Abstract Nonsense

Crushing one theorem at a time

The Group Algebra (Pt. II)


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

Our next theorem is more of an observation but is worth mentioning

Theorem: Let G be a finite group and \mathcal{A}\left(G\right) the group algebra over G. Then,  \left\{\delta_g\right\}_{g\in G} is a basis for \mathcal{A}\left(G\right).

Proof: It’s evident that \{\delta_g\}_{g\in G} is linearly independent  since if

 

\displaystyle \sum_{g\in G}\alpha_g \delta_g=\bold{0}

 

implies in particular that

 

\displaystyle \alpha_h=\sum_{g\in G}\alpha_g\delta_g(h)=0

 

for each h\in G from where linear independence follows. The fact that \text{span }\{\delta_g\}_{g\in G}=\mathcal{A}\left(G\right) follows immediately from the evident formula

 

\displaystyle a=\sum_{g\in G}a(g)\delta_g

 

for any a\in\mathcal{A}\left(G\right). The conclusion follows. \blacksquare

Corollary: Let G be a finite group and \mathcal{A}\left(G\right) the group algebra over G. Then, \dim_{\mathbb{C}}\left(\mathcal{A}\left(G\right)\right)=|G|.


Our next theorem just has to do with a couple of the menial, but convenient algebraic properties of the group algebra. In particular:

Theorem: Let G be a finite group and \mathcal{A}\left(G\right) the group algebra over G. Then, the following properties hold for all g,g'\in G, \alpha\in\mathbb{C},  and a,b\in\mathcal{A}\left(G\right):


\begin{aligned}&\mathbf{(1)}\quad \delta_g\ast\delta_{g'}=\delta_{gg'}\\ &\mathbf{(2)}\quad \left(a\ast b\right)^\ast=b^\ast\ast a^\ast\\ &\mathbf{(3)}\quad \left(a^\ast\right)^\ast=a\\ &\mathbf{(4)}\quad \left(a+b\right)^\ast=a^\ast+b^\ast\\ &\mathbf{(5)}\quad \left(\alpha a\right)^{\ast}=\overline{\alpha}a^{\ast}\end{aligned}

 

 

Proof:


\mathbf{(1)}: We merely note that for any h\in G we have that

\displaystyle \left(\delta_g\ast \delta_{g'}\right)(h)=\sum_{k\in G}\delta_g\left(h k^{-1}\right)\delta_{g'}\left(k\right)

 

and that this sum will be zero if there does not exists a k\in G such that g'=hk^{-1} and g=k and 1 if it does. In other words, the sum will be zero if g'g\ne h and 1 if gg'=h. But, this implies that for any h\in G We have that \left(\delta_g\ast\delta_{g'}\right)(h)=\delta_{gg'}(h) from where it follows that \delta_g\ast\delta_{g'}=\delta_{gg'}.

\mathbf{(2)}: We merely note for any h\in G we have

\displaystyle \begin{aligned} \left(a\ast b\right)^{\ast}(h) &= \overline{\left(a\ast b\right)\left(h^{-1}\right)}\\ &= \overline{\sum_{g\in G}a\left(h^{-1}g^{-1}\right) b(g)}\\ &= \sum_{g\in G}\overline{b\left(ghh^{-1}\right)}\;\overline{a\left(\left(gh\right)^{-1}\right)}\\ &= \sum_{g\in G}\overline{b\left( g h^{-1}\right)}\;\overline{a\left(g^{-1}\right)}\\ &= \sum_{g\in G}\overline{b\left(\left(hg^{-1}\right)^{-1}\right)}\;\overline{a\left(g^{-1}\right)}\\ &= \sum_{g\in G}b^{\ast}\left(h g^{-1}\right) a^{\ast}\left(g\right)\\ &= \left(b^{\ast}\ast a^{\ast}\right)(h)\end{aligned}

and since h\in G was arbitrary the conclusion follows.

\mathbf{(3)}: This is fairly clear since for any h\in G we have that

\left(a^{\ast}\right)^{\ast}\left(h\right)=\overline{a^{\ast}\left(h^{-1}\right)}=\overline{\overline{a\left(\left(h^{-1}\right)^{-1}\right)}}=a(h)

since h was arbitrary the conclusion follows.

\mathbf{(4)}: This is also fairly clear since for any h\in G we have that

\displaystyle \begin{aligned}\left(a+b\right)^{\ast}(h) &=\overline{(a+b)\left(h^{-1}\right)}\\ &=\overline{a\left(h^{-1}\right)+b\left(h^{-1}\right)}\\ &=\overline{a\left(h^{-1}\right)}+\overline{b\left(h^{-1}\right)}\\ &=a^{\ast}(h)+b^{\ast}(h)\end{aligned}

since h\in G was arbitrary the conclusion follows.

\mathbf{(5)}: Lastly we merely note that for any h\in G we have that

\left(\alpha a\right)^{\ast}\left(h\right)=\overline{\left(\alpha a\right)\left(h^{-1}\right)}=\overline{\alpha}\;\overline{a\left(h^{-1}\right)}=\overline{\alpha}a^{\ast}\left(h\right)

since h\in G was arbitrary 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 | , , ,

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