# Abstract Nonsense

## 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