sin2 tk + 2

= cos tk + (cos tk cos t + sin tk sin t )

6n

k=1 k=1 16k<

=n+2 cos(tk ’ t )

6n

16k<

n

= n + n(n ’ 1) = n2 .

n+2

2

with equality if and only if cos(tk ’ t ) = 1 whenever 1 k < n. But if |χρ (g)| = χρ (e) = n,

2 2 with equality if and only if cos(t ’ t ) = 1 for all k, . Assuming

then we must have n n k

that tj ∈ [0, 2π) for each j, we must have t = tk , since we do indeed have equality here. Hence

ρ(g) = »g IdV . In fact we have |»g | = 1 since eigenvalues of ρg are roots of unity.

If g1 , g2 ∈ ker |χρ |, then

ρg1 g2 = »g1 »g2 IdV

and so g1 g2 ∈ ker |χρ |, hence ker |χρ | is a subgroup of G. Normality is also easily veri¬ed.

2. A result on representations of simple groups

Let G be a ¬nite non-abelian simple group (hence of order |G| > 1). We already know that

G has no non-trivial 1-dimensional representations.

Theorem 5.7. An irreducible 2-dimensional representation of a ¬nite non-abelian simple

group G is trivial.

Proof. Suppose we have a non-trivial 2-dimensional irreducible representation ρ of G. By

choosing a basis we can assume that we are considering a representation ρ : G ’’ GLC (C2 ). We

can form the composite det —¦ρ : G ’’ C— which is a homomorphism whose kernel is a proper

normal subgroup of G, hence must equal G. Hence ρ : G ’’ SLC (C2 ), where

SLC (C2 ) = {A ∈ GLC (C2 ) : det A = 1}.

Now notice that since ρ is irreducible and 2-dimensional, Proposition 4.21 tells us that |G|

is even (this is the only time we have actually used this result!) Now by Cauchy™s Lemma,

Theorem 2.13, there is an element t ∈ G of order 2. Hence ρt ∈ SLC (C2 ) also has order 2 since

ρ is injective. Since ρt satis¬es the polynomial identity

ρ2 ’ I2 = O2 ,

t

its eigenvalues must be ±1. But by Theorem 4.5 we know that we can diagonalise ρt , hence at

least one eigenvalue must be ’1. If one eigenvalue were 1 then for a suitable invertible matrix

P we would have

10

P ρt P ’1 =

0 ’1

implying det ρt = ’1, which contradicts the fact that det ρt = 1. Hence we must have ’1 as a

repeated eigenvalue and so for suitable invertible matrix P ,

’1 0

P ρt P ’1 = = ’I2

0 ’1

56 5. SOME APPLICATIONS TO GROUP THEORY

and hence

ρt = P ’1 (’I2 )P = ’I2 .

For g ∈ G,

ρgtg’1 = ρg ρt ρ’1 = ρg (’I2 )ρ’1 = ’I2 = ρt ,

g g

and since ρ is injective, gtg ’1 = t. Thus e = t ∈ Z(G) = {e} since Z(G) G. This provides a

contradiction.

3. A Theorem of Frobenius

Let G be a ¬nite group and H G a subgroup which has the following property:

For all g ∈ G ’ H, gHg ’1 © H = {e}.

Such a subgroup H is called a Frobenius complement.

Theorem 5.8 (Frobenius™s Theorem). Let H G be a Frobenius complement and let

gHg ’1 ⊆ G,

K =G’

g∈G

the subset of G consisting of all elements of G not conjugate to elements of H. Then N = K∪{e}

is a normal subgroup of G which is the semidirect product G = N H.

Such a subgroup N is said to be a Frobenius kernel of G.

The remainder of this section will be devoted to giving a proof of this theorem using Char-

acter Theory. We begin by showing that

|G|

(3.1) |K| = ’ 1.

|H|

First observe that if e = g ∈ xHx’1 © yHy ’1 , then e = x’1 gx ∈ H © x’1 yHy ’1 x; the latter

can only occur if x’1 y ∈ H. Notice that the normalizer NG (H) is no bigger than H, hence

NG (H) = H. Thus there are exactly |G|/| NG (H)| = |G|/|H| distinct conjugates of H, with

only one element e in common to two or more of them. So there are exactly

|G|

(|H| ’ 1) + 1

|H|

elements of G which are conjugate to elements of H. Hence,

|G| |G|

|K| = |G| ’ (|H| ’ 1) ’ 1 = ’ 1.

|H| |H|

Now let ± ∈ C(H) be a class function on the group H. Then we can de¬ne a function

± : G ’’ C by

˜

±(xgx’1 ) if xgx’1 ∈ H,

±(g) =

˜

±(e) if g ∈ K.

This is well de¬ned and also a class function on G. We also have

± = ± ‘G ’±(e)(χ ‘G ’χG ),

(3.2) ˜ H H 1

where we use the notation of Ex. Sh. 4 Qu. 3. In fact, χ ‘G ’χG is the character of a

1

H

representation of G.

3. A THEOREM OF FROBENIUS 57

Given two class functions ±, β on H,

«

1

±˜ ˜˜

±(g)β(g)

(˜ |β)G =

|G|

g∈G

«

1

˜˜

±(g)β(g)

= (|K| + 1)±(e)β(e) +

|G|

g∈G’N

«

1 |G| |G| ˜

±(h)β(h) (by Equation (3.1))

= ±(e)β(e) + ˜

|G| |H| |H|

e=h∈H

1 ˜

= ±(h)β(h)

˜

|H|

h∈H

±| ˜

= (˜ β)H .

