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From Groupprops

Contents 1 Statement 2 Applications 3 Results used in proof 4 Proof 5 References 5.1 Textbook references 5.2 Journal references

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 that order has only two prime factors implies solvable -- the result commonly called 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 whose conjugacy class is C, the following number is an algebraic integer:

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 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 . 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:

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:

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

Thus G cannot be simple.

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}

Journal references

• On groups of order p^αq^β by William Burnside, Proceedings of the London Mathematical Society, ISSN 1460244X (online), ISSN 00246115 (print), Volume 1, Page 388 - 392(Year 1904): ^{More info}