a normal p-complement in G, we easily get K = Op(G) =Op′(G).

(3) Write again |G| = pam. If Op(G) = Op′(G) = P, then P is a normalp-Sylow group in G : As |G : Op′(G)| | m, we have pa | |Op′(G)| = |P |.Since P = Op(G) is a p-group, we get |P | = pa. Conversely, if P is a normalp-Sylow group in G, then P = Op(G) = Op′(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)

transversal to M in G if G =⋃

x∈XxM. Thus X contains exactly one element

11

from every coset of M in G. Remark that G = XM and |X| = |G : M |. IfM has a complement in G, then this complement is also a transversal to Min G. We later need the following:

(2N) Theorem: Let N ⊆ M ⊆ G be subgroups. If X is a transversal forM in G and Y a transversal for N in M , then XY is a transversal for N inG.

Proof: We have M =⋃

y∈Y yN and G =⋃

x∈XxM , whence

G =⋃

x∈XxM =

⋃x∈X

x(⋃

y∈YyN) =

⋃t∈XY

tN.

You may also consider right transversals to a subgroup of G. X ⊆ G is aright transversal to the subgroup M ⊆ G if G =

⋃x∈XMx.

In analandy with (2N) we have

(2O) Theorem: Let N ⊆ M ⊆ G be subgroups. If X is a right transversalfor M in G and Y a right transversal for N in M , then Y X is a righttransversal for N in G. �

Remarks: 1◦. X is a (left) transversal to M in G⇔ X−1 is a right transver-sal to M in G, since we have (xM)−1 = Mx−1.

2◦. IfM�G, then a right transversalX toM in G is also a left transversaland vice versa. If the subgroup M is not normal in G, then you may alwaysfind a left transversal til M in G, which is not a right transversal to M in G.(Exercise).

12

3. Hall subgroups and complementsG3 - 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. Alsothe 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 writen = p1

a1 · · · pkak , where p1, · · · , pk are different primes, and a1, · · · , ak ∈ N.

We definenπ =

∏{i|pi∈π}

piai , nπ′ = n/nπ .

(Thus if n = 60, π = {2, 3}, then nπ = 12, nπ′ = 5.) Remark, that if π is theset of primes, which are not in π, then nπ′ = nπ. We define n∅ = 1.

A π–Hall subgroup in the groupG is a subgroup of order |G|π. For examplethe p–Sylow groups are also π–Hall subgroups, where π = {p}. A π′–Hallsubgroup in G is a subgroup of order |G|π′ .

Eksemple: The alternating group A5 has no {3, 5}–Hall subgroup. Sucha 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 thecycle–structure of such an element).

Remark: If M is a π–Hall subgroup of G and there exist also a π′–Hallsubgroup N i G, then N is a complement to M in G. (See the previousChapter).

Generally Sylow’s Theorems may thus not be extended to statementsabout the existence of Hall subgroups. But a famous Theorem of Philip Hallshows 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 thisway, 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 subgroupof order m.

Remark: There is also a statement on the number of Hall subgroups of Gof order m (corresponding to Sylow’s 3. Theorem.) The statement is thatthis number is a product of factors ai, where for each ai there exists a primedivisor 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 fulfilledsolvable groups. The proof of this is based on the following Theorem, whichmay be proved using representation theory. There is also a purely grouptheoretic 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 pand 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 containssubgroups 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 allprimes p with p | |G|. Then G is solvable.

14

Remark: A p–complement is a π′–Hall subgroup, where π = {p} and is alsocalled a “p′–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 orderm for enhver factorization |G| = mn, (m,n) = 1 of G’s order. Then G issolvable.

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 allchoices of the set of primes π, when G is solvable. But a non-solvable groupmay under certain conditions contain Hall subgroups for special choices ofπ. An example of this is Burnside’s Theorem on the existence of a normaltp–complement (shown in Chapter 6). Here we prove the Schur–ZassenhausTheorem, stating that if G has a normal π–Hall subgroup, then this hascomplement (not necessarily normal) which is then a π′–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 laterTheorem (3I) er “cohomolandical” and the map t in the proofs is a “2-cocycle”. You may find much more about cohomolandy theory for groups inmany 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, HansZassenhaus 1912–1991. Schur proved (3F) which ie really the hard part, andZassenhaus then proved (3G) on the basis of (3F)).

Theorem (3G): (Schur–Zassenhaus Theorem) Let M be a normal Hall sub-group 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 toM in G are conjugate in G. This is the case, but we cannot prove it here. Youmay prove without much difficulty that if M �G is a Hall subgroup, and ifeither M or G/M is solvable, then all complementer to M in G are conjugatein 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 issolvable. But if M � G is a Hall subgroup, then (|G : M |, |M |) = 1 andthus |G : M | or |M | must be odd. Thus according to Feit–Thompson, eitherG/M or M be must be solvable. Feit–Thompson’s proof is more than 250pages long (Pacific Journal of Mathematics (1963)). There is a monandraphwhich treats an essential part of the proof (H. Bender–G. Glauberman: Localanalysis for the odd–order theorem).

The next quite remarkable theorem does not appear to have anything todo with Theorem (3F), but the proofs are quite similar. They have the exis-tence of a normal abelian subgroup of G in common and both are containedin a more general Theorem of Gaschutz, see for example Hauptsatz 17.4 iHuppert: Endliche Gruppen I, p. 121.

(3I) Theorem: (Gaschutz) 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 examplesG4 - 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 GoA, (the semidirect product of G with/by A).The underlying set is G× A. The composition is defined by

(g, ϕ)(g′, ψ) = (gϕ(g′), ϕψ)

for g, g′ ∈ 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 GoAut(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 homomorphismα : H → Aut(G). We form a new group calledGoαH (the semidirect productof G by H, relative to α). The underlying set is G×H, and the compositionis defined by

(g, h)(g′, h′) = (gα(h)(g′), hh′).

