## A ‘Lemma’ (Pt. I)

**Point of post: **In this post we will prove a lemma which shall prove to be absolutely necessary in proving Burnside’s lemma.

*Motivation*

The basic idea of this post is to use a variant of the left -representation of generated by an irrep to create a mapping . We shall then use this to show that the number is algebraic. This will prove to be surprising important in our proof of Burnside’s theorem.

*The Preamble*

Let be a finite group and be an irreducible character of . Let then be any irrep which admits as its character. Recall that we may then define the induced -representation by . Our first claim is that is irreducible. Indeed:

**Theorem:** *Let be defined as above. Then, is irreducible.*

**Proof: **Suppose that is -invariant then in particular we must have that but so that . Since this was true for every it follows that is -invariant and thus is either or . Since was abritrary the conclusion follows.

Note though from basic properties of -representations we have that for any (i.e. a class function) one has that for any . Thus, by Schur’s lemma we may conclude that for some . Since this mapping is conjugation invariant (i.e. for every for some isomorphism ) and every irrep which admits as its character is conjugate it follows that for every which admits as its character. Thus, we can define the function by for any where is an irrep which admits as its character. Our first theorem says that is a homomorphism Indeed:

*Remark: *We now begin using to denote .

**Theorem:** *Let , and let be defined as above. Then is a -homomorphism (i.e. it’s an associative algebra homomorphism)*

**Proof:** Let and . Then, for any representation admitting as it’s character we have that

and so by definition . Moreover, let then one has that

Furthermore, let then one has that

and so from where the conclusion follows.

In particular this tells us that is entirely determined on the canonical basis for where are the conjugacy classes of .

Note though that there is a natural way to relate and any . Indeed:

**Theorem: ***Let be a finite group with conjugacy classes then for any and any one has that*

**Proof:** We note that by definition one has for any character which admits as its character

from where the conclusion follows.

**References:**

1. Isaacs, I. Martin. *Character Theory of Finite Groups*. New York: Academic, 1976. Print.

[…] Point of post: This post is a continuation of this one. […]

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