# Conjugacy class of prime power size implies not simple

## Statement

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

## Applications

This is used to prove that order has only two prime factors implies solvable -- the result commonly called Burnside's -theorem.

## Facts used

The table below lists key facts used directly and explicitly in the proof. Fact numbers as used in the table may be referenced in the proof. This table need not list facts used indirectly, i.e., facts that are used to prove these facts, and it need not list facts used implicitly through assumptions embedded in the choice of terminology and language.
Fact no. Statement Steps in the proof where it is used Qualitative description of how it is used What does it rely on? Difficulty level Other applications
1 Column orthogonality theorem which states that any two distinct columns in the character table of a finite group are orthogonal. Step (1) Provides the basic equation setup whose terms we then deduce more facts about. Basic linear representation theory click here
2 Zero-or-scalar lemma (which follows from the fact that size-degree-weighted characters are algebraic integers): If the size of the conjugacy class of  and the degree of an irreducible representation  with character  are relatively prime, then either  is a scalar matrix (where  is the linear representation associated to ) or . Step (6) Shows that many terms are either zero or divisible by the prime . Algebraic number theory + linear representation theory 4 click here
3 Characters are algebraic integers Step (9) Build-up to contradiction from assumption that  is simple. Basic linear representation theory 2 click here

## Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

Given: A finite group , a prime number , and a conjugacy class  in  of size  where  is a prime number and .  is a representative element.

To prove:  is not a simple group.

Proof: Note that when the symbol  appears as an input to a representation or a character, it refers to the identity element of . When it appears as the output of a character, or in another context, it refers to the real number .

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation Commentary
1 We have  where the summation is over all irreducible characters of  over the complex numbers. Note that this summation includes the trivial character, for which the corresponding summand is . Fact (1)  is a nontrivial conjugacy class (since its size is , it cannot be the trivial conjugacy class) [SHOW MORE] Use column orthogonality to get the equation with which we'll play havoc.
2 For any faithful linear representation of  and any  such that  is scalar,  must be in the center of . [SHOW MORE] Building toward step (5)
3 If  is simple, then it is centerless.  is non-abelian (on account of having a conjugacy class of size greater than 1) [SHOW MORE] Building toward step (5)
4 If  is simple, then any nontrivial linear representation of  is faithful. [SHOW MORE] Building toward step (5)
5 If  is simple, then, for any nontrivial linear representation  of ,  is not a scalar matrix 
 is a nontrivial conjugacy class
Steps (2), (3), (4) We combine the steps, setting  in Step (2). Obtain that the image of a particular element under a nontrivial linear representation is not a scalar.
6 Suppose  is a nontrivial irreducible linear representation of  over  whose degree is not a multiple of , with character , then either  or  is a scalar. Fact (2) The conjugacy class  of  in  has size a power of  [SHOW MORE] Combine the prime power size of conjugacy class with the zero-or-scalar lemma. (Note that nontriviality is not essential here, but it is essential for combining with Step (5)).
7 If  is simple, then for every nontrivial irreducible character  of , either  divides  or . Steps (5), (6) [SHOW MORE] Force character values for -degrees to be zero.
8 If  is simple, we get the following:  with the sum taken over all nontrivial irreducible characters  of  over  for which . Steps (1), (7) [SHOW MORE] Bring back the column orthogonality relation from Step (1)
9 Let  be the sum , the sum being taken over all nontrivial irreducible characters  of  over  for which . Then,  is an algebraic integer. Fact (3) [SHOW MORE] Now it is an algebraic integer ...
10 If  is simple,  Steps (8), (9) [SHOW MORE] ... and now it isn't.
11 If  is simple, we have a contradiction. Hence,  is not simple. Steps (9), (10) [SHOW MORE] contradiction clincher.

## References

### Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 890, Lemma 7, Section 19.2 (Theorems of Burnside and Hall), (The proof of this builds on lemmas and definitions ranging across pages 887-890. The proof is immediately followed by the proof of Burnside's p^aq^b-theorem.)More info