(You may here consider “conjugation of g′ by h”, ie. hg′h−1 in this groupis being replaced by α(h)(g′). (For elements in an arbitrary group we knowthat ghg′h′ = g(hg′h−1)hh′.)

In Goα 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 specialcase of the second.

17

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

If α is clear from the context we often just write GoH instead of GoαH.

You may want to realize a given group as a semidirect product:On the one hand we have: If X = Goα 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 inverseelements. 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 thatY 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 byH: For h ∈ Hwe let α(h) be the restriction of the inner automorphism κh of Y to G. Wethus 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 formX = Goα H.

How are X and Y related? Here is the answer:

(4A) Theorem: Let Y = GH and X = Goα 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, g′ ∈ G, h, h′ ∈ H. We have

ρ((g, h)(g′, h′)) = ρ(gα(h)(g′), hh′) (multiplication in X)

= ρ(g(hg′h−1), hh′) (definition of α(h))

= g(hg′h−1)hh′ (definition of ρ)

= ghg′h′ (cancel h−1h)

= ρ(g, h)ρ(g′, h′) (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 whenx 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 = 〈(1, 2)〉 n ≥ 2.

Another example we have seen is the result of (3G). If G has a normalHall subgroup M, then there exists a complement L to M in G. Thus G isa semidirect product of M with L.

Let us consider various other examples.

(4B) Example: Dihedral groups. If G = 〈g〉 is a cyclic group, then the mapι : g 7→ g−1 is an automorphism of G. In this situation Go 〈ι〉 is a dihedralgroup. If |G| = n ≤ ∞, we denote Go 〈ι〉 by Dn. We then have that |Dn| =2n. The group Dn is generated by two elements (g, 1) and (1, ι). Let usremark that |(1, ι)| = |(g, ι)| = 2 since (g, ι)2 = (gι(g), 1) = (gg−1, 1) = 1. AsDn is generated by (g, ι) and (1, ι), we see that a dihedral group is generatedby 2 elements of order 2 (so-called “involutions”).

On the other hand a group generated by 2 involutions is isomorphic toa dihedral group. This is seen as follows. Assume that D = 〈x, y〉, wherex2 = y2 = 1, x 6= y. We then have that x−1 = x and y−1 = y. Put g = xy.

19

Then G := 〈g〉 is cyclic and D = 〈g, y〉. Furthermore ygy−1 = yxyy−1 =yx = y−1x−1 = (xy)−1 = g−1. Thus conjugation by y correponds to the mapι above.

When n ∈ N, n ≥ 3, then Dn may be realized as a subgroup of Sn, if weconsider the subgroups

D∗n = 〈(1, 2, · · · , n), τ = (1, n)(2, n− 1) · · ·〉

of Sn. If G = 〈(1, 2, · · · , n)〉, H = 〈τ〉 then D∗n = GH and G� D∗

n, 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∼= D∗

n (use (4A)). We also have that D∗3∼= S3, as they

have the same order. Let us remark that for n = 4 we have |D∗4| = 8, so that

D∗4 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 mayform a “di-abelian” group Go 〈ι〉 of order 2|G|. In the following we do notdistinguish between the abstract group Dn and the concrete permutationgroup D∗

n.

(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 calledGL(n,R) (= {A ∈ Rn

n | detA invertible in R}). When π ∈ Sn we define amatrix P (π) ∈ Rn

n by

P (π) = [aij] where aij = δiπ(j)

(δ is “Kronecker delta”). Clearly P (π) has exactly one element 6= 0 in eachcolumn and in each row. Repeated application of the column rule for deter-minants show that detP (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

cij =∑

k

δiπ(k)δkρ(j) 6= 0 ⇔

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−) permutationmatrix. We have:

detP (π) = sign(π), (π′s sign)

P (π)t = P (π−1) 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 col-umn. Thus detP (τ) = −1, when τ is a transposition. Clearly we also haveP (τ) = P (τ)t, when τ is a transposition. If π = τ1τ2 · · · τk, where all τi aretranspositions then

det(P (π)) =n∏

i=1

det(P (τi)) = (−1)k = sign(π)

andP (π)t = [P (τ1) · · ·P (τk)]

t = P (τk)t · · ·P (τi)

t

= P (τk) · · ·P (τ1) = P (τk · · · τ1) = P (π−1).

A permutation matrix consists of zeros except exactly one 1 in each row andeach column. You may now replace the numbers 1 in a permutation matrixP (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” matrixwith elements from the set G ∪ {0} (when we define that δiig = g, δijg = 0for 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 com-

position and the usual rule for matrix multiplication induces a compositionon 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

cij =∑

k

δiπ(k)giδkρ(j)hk = δiπρ(j)gihρ(j) = δiπρ(j)gihπ−1(i)

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 withP (1, · · · , 1; (1)) as neutral element and where

P (g1, · · · , gn; π)

hasP (g−1

π(1), · · · , g−1π(n); π

−1)

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 toG× · · · ×G︸ ︷︷ ︸n

) with the subgroupA∗ = {P (1, · · · , 1;π) |

π ∈ A} (which is isomorphic to A.)

(4D) Example: Let us look at the monomial group Mon(G,A) (as a semi-direct product) from the“outside”.

If A ⊆ Sn and G∗ = G× · · · ×G︸ ︷︷ ︸n

(= Gn), we may define a homomorphism

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 anantihomomorphism. 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 thatki = 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

G∗ oα Aϕ∼= 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 indicesof the hi’s. This may be avoided by “interchanging ” G and A, as we do inthe next example.