If χ is an irreducible character of H, then by Proposition 4.17 we have

(χ|χ)G = (χ|χ)H = 1.

˜˜

Also, Equation (3.2) implies that

mj χG ,

χ=

˜ j

j

where mj ∈ Z and the χG are the distinct irreducible characters of G. Using Frobenius Reci-

j

procity (Theorem 4.32), these coe¬cients mj are given by

mj = (χ|χG )G = (χ|χG “G )H

˜j 0

j H

since χ, χG “G are characters of H. As χ(e) = χ(e) > 0, χ is itself the character of some

˜ ˜

j H

representation ρ of G, i.e., χ = χρ . Notice that

˜

N = {g ∈ G : χρ (g) = χρ (e)} = ker ρ.

Hence, by Proposition 5.1, N is a normal subgroup of G.

Now H © N = {e} by construction. Moreover,

|N H| |H|| |N | = |G|,

hence N H = N H = G. So G = N H. This completes the proof of Theorem 5.8.

An equivalent formulation of this result is the following which can be found in Chapter 6 of

[A&B].

Theorem 5.9 (Frobenius™s Theorem: group action version). Let the ¬nite group G

act transitively on the set X, and suppose that each element g = e ¬xes at most one element of

X, i.e., |X g | 1. Then

N = {g ∈ G : |X g | = 0} ∪ {e}

is a normal subgroup of G.

Proof. Let x ∈ X be ¬xed by some element of G not equal to the identity element e, and

let H = StabG (x). Then for k ∈ G ’ H, k · x = x has

StabG (k · x) = k StabG (x)k ’1 = kHk ’1 .

If e = g ∈ H © kHk ’1 , then g stabilizes x and k · x, but this contradicts the assumption on the

number of ¬xed points of elements in G. Hence H is a Frobenius complement. The result then

follows from Theorem 5.8.

58 5. SOME APPLICATIONS TO GROUP THEORY

Example 5.10. The subgroup H = {e, (12)} S3 satis¬es the conditions of Theorem 5.8.

Then

gHg ’1 = {e, (12), (13), (23)}

g∈S3

and N = {e, (123), (132)} is a Frobenius kernel.

CHAPTER 6

Automorphisms and extensions

In this chapter we will study how a ¬nite group G with a normal subgroup N G is ˜built

up™ from N and the quotient group Q = G/N . We will also study actions of one group on

another by automorphisms.

1. Automorphisms

Let G be a group. A homomorphism • : G ’’ G is called an endomorphism of G, and an

automorphism if it is invertible. The set of all automorphisms of G is denoted Aut(G), and

forms a group under composition of functions.

If g ∈ G, then conjugation by g gives rise to a function

conjg : G ’’ G; conjg (x) = gxg ’1 ,

which satis¬es the conditions

conjg (xy) = conjg (x) conjg (y),

conjg —¦ conjg’1 = IdG = conjg’1 —¦ conjg ,

conjg —¦ conjh = conjgh .

Hence, conjg ∈ Aut(G) with inverse conjg’1 . Such automorphisms are called inner and form a

subgroup Inn(G) Aut(G). There is also the conjugation homomorphism

conj : G ’’ Aut(G); conj(g) = conjg .

The image of conj is Inn(G), while its kernel is the centre of G,

Z(G) = {c ∈ G : ∀g ∈ G, cg = gc}.

Theorem 6.1. For a group G, and its conjugation homomorphism conj : G ’’ Aut(G), we

have the following.

a) There is an isomorphism

G/ Z(G) ∼ Inn(G).

=

b) Inn(G) is a normal subgroup of Aut(G).

Proof. a) This follows from the above observations that ker conj = Z(G) and im conj =

Inn(G).

b) Let θ ∈ Aut(G) be any automorphism and g ∈ G. Then

θ —¦ conjg —¦θ’1 (x) = θ(gθ’1 (x)g ’1 )

= θ(gθ’1 (x)g ’1 )

= θ(g)θ(θ’1 (x))θ(g ’1 )

= θ(g)(x)θ(g)’1

= conjθ(g) (x),

Hence

θ —¦ conjg —¦θ’1 = conjθ(g) .

59

60 6. AUTOMORPHISMS AND EXTENSIONS

The quotient group Aut(G)/ Inn(G) is called the group of outer automorphisms of G, and is

denoted Out(G). An automorphism θ ∈ Aut(G) ’ Inn(G) is often referred to as outer, although

strictly speaking we should refer to its coset θ ∈ Inn(G) as an outer automorphism.

Example 6.2. Let G = γ ∼ Z/n, a cyclic group of order n 1. Then as G is abelian,

=

Inn(G) = {IdG } since Z(G) = G. On the other hand, we have many outer automorphisms of G

as is shown in the following result which may be familiar; proofs can be found in many books,

including [JBF].

Theorem 6.3. Let G = γ ∼ Z/n be cyclic of order n 1.

=

a) If n = ab where a, b 1 and (a, b) = 1, then

G ∼ Z/a — Z/b.

=

Hence

Aut(G) ∼ Aut(Z/a) — Aut(Z/b).

=

b) If n = pk where p is a prime and k 1, then

±

Z/(p ’ 1) — Z/pk’1 if p is odd,

Z/2 — Z/2k’2 if p = 2 and k 2,

Aut(G) ∼ =

Z/2 if n = 4,

{0} if n = 2.

c) In general, Aut(Z/n) ∼ (Z/n)— , the group of units in Z/n, and

=

| Aut(G)| = •(n),