Conjugacy class of prime power order implies not simple

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Statement

If a finite group contains a conjugacy class whose order is a power of a prime (and also, greater than 1), then the finite group cannot be simple.

Applications

This is used to prove Burnside's p^aq^b theorem

Results used in proof

Column orthogonality theorem which states that any two distinct columns in the character table of a finite group, are orthogonal.
The fact that for any character $χ$ over complex numbers and any $g \in G$ whose conjugacy class is $C$ , the following number is an algebraic integer:

$\frac{\chi(g)|C|}{\chi(1)}$

For full proof, refer: degree-class size-normalized character is algebraic integer

We shall actually use a corollary of this result, the zero-or-scalar lemma, namely:

If the size of the conjugacy class of $g \in G$ and the degree of a representation $φ$ with character $χ$ are relatively prime, then either $φ(g)$ is a scalar matrix (where $φ$ is the linear representation associated to $χ$ ) or $χ(g) = 0$ .

For full proof, refer: Zero-or-scalar lemma

Proof

Let $C$ be a conjugacy class in $G$ of order $q d$ where $q$ is a prime power. Take $g \in C$ . Now consider the inner product of the columns for $C$ and the identity element, in the character table. By the column orthogonality theorem, this inner product is zero, so we get:

∑	χ(g)χ(1) = 0
χ

where the summation is over all irreducible characters over the complex numbers.

Now suppose $G$ is a simple group. Clearly $G$ is not Abelian (otherwise there would not be a conjugacy class of size greater than one). So $G$ is a simple non-Abelian group. In particular:

Any nontrivial representation of $G$ is faithful
The center of $G$ is trivial, viz $G$ is a centerless group

Combining these facts, $φ(g)$ cannot be a scalar for any nontrivial linear representation $φ$ and nontrivial element $g$ of $G$ .

Now suppose $φ$ is a nontrivial irreducible representation whose degree is not a multiple of $q$ . Then, the zero-or-scalar lemma tells us that either $χ(g) = 0$ or $φ(g)$ is a scalar. $φ(g)$ cannot be a scalar, so $χ(g) = 0$ whenever the degree of $χ$ is not divisible by $q$ .

The upshot is that the sum above transforms to:

$1 + q\sum \chi(g)\chi(1)/q = 0$

with the sum taken over all nontrivial irreducible characters $χ$ for which $q | χ(1)$ .

Now $χ(g)$ is an algebraic integer for every $χ$ so the expression under summation is an algebraic integer. Call it $α$ . Then:

$1 + q\alpha = 0 \implies \alpha = -1/q$

This tells us that $- 1 / q$ is an algebraic integer, which is a contradiction.

Thus $G$ cannot be simple.

Conjugacy class of prime power order implies not simple

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Statement

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Results used in proof

Proof

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