(4E) Example: (Wreath product, Kransprodukt). As in (4D) A ⊆ Sn andG∗ = G× · · · ×G︸ ︷︷ ︸

n

. We define a composition on A×G∗ by

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

ThenA×G∗ is a group with ((1); 1, · · · , 1) as a neutral element, and (π; g1, · · · , gn)has (π−1;h1, · · · , hn) as inverse element, where hi = g−1

π−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 replacedby “gj”. If we “multiply” P ∗(π; g1, · · · , gn) and P ∗(ρ;h1, · · · , hn) we getP ∗(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 usingG 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 mayform 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 isclear that Gop is a group with the same neutral element and same inverseelements. Moreover the map ι : g 7→ g−1 is an isomorphism between G andGop, 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

ψ : (π; g1, · · · , gn) 7→ (g−11 , · · · , g−1

n ; π−1)

is defined an isomorphism between G oA and (G∗ oαA)op. Thus we also have

G o A ' G∗ oα A.

Proof: We have

ψ(π; g1, · · · , gn) ◦ ψ(ρ;h1, · · · , hn) = (h−11 , · · · , h−1

n ; ρ−1)(g−11 , · · · , g−1

n , π−1)

24

= (h−11 g−1

ρ(1), · · · , h−1n g−1

ρ(n); ρ−1π−1) = 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) asubgroup 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 bea 6 × 6-permutation matrix. Let us calculate this and the correpondingpermutation. First we consider the permutation matrix P (π)

P (π) =

0 0 11 0 00 1 0

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

P ∗(π;P (g1), P (g2), P (g3)) =

0 0 0 0 0 10 0 0 0 1 00 1 0 0 0 01 0 0 0 0 00 0 1 0 0 00 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. Weget

ρ =

(1 2 3 4 5 64 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}. Theng1g2g3 = (1, 2)(3)(4)(5, 6) = (1, 2)(5, 6). Next you consider π = (1, 2, 3) asa permutation ∆(π) of 6 which permutes the sets {1, 2}, {3, 4}, {5, 6} by1 → 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) meansintuitively that we are dealing with a “duplicartion” of π. If ρ = (1, 3) ∈ S3

then ∆(ρ) = (1, 5)(2, 6).)

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

number such that pν

p(n) | n. For example ν3(72) = 2 as 72 = 32 · 8. Letus consider νp(|Spa|), a ≥ 1. We have νp(|(Spa|) = νp(p

a!). It is clear thatpa−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. Bycontinuing with p3 etc. we get

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

For a = 1, a p–Sylow group in Sp is cyclic, generated by for example(1, 2, · · · , p), ie. ' Zp. By (4J) |Zp o Zp| = pp+1. Thus Zp o Zp has thesame 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 inductionover a we see that Xa is isomorphic to a subgroup of Spa (use (4H)) andthat νp|Xa| = νp(|Spa|) = νp(p

a!) (use (4J)), such that the p–Sylow groupof 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 hasan elementary abelian subgroup of order pp, namely

〈(1, 2, · · · , p)〉 × 〈(p+ 1, · · · , 2p)〉 × · · · × 〈(· · · , p2)〉.

This corresponds to the subgroup G×· · ·×G in the generel case. The groupA becomes in this example the group generated by

〈(1, p+1, 2p+1, · · · , (p−1)p+1)(2, p+2, · · · , (p−1)p+2) · · · (p, 2p, · · · , p2)〉

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 〈(1, 2)〉×〈(3, 4)〉 andA = 〈(1, 3)(2, 4)〉.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 elementsis 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.

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

Then in Sn the p–Sylow group is isomorphic to

Xa11 ×Xa2

2 × · · · ×Xakk ,

where Xiai = Xi × · · · ×Xi︸ ︷︷ ︸

ai

and Xi is as in the previous example.

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

CG(a) = {g ∈ G |gag−1 = a}.

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

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

27

“homocyclic” elements, ie. elements which is a product of cycles of the samelength.

Let k, ` ∈ N and assume that κ ∈ Sk` is a product of k disjoint cycles ofthe same length `. We explain that

CSk`(κ) ∼= Z` o Sk,

where Z` ıs a 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`

ϕκϕ−1 = (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 alsoϕ(ai2), · · · , ϕ(ai`) are determined since the order in cycles must be respected:If ϕ(ai1) = ai′j′ where 1 ≤ i′ ≤ k and 1 ≤ j′ ≤ `, then also

(∗) ϕ(aij) = ai′(j′+j−1) for 1 ≤ j ≤ `,

where the second index is calculated modulo `. An element ϕ ∈ C(κ) is thuscompletely 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 differentcycles 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 apermutation of 1, · · · , k, ie. π ∈ Sk. By definition of ∆(π−1) we then getfrom (∗∗) and (∗ ∗ ∗)

∆(π−1)ϕ(ai1) = aiti .

It is clear that ∆(π−1) = ∆(π)−1, 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. Thusalso the element ψ′ defined by

ψ′ = zt1−11 zt2−1

2 · · · ztk−1k ∈ C(κ).

Now for 1 ≤ i ≤ k

ψ′(ai1) = zti−1i (ai1)u (why?)

= aiti (why??)

We conclude that ψ = ψ′ and get that

ϕ = ∆(π) · zt1−11 · · · ztk−1

k .

If we define a mapα : Zk o Sk → C(κ)

byα(π; s1, · · · , sk) = ∆(π) · zs1

1 · · · zskk ,

where the si’s are calculated modulo ` = |zi|, then α is surjective by theabove. It is easy to see that α is a homomorphism, and that the kernel of αis 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 ofG inG∗ = G×· · ·×G(n gange) operate on pairwise disjoint subsets of {1, · · · ,mn}, where each ofthese subsets have m elements. By “duplication” the elements of A are blownup 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 sub-groups of the direct product of two groups. This parametrization of thesubgroups is usually not found in textbooks on group theory although it isfairly 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, andϕ 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:If (g, h), (g1, h1) ∈ UT we have ϕ(gG2) = hH2 and ϕ(g−1

1 G2) = h−11 H2, since

ϕ is a homomorphism. Thus we get that

ϕ(gg−11 G2) = ϕ(gG2)ϕ(g−1

1 G2) = hH2h−11 H2 = hh−1

1 H2

ie. (gg−11 , hh−1

1 ) = (g, h)(g1, h1)−1 ∈ UT .

