Abstract Nonsense

Crushing one theorem at a time

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


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January 20, 2011 - Posted by | Algebra, Representation Theory | , ,

4 Comments »

  1. […] be a finite group and be as before . We claim that in the usual inner product on the group algebra   is (up to a scalar factor) orthonormal. More […]

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  2. […] Note that this gives an alternate proof to our previous one that the group algebra is a commutative algebra if and only if is abelian. Indeed, $latex […]

    Pingback by Representation Theory: Class Functions « Abstract Nonsense | March 21, 2011 | Reply

  3. […] shows up in the decomposition of is equal to where the inner product is taken in the group algebra. But, the above shows […]

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  4. […] We need only compute that where the inner product is taken on the group algebras and respectively. But, this is clear since by the above we have that and […]

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