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Lecture notes for Introduction to Representation Theory

Lecture notes for Introduction to Representation Theory

Lecture notes for Introduction to Representation Theory

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The proof will be based on the following lemma.<br />

Lemma 4.22. If π 1 , π 2 . . . π n are roots of unity such that<br />

integer, then either π 1 = . . . = π n or π 1 + . . . + π n = 0.<br />

1<br />

(π 1 + π 2 + . . . + π n ) is an algebraic<br />

n<br />

Proof. Let a = 1 n<br />

(π 1 + . . . + π n ). If not all π i are equal, then | a | < 1. Moreover, since any algebraic<br />

conjugate of a root of unity is also a root of unity, | a | ∗ 1 <strong>for</strong> any algebraic conjugate a of a. But<br />

the product of all algebraic conjugates of a is an integer. Since it has absolute value < 1, it must<br />

equal zero. There<strong>for</strong>e, a = 0.<br />

Proof of theorem 4.21.<br />

Let dim V = n. Let π 1 , π 2 , . . . π n be the eigenvalues of δ V (g). They are roots of unity, so<br />

1<br />

ν V (g) is an algebraic integer. Also, by Proposition 4.17, n<br />

|C|ν V (g) is an algebraic integer. Since<br />

gcd(n, | C | ) = 1, there exist integers a, b such that a| C | + bn = 1. This implies that<br />

ν V (g) 1<br />

= (π 1 + . . . + π n ).<br />

n n<br />

is an algebraic integer. Thus, by Lemma 4.22, we get that either π 1 = . . . = π n or π 1 + . . . + π n =<br />

ν V (g) = 0. In the first case, since δ V (g) is diagonalizable, it must be scalar. In the second case,<br />

ν V (g) = 0. The theorem is proved.<br />

Theorem 4.23. Let G be a finite group, and let C be a conjugacy class in G of order p k where p<br />

is prime and k > 0. Then G has a proper nontrivial normal subgroup (i.e., G is not simple).<br />

Proof. Choose an element g C. Since g ⇒ = e, by orthogonality of columns of the character table,<br />

<br />

dim V ν V (g) = 0. (4)<br />

We can divide IrrG in<strong>to</strong> three parts:<br />

1. the trivial representation,<br />

V IrrG<br />

2. D, the set of irreducible representations whose dimension is divisible by p, and<br />

3. N, the set of non-trivial irreducible representations whose dimension is not divisible by p.<br />

Lemma 4.24. There exists V N such that ν V (g) ⇒= 0.<br />

Proof. If V D, the number 1 dim(V )ν V (g) is an algebraic integer, so<br />

p<br />

is an algebraic integer.<br />

Now, by (4), we have<br />

1<br />

a = dim(V )ν V (g)<br />

p<br />

V D<br />

0 = ν C (g) + dim V ν V (g) + dim V ν V (g) = 1 + pa + dim V ν V (g).<br />

V D V N V N<br />

This means that the last summand is nonzero.<br />

53

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