It is also clear that if T, T ′ ∈ U(G,H) and T 6= T ′ (ie. if at least one ofthe five “coordinates” in T and T ′ is different ), then UT 6= UT ′ .

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, sa g, h) ∈ U}

G2 = {g ∈ G | (g, 1) ∈ U}

H1 = {h ∈ H | There exists g ∈ G, sa (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 Hand 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,

ie. hxh−1 ∈ 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(g, h)−1(g, h1) = (1, h−1h1) ∈ U so that h−1h1 ∈ H2. Thus hH2 = h1H2. Wesee that ψ : g 7→ hH defines a map from G1 → H1/H2. It is clear that thismap 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 isomorphismϕ : 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 thanhave

|UT | = |G1||H2| = |G2||H1|,

if the groups are finite. (Why?)

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

(G1, G1, H1, H1, 1).

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

31

5. Subgroups and VerlagerungG5 - 2007-version

Some parts of this Chapter are also discussed in GT3, Chapter 1.16.

(5A) Remark: Let us consider a wreath product G oA as in (4E). The map

(π; g1, · · · , gn) 7→ π

is a en homomorphism from G oA onto A with G∗ as a kernel. But generallynone 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)} whichis uninteresting. However the third map may be a homomorphism, when Gis 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 bypG/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 .

ThereforepG/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 Chapter2),

(∗) G =•⋃

gi∈T

giH.

Let x ∈ G. By (∗) there exists for 1 ≤ i ≤ n a uniquely determinedinteger αT

x (i) ∈ {1, · · · , n}, and a uniquely determined element hTxi ∈ H such

thatxgi = gαT

x (i)hTxi .

�

Let us try to replace T by another transversal T ′ = {g1h1, · · · , gnhn}where hi ∈ H, and for x ∈ G compare αT

x with αT ′x , and hT

xi with hT ′xi . We

have

(5C) Lemma: In the above notation we have

αTx = αT ′

x .

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

αTx = αx).

MoreoverhT ′

xi = h−1αx(i)h

Txihi .

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

αTx (i) = j = αT ′

x (i) ,

ie. αTx = αT ′

x . As xgi = gαx(i)hTxi, we get

x(gihi) = gαx(i)hTxihi

= (gαx(i)hαx(i))(h−1αx(i)h

Txihi) ,

33

and thushT ′

xi = h−1αx(i)h

Txihi .

�

(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 Gon the cosets of H) with kernel

K =⋂g∈G

gHg−1 .

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) =j = αx(i

′) then xgi ∈ gjH and xgi′ ∈ gjH, so that

g−1i gi′ = (xgi)

−1(xgi′) ∈ H .

This means giH = gi′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

⇔ x ∈ giHg−1i for 1 ≤ i ≤ n

⇔ x ∈n⋂

i=1

giHg−1i =

⋂g∈G

gHg−1 .

�

(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 kerα, where α is as in (5D), since α induces aninjective homomorphism from G/ kerα into Sn. Thsu |G/ kerα| = |G/N | ||Sn| = n! �

(5F) Example: Let n ≥ 5. Then the alternating group An has no subgroupof index k, when 1 < k < n. Otherwise the simple group An would containa 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 indexn). �

(5H) Example: If G has a subgroup H of index 3, then G has also a normalsubgroup of index 2 or 3. Either H �G or G has a normal subgroup N withH ⊆ N , |G : N | = 3! = 6. In that case G/N ' S3 which has A3 as a normalsubgroup of index 2. Thus G also has a normal subgroup of index 2. �

(5I) Remark: If we chooseH = {1} in (5D), and G is finite then αG/H = αG

is a monomorphism G → S|G|, which is called the regular representation ofG.

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

VT (x) = (αx;hTx1, · · · , hT

xn)

is defined a monomorphism VT : G→ H o Sn.

Proof: Let x, y ∈ G. As ygi = gαy(i)hTyi we get

(xy)gi = x(ygi) = (xgαy(i))hTyi = (gαx(αy(i))h

Txαy(i))h

Tyi

such thatVT (xy) = (αxαy;h

Txαy(1)h

Ty1, · · · , hT

xαy(n)hTyn)

= (αx;hTα1, · · · , hT

xn)(αy;hTy1, · · · , hT

yn) = 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 suchthat 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 ′ is another transversal it must beshown that pH/K ◦ VT = pH/K ◦ VT ′ . This follws easily from (5C). �

We use the German term “Verlagerung” in these notes. In English text-books it is often called “transfer”.

36

6. Focal subgroups, Grun’s Theorems,Z–groups

G6 - 2007-version

First a couple of general remarks. As usual G′ = [G,G] denotes thecommutator subgroup of the group G.

(6A) Lemma: Let H be a subgroup of G. We have

G′ ⊆ 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

D20 = 〈x, y | x20 = 1, y2 = 1, yxy−1 = x−1〉 .

It is easy to see that D40’s commutator group is D′40 = 〈x2〉, |D′

40| = 10. Thecommutator factor group D40/D

′40 is Klein’s four group Z2 × Z2. By (6A)

the list of normal subgroups N / D20, with D20/N abelian is as follows:

D20, 〈x〉, 〈x2, y〉, 〈x2, xy〉, 〈x2〉 .

In this list the 1., 3. and 4. group are dihedral groups and the other arecyclic. (Why?)

In this Chapter G is a finite group.

A p–focal subgroup in G is a p–Sylow group i G′. By (2F) the p–focalsubgroups 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 order5. The p–focal subgroups are important because they may under certaincircumstances be described “locally” using certain “proper” subgroups in G.This is particularly important in the study of non-abelian simple groups. Insuch a group G′ = G, and the p–focal subgroup are exactly the p-Sylowgroups in G. Generally we have

(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 ∩G′ .

