# Abstract Nonsense

## 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 $\mathcal{A}\left(G\right)$ is a commutative algebra, but first we need  a small lemma:

Lemma: Let $G$ be a finite group and $\mathcal{A}\left(G\right)$ the group algebra on $G$. Define $\tilde{G}=\{\delta_g\}_{g\in G}$. Then $\tilde{G}$ is a multiplicative group and $\tilde{G}\cong G$.

Proof: We have from the first theorem in this chapter that $\ast$ is an associative binary operation and that $\delta_e$ is an identity. Thus, it suffices to show that each element of $\tilde{G}$ has a two-sided inverse. But, this follows immediately from the previous theorem since if $\delta_g\in\tilde{G}$ then $\delta_{g^{-1}}\in\tilde{G}$ and

$\delta_{g}\ast \delta_{g^{-1}}=\delta_{gg^{-1}}=\delta_e=\delta_{g^{-1}g}=\delta_{g^{-1}}\ast\delta_g$

To show that $G\cong\tilde{G}$ we merely note that the canonical association is in fact an isomorphism. More directly define

$\phi:G\to\tilde{G}:g\mapsto \delta_g$

This is clearly surjective and since $|G|=\left|\tilde{G}\right|<\infty$ it must be a bijection. The fact that it’s a homomorphism follows almost immediately from the preceding theorem, namely:

$\phi\left(gh\right)=\delta_{gh}=\delta_g\ast\delta_h=\phi\left(g\right)\ast\phi\left(h\right)$

the conclusion follows. $\blacksquare$

With this we are now able to state explicitly when $\mathcal{A}\left(G\right)$ is a commutative algebra:

Theorem: Let $G$ be a finite group and $\mathcal{A}\left(G\right)$ the group algebra over $G$. Then, $\mathcal{A}\left(G\right)$ is a commutative algebra if and only if $G$ is abelian.

Proof: Suppose first that $G$ is abelian. Then, by the lemma we know that $\tilde{G}$ is abelian and so in particular they all commute with each other. The rest is a simple calculation. Namely, if $a,b\in\mathcal{A}\left(G\right)$ then

\displaystyle \begin{aligned}a\ast b &= \left(\sum_{g\in G}a(g)\delta_g\right)\ast\left(\sum_{h\in G}b(h)\delta_h\right)\\ &= \sum_{g\in G}\left(a(g)\delta_g\ast\left(\sum_{h\in G}b(h)\delta_h\right)\right)\\ &= \sum_{g\in G}\sum_{h\in G}\left(a(g)\delta_g\ast b(h)\delta_h\right)\\ &= \sum_{g\in G}\sum_{h\in G}\left(a(g)b(h)\right)\left(\delta_g\ast\delta_h\right)\\ &= \sum_{g\in G}\sum_{h\in G}\left(b(h)a(g)\right)\left(\delta_h\ast\delta_g\right)\\ &= \sum_{g\in G}\sum_{h\in G}b(h)\delta_h\ast a(g)\delta_g\\ &= \sum_{g\in G}\left(\left(\sum_{h\in G}b(h)\delta_h\right)\ast a(g)\delta_g\right)\\ &= \left(\sum_{h\in G}b(h)\delta_h\right)\ast\left(\sum_{g\in G}a(g)\delta_g\right)\\ &= b\ast a\end{aligned}

and since $a,b\in\mathcal{A}\left(G\right)$ is arbitrary the conclusion follows.

Conversely, if $\mathcal{A}\left(G\right)$ is a commutative algebra then it easily follows that $\tilde{G}$ is an abelian group and since $\tilde{G}\cong G$ we may conclude that $G$ is abelian. The conclusion follows. $\blacksquare$

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

January 20, 2011 -

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