## The Group Algebra (Pt. III)

**Point of post: **This post is a continuation of this one.

As one last tid-bit we decide precisely when is a commutative algebra, but first we need a small lemma:

**Lemma:*** **Let be a finite group and the group algebra on . Define . Then is a multiplicative group and .*

**Proof:**** **We have from the first theorem in this chapter that is an associative binary operation and that is an identity. Thus, it suffices to show that each element of has a two-sided inverse. But, this follows immediately from the previous theorem since if then and

To show that we merely note that the canonical association is in fact an isomorphism. More directly define

This is clearly surjective and since it must be a bijection. The fact that it’s a homomorphism follows almost immediately from the preceding theorem, namely:

the conclusion follows.

With this we are now able to state explicitly when is a commutative algebra:

**Theorem:**** ***Let be a finite group and the group algebra over . Then, is a commutative algebra if and only if is abelian.*

**Proof:**** **Suppose first that is abelian. Then, by the lemma we know that is abelian and so in particular they all commute with each other. The rest is a simple calculation. Namely, if then

and since is arbitrary the conclusion follows.

Conversely, if is a commutative algebra then it easily follows that is an abelian group and since we may conclude that is abelian. The conclusion follows.

**References:**

1.Simon, Barry. *Representations of Finite and Compact Groups*. Providence, RI: American Mathematical Society, 1996. Print.

2. Serre, Jean Pierre. *Linear Representations of Finite Groups*. New York: Springer-Verlag, 1977. Prin

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