37

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

If x, y ∈ G and H ⊆ G is a subgroup we write x ∼H y, if there existsa h ∈ H so that hxh−1 = y, and we call x and y conjugate i H. It is clearthat ∼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 G′ = 〈[x, y] | x, y ∈ G〉.

The next Theorem is very useful.

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

P ∩G′ = 〈xy−1 | x, y ∈ P , x ∼G y〉 .

Proof D. Gorenstein: Finite groups, 7.3.3-7.3.4

Definition: Let K ⊆ L ⊆ G be subgroups. We say that L controls fusionin 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 thatgCG(a)g−1 = CG(b). As P is abelian, we have P ∈ Sylp(CG(a)) and P ∈Sylp(CG(b)). By the above we also have

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

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

38

(6F) Theorem: If P ∈ Sylp(G), and if L ⊇ P controls fusion in P , then

P ∩G′ = P ∩ L′ .

Proof: By assumption

∀ x, y ∈ P : x ∼G y ⇔ x ∼L y .

Thus by (6D) P ∩G′ and P ∩ L′ have the same generators so that

P ∩G′ = P ∩ L′ .

�

(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 = 〈(1, 2, 3, 4), (1, 4)(2, 3)〉 ∈ 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).

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 elementin (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

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

39

Corollary (6G) may be generalized as follows:

(6I) Theorem: (Grun’s Theorem) Let P ∈ Sylp(G). We have

P ∩G′ = 〈P ∩NG(P )′ , P ∩ g−1P ′g | g ∈ G〉 .

Remarks: If P is abelian then P ′ = {1} and thus P ∩ g−1P ′g = 1 for allg ∈ G. Therefore (6I) implies (6G). We again have to consider Verlagerunginto an abelian factor group of P , and need to choose coset representatives forP in G in a special way. You start by dividing G into double cosets moduloP so we need to add some facts about double cosets. This is done generallyhere.

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 thefollowing 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 aresult 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| rightcosets 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(2A) to the subgroups H and gKg−1 of G. �

When u ∈ P , P ∈ Sylp(G), g∗ ∈ G we give a description of some cosetrepresentatives in Pg∗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) Pg∗P = PgP .

(2) PgP =r⋃

i=0

pti−1⋃j=0

ujaigP .

40

(3) For all i we have g−1a−1i uptiaig ∈ P

(4) For all i we have g−1uptig ∈ P .

(5)r∑

i=0

pti = |P : P ∩ gPg−1|.

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 iscalled 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 direc-tion.

(6L) Theorem: (Grun’s 2. Theorem) Let G be p–normal. Let P ∈ Sylp(G)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 Ghas a normal p–complement K.

Proof: We show that P ∩G′ = {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 ∩G′ = P ∩N ′ .

But P ∩N ′ may be calculated by (6D). As P ⊆ Z(N) we get for x, y ∈ P thatx ∼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 faktorgroup NG(L)/CG(L) is isomorphic to a subgroup of Aut(L).

Proof: If x ∈ NG(L)αx : ` 7→ x`x−1

defines an automorphism of L. The map x → αx is a homomorphism fromNG(L) into Aut(L) with CG(L) as kernel. �

(6P) Remark: If L is cyclic then Aut(L) is abelian. If L is an elementaryabelian p–group of order pn then L may be considered as a vector space overthe field Zp with p elements, dimL = n. The homomorphismes from thegroup L to itself “correspond” then to linear maps of the vector space L. Itfollows that Aut(L) ' GL(n,Zp), the set of invertible n × n–matrices withcoefficients from Zp.

(6Q) Theorem: Let p be smallest prime dividing |G|. If P ∈ Sylp(G) iscyclic, then G has a normal p–complement. In particular a group with acyclic 2–Sylow group always has a normal 2–complement.Proof: By (6N) NG(P )/CG(P ) is isomorphic to a subgroup of Aut(P ). ButAut(P ) is an abelian group of order pn−1(p − 1), if |P | = pn. (See eg. GT3,Chapter 1.13) As pn | |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 alternatinggroup 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 haven2(G)|15 = |G : P |. By (5E) we have n2(G) ≥ 5, so that n2(G) ∈ {5, 15}.If n2(G) = 15, then P = NG(P ) as |NG(P )| = 60

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, orwhat G(j) is for j ≥ i + 2. We may therefore assume that i = 1, and thatG(i+2) = G(3) = {1}. We then know that G(2)/G(3) = G(2) = 〈a〉 is cyclic. AsG(2) = G′′ = 〈a〉�G, we have NG(〈a〉) = G. Let X = CG(a). By (6N) G/Zis isomorphic to a subgroup of Aut(〈a〉). Also Aut(〈a〉) is abelian ((6P)).Thus G/X is abelian. This means that G′ ⊆ X. We then have

G′′ = 〈a〉 ⊆ G′ ⊆ X = CG(a) .

As G′ ⊆ CG(a), we get G′′ = 〈a〉 ⊆ Z(G′). Thus, by an isomorphismtheorem,

G′/Z(G′) '(G′

/〈a〉

) /(Z(G′)/〈a〉) .

As G′/〈a〉 = G′/G′′ is cyclic, we get that G′/Z(G′) is cyclic. By (6S) G′ isabelian, so that G(2) = (G′)′ = {1}. Thus G(2) = G(3)(= {1}), as desired. �

We can now give a description of the finite groups, where all Sylow groupsare cyclic (for all primes). Such a group is called a Z-group. (Z is anabreviation for Zassenhaus. Hans Zassenhaus 1912–1991. (See also Chapter3)).

(6U) Theorem: A Z–group is solvable.

Proof: Let G be a Z–group. The proof is by induction by |G|. Let p bethe smallest prime divisor in |G|. By (6Q) G has a normalt p–complementK. Thus G/K is a (cyclic) p–group, ie. solvable, and K is solvable bythe induction hypothesis (See also Theorem (2F), which shows that K is aZ–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 = AoB ,

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/G′ and G′ are cyclic.Let |G′| = m, |G : G′| = n, so that |G| = mn. Let G′ = 〈a〉 and choose b ∈ Gwith G/G′ = 〈bG′〉. We have G = 〈a, b〉. As 〈a〉 � G, we get b−1ab = ar,where 〈ar〉 = 〈a〉 (so that (r,m) = 1). Furthermore

a−1b−1ab = ar−1 .

If X = 〈ar−1〉 then G/X = 〈aX, bX〉. As [a, b] ∈ X, G/X is abelian, so thatG′ ⊆ X. Hence G′ = X = 〈ar−1〉. Since |a| = |ar−1|, we have (r− 1,m) = 1.As G/G′ = 〈bG′〉 has order n, we have bn ∈ G′, 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 getas = 1. Thus bn = as = 1 and |b| = n. We can now show (m,n) = 1: Ifp|m, p|n, p a prime and m = m1p, n = n1p, then am1 and bn1 are elements oforder p, and as 〈a〉 ∩ 〈b〉 = {1} we see that 〈am1 , bn1〉 is an abelian group oforder p2, which is not cyclic. This is impossible as G’s p–Sylow groups arecyclic. Thus (m,n) = 1. �

We have shown that G is a semidirect product of A = G′ = 〈a〉 andB = 〈b〉. Let us remark that as (m,n) = 1, A and B are Hall subgroups inG.

In the case where all Sylow groups have prime order there is a strongerstatement than (6W):

(6X) Theorem: Assume that |G| = p1p2 · · · pr, where p1 < p2 < . . . < pr

are primes. ThenG = AoB ,

where there exists an s, 1 ≤ s ≤ r such that

|B| = p1 · · · ps , |A| = ps+1 · · · pr .

44

7. Groups of a given finite orderG7- 2007-version

This chapter is essentially different from the other chapters. It is inspireredby 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 mayfor instance want to show that such a group cannot be simple.We use Sylow’s Theorems and results from the previous 2 Chapters. First wemake a number of Remarks, which might be useful for this type of problemsand 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 groupsof 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 orderp.

(This follows from Sylow’s Theorem.)

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

(We know that as |P | = p, none of the np(G) p-Sylow groups can haveelements 6= 1 in common. Each Sylow group contais (p − 1) elements6= 1.

This counting argument does not work, if |P | > p, unless you can showthe 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. “TrivialIntersection.” 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 countp-elements, see (II)), you may do this indirectly by considering a Sylowgroup 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) thesubgroup R of P (or P ∗) cannot be equal to its own normalizer in P (orP ∗.) You then know that NG(R) must contain at least (p+ 1) p-Sylowgroups, since the p-subgroups NP (R) and NP ∗(R) of NG(R) cannot becontained in the same p-Sylow group T ofNG(R), due to the maximalityof 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 thereis a chance that NG(R) may be a very large subgroup of G. See alsoRemark (XI).

(IV) If G is simple and p is the largest prime dividing |G|, then G has nosubgroup U of index k, 1 < k < p.

(If it did then G would have a normal subgroup N N ⊆ U ⊆ G,with G/N isomorphic to a subgroup of Sk. (Corollary (5E)) It is clearthat N = 1. As p | |G|, but p - k!, because k < p, we would get acontradiction.)

(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 nosubgroup 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 isisomorphic to a subgroup of the alternating Ak.

(We know that G is isomorphic to a subgroup H of Sk. But actually Hmust 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

|NAk(P )| = 1

2|NSk

(P )|.

(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|NAk

(P )| = 12p(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 knowthat every group of order pq is cyclic). Assume that P ∈ Sylp(G) andQ ∈ 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. ThenP ∗Q is a subgroup of order pq in G, which then is cyclic. Therefore Qand P ∗ centralizes each other so that Q ⊆ CG(P ∗) ⊆ NG(P ∗). As Pand 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 thanthe one in (IX). Assume that P is a p-subgroup and at the primeq 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 thatnq(NG(P )) = 1, then NG(P ) ⊆ NG(Q), where Q ∈ Sylq(NG(P )). Theremy be other subgroups in NG(Q), eg. if Q 6∈ Sylq(G), which may makethis subgroup very large. Then you may apply for example (IV)-(VI).

(XI) Assume that np(G) 6≡ 1(mod p2). Then there exist p-Sylow groups P, P ∗

in G, such that R = P ∩ P ∗ has index p in P and in P ∗. In particularthen NG(R) contains at least (p+ 1) p-Sylow groups.

(This is seen as follows: Choose P ∈ Sylp(G). Consider for P ∗ ∈Sylp(G) the set {P ∗x|x ∈ P}, ie. the set of P -conjugate subgroupsto 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 ∗| Sylowgroups. The assumption np(G) 6≡ 1(mod p2) shows then that thereexists a p-Sylow group P ∗ 6= P, such that p2 - |P : P ∩ P ∗|. In thatcase 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 acontradiction, you may here ’count elements’. It is usually good to startwith the largest prime. Remark that if G is simple, then np(G) 6= 1 forp = 2, 3, 11. As n11(G) | 12, we egt n11(G) = 12, so G has 12 · 10 = 120elements of order 11 by (I). As n3(G) ≥ 4, G has at least 4 · 2 = 8 element´sof order 3 by (I). In total you have at least 128 elements of order 11 and 3 inG. Only 4 elements remains. Among these 4 elements must be all elementsin 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 notsimple: 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.We get np(G) = q, so that |NG(P )| = p2. This shows that NG(P ) = P, and asP 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 musthave order q.Alternatively you may by elementary number theoretic considerations showthat 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 Gis simple, G cannot contain a subgroup U of index 1 < k < 29, by (IV). Butn3(G) = 13, as all other divisors 6= 1 of 13 · 31 are not congruent to 1 (mod3). 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, whichis congruent to 1 modulo 11, we have n11(G) = 12. G and its subgroupsare 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. Anelement of order 33 in a symmetric group needs at least 11+3=14 elementsto 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 asubgroup of A7, by (VI). But 25 - |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 = 3·7·13and |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(I), so that n3(G) 6≡ 1 (mod 32). By (XI) there exists P ∗ ∈ Syl3(G), sothat R = P ∩ P ∗ has order 33. Let X = NG(R). Then X contains at least 23-Sylow groups, namely P and P ∗. Thus |P | is a proper divisor of |X|, sinceotherwise 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.

Proof: Assume G simple. We get n7(G) = 15 6≡ 1 (mod 72). Choose by(XI) P, P ∗ ∈ Syl7(G), so that R = P ∩ P ∗ has order 7. Thus we have forX = 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 generatedby 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 setof maximal subgroup in G. If G 6= {1} is a finite group, then Max(G) 6= ∅,and we put

Φ(G) =⋂

M∈Max(G)

M.

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 〈X〉 is the subgroup of G generated byX, ie. the smallest subgroup i G, containing X. If 〈X〉 = G, we call X agenerating set for G.

Consider elements g ∈ G satisfying:

∀ X ⊆ G : 〈X, g〉 = G⇒ 〈X〉 = G .

Such elements are called non-generators, becuse they are superfluous in allgenerating 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 = 〈Φ(G), NG(P )〉 = 〈I(G), NG(P )〉 = 〈NG(P )〉 = NG(P ) ,

ie. P �G. �

Remark: This shows that whenG is finite then all Sylow groups in the groupΦ(G) are normal in G and therefore also in Φ(G). This actually means thatΦ(G) is an (inner) direct product of its Sylow groups (ie. a finite nilpotentgroup, see Theorem (8J)).

We give in the following a description of nilpotent groups. Here the groupG is not necessarily finite.

When H and K are subsets of G, then

[H,K] = 〈[h, k] | h ∈ H , k ∈ K〉

is the subgroup of G generated by all commutators of elements from H andK. 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], seeChapter 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 surjektivethere exists a g ∈ G such that ϕ(g) = h. Then [a, g] = 1, as a ∈ Z(G) andthus

1 = ϕ([a, g]) = [ϕ(a), ϕ(g)] = [ϕ(a), h] .

�

In connection with the definitione of nilpotent group central series areimportant.

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} , Z i(G)/Zi−1(G) = Z(G/Zi−1(G)

)for i ≥ 2 .

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

(8E) Examples: (1) Let G = S4. We have Z(G) = {1}, and thus Z0(G) =Z1(G) = Z2(G) = · · · = {1}. We have that G′ = A4. Furthermore[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 = 〈x, y | x8 = y2 = 1 , y−1xy = x−1〉 .

Here we have Z(G) = 〈x4〉 and that G/Z(G) ∼= D4. The group D4 has alsoet center of order 2, so we get

Z(G/〈x4〉) = 〈x2〉/〈x4〉 .

52

Now G/〈x2〉 is a four group. We get

Z0(G) = {1} , Z1(G) = 〈x4〉 , Z2(G) = 〈x2〉

Zi(G) = G for i ≥ 3.

Furthermore G′ = 〈x2〉 (as we have seen earlier). Now [G′, G] = 〈[x2, y]〉 =〈x4〉 and [〈x4〉, G] = 1, da x4 ∈ Z(G). We get

L1(G) = G , L2(G) = 〈x2〉 , L3(G) = 〈x4〉

Li(G) = {1} for i ≥ 4 .

The first example shows that den lower (respectively upper) central se-ries need not finish in {1} (respectively G). In the second example we hadZ2(G) = G and L3(G) = {1}. The connection is explained here:

(8F) Theorem: For any group G and any m ≥ 0 we have

Zm(G) = G⇔ Lm+1(G) = {1} .

Furthermore in this case we have

(∗) Li+1(G) j Zm−i(G) for alle i

Proof: ...

Definition: A group G is called nilpotent if there exists an m ≥ 0 suchthat Zm(G) = G. The smallest integer m with Zm(G) = G is called thenilpotency 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 nilpotentof class 1, if and only if it is abelian. A non-abelian group G is nilpotent ofklasse 2 if and only if G′ ⊆ 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 (andthus nilpotent), but G is not nilpotent. (Ln(G) = A3 for n ≥ 2.) Thus thegroup S3 is solvable, but not nilpotent.

(8H) Theorem: Assume that G is nilpotent, and that H ⊂ G is a subgroup6= G. Then H 6= NG(H).

Proof: Assume that Lm(G) = {1}. As L1(G) = G and H 6= G, there existsan 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?) AsLi(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 thatany finite p-group 6= {1} has a center 6= {1}. (See Theorem (9D) below orChapter 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 arepermutable. As every element may in a unique way be written as a productof 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 hasorderm. 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, asxn = 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) weget

(8K) Theorem: Let G be finite. Then Φ(G) is a nilpotent and characteristicsubgroup of G. �

Finally we consider the Fitting subgroup of a finite group G. In the restof this Chapter we thus consider only finite groups.

(8L) Lemma: Assume that H and K are normal nilpotent subgroups of thegroup 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 thatP1P2 �HK. It is easily seen that P1P2 ∈ Sylp(HK). Thus in HK all Sylowgroups are normal, so that HK is nilpotent by (8J). �

(8M) Theorem: Let

F (G) = 〈H | H �G , H nilpotent〉 .

55

Then F (G) is a characteristic nilpotent subgroup of G, containing all nilpo-tent 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) ' His nilpotent. Therefore F (G) char G, by (1C). That F (G) is nilpotent is easilyseen from (8L). �

The next Theorem may look rather technical, but it is useful in variousconnections:

(8N) Theorem: Let M ∈ Max(G). Put L =⋂

x∈G xMx−1, so that L is

the largest normal subgroup of G contained in M . Put G = G/L, and letF = 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 subgroup6= 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 permutedby automorphisms of Z(F ). P is abelian as P ⊆ Z(F ), and therefore itis 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).

But P * M : If P j M , then P ′s inverse image inG (called P ) is a normalsubgroup in G, contained in M . As L is the largest normal subgroup in Gcontained 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 �Gso that C ∩M �M . As P centralizes C, and therefore also C ∩M , we haveP ⊆ NG(C∩M). We get C∩M�G as G = P M . In inverse image of C∩M inG 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 = CM = 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 = Cwe 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.

Proof: Let L =⋂

x∈G xMx−1 be as in (8N). As G is solvable, G = G/L is

also solvable. Therefore F (G) 6= {1}, since for example a minimal normalsubgroup 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 role in the following interestingresult describing connections between the Frattini group and the Fittinggroup 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[F, F ] ⊆ Φ we choose M ∈ Max(G), and put L =

⋂x∈G xMx−1 as in (8N).

Since xMx−1 ∈ Max(G) for all x ∈ G, we must have Φ ⊆ L by definition ofΦ. Put G = G/L and F (G) = K/L, where then K � G is the invere imageof F (G) in G.

We have F ⊆ K: By an isomorphism Theorem F/F ∩L ∼= FL/L. There-fore FL/L is a normal nilpotent subgroup in G, as F/F ∩ L is nilpotent by(8F) (2). We get that FL/L ⊆ K/L, ie. F ⊆ K. By (8N) K/L is abelian(also if F (G) = 1, ie. K = L!) As FL/L ⊆ K/L FL/L is abelian; therefore[F, F ] ⊆ [FL, FL] j L, (see (6A)).

This holds for all intersections L of conjugate maximal subgroups. LetM1,M2, · · · ,Mt be a set of representatives for the G-conjugacy classes ofmaximal subgroups. Put Li =

⋂x∈G

xMix−1. Then Φ = L1 ∩L2 ∩ · · · ∩Lt. By

the above [F, F ] ⊆ Li for all i. We get [F, F ] ⊆ Φ. Thus (1) is proved.

We put G = G/Φ and F (G) = H/Φ, where H is the inverse image of

F (G) in G. Then F ⊆ H because F/Φ (by (8F)(2)) is a nilpotent normal

subgroup in G/Φ = G, ie. F/Φ ⊆ H/Φ. To show the inclusion H ⊆ F we

show that H is nilpotent. Let P ∈ Sylp(H). Then PΦ/Φ ∈ Sylp(G), and

57

therefore PΦ/Φ�G/Φ by (8J), ie. PΦ�G. As Φ ⊆ H and P ∈ Sylp(H), weget P ∈ Sylp(PΦ). By the Frattini argument (1E) we get G = PΦNG(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 acontradiction. We have that Z(F ) = C ∩F ⊂ C. Choose a normal series forG, 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 furthermoreGs ⊆ 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–groupsG9 - 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 alli, 2 ≤ i ≤ r. This means by (∗), that p | |N | − 1. As p | |N |, becauseN 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 ) iccyclic, 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|Z3(P )| > |Z2(P )|, etc. Finally we get Zm(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 elementx of order p in Z(Q/U). (x exists by (9A), as Q/U 6= {1}).

The inverse image V of 〈x〉 in Q has the properties U ⊆ V , |V : U | =|〈x〉| = |x| = p, and V �Q as 〈x〉�Q = Q/U .

Thus |V | = |V : U | |U | = p|U | = pa+1 and U � V , as V ⊆ NP (U) = Q.This shows (1). As furthermore V � Q, we see that if U � P, then Q = Pand then V � P showing (2). �

(9G) Corollary: If |P | = pn then P contains a en (normal) subgroup Nmed |N | = pa 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 followsfrom 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 ′ ⊆M

and

(ii) For all x ∈ P we have xp ∈ M (since x = xM has order 1 or p inG/M).

60

As (i) and (ii) hold for allM ∈ Max(P ) we get from the definition of Φ(P ) =⋂M∈Max(P )

M that

(iii) P ′ ⊆ 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 forall x ∈ P/Φ(P ) we have xp = 1. Thus P/Φ(P ) is elementary abelian. �

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

〈x1, · · · , xk〉 = P

m

〈x1, · · · , xk〉 = P ,

where xi is the image of xi in the Frattini factor group P = P/Φ(P ).

Proof: ⇓ er trivial.⇑: Assume 〈x1, · · · , xk〉 = P . Put

U := 〈x1, · · · , xk,Φ(P )〉 ⊆ P .

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

x = x1a1 · · ·xk

ak where 0 ≤ ai ≤ p− 1 ,

ie.x = x1

a1 · · ·xkak

or u := x(x1a1 · · ·xk

ak)−1 ∈ Φ(P ). But then

x = x1a1 · · ·xk

aku ∈ 〈x1, · · · , xk,Φ(P )〉 = U .

We have shown P = 〈x1, · · · , xk,Φ(P )〉. From (8A) we then get P =〈x1, · · · , xk〉, 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 gen-erating 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” theoremfor finite-dimensional vector spaces states that a set of spanning vectors forthe 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 thusbe considered as a vector space of dimension r over the field Zp med p ele-menter. (See (6P)). If 〈x1, · · · , xs〉 = P , then 〈x1, · · · , xs〉 = P and therefore{x1, · · · , xs} is a generating set of vectors for the vector space P . Thereforeit contains at least r = dimP elements. This shows (i). (ii) follows from thecontraction theorem for vector spaces and (9J), and then (iii) is trivial. �

62