# Abstract Nonsense

## The Irreps of the Product of Finitely Many Finite Groups (Pt. II)

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

Corollary: Let $G_1,\cdots,G_n$ be finitely many finite groups then,

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$\displaystyle \widehat{\prod_{j=1}^{n}G_j}=\left\{\rho^{(\alpha_1)}\boxtimes\cdots\boxtimes\rho^{(\alpha_n)}:\alpha_k\in\widehat{G_k},\;k\in[n]\right\}$

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From the above it’s clear that we may index the elements of $\displaystyle \widehat{\prod_{j=1}^{n}G_j}$ as $\alpha_1\boxtimes\cdots\boxtimes\alpha_n$ and so we have the characters of $\displaystyle \widehat{\prod_{j=1}^{n}G_j}$ may be written as $\chi^{(\alpha_1\boxtimes\cdots\boxtimes\alpha_n)}$ and we have that

$\displaystyle \chi^{(\alpha_1\boxtimes\cdots\boxtimes\alpha_n)}(g_1,\cdots,g_n)=\chi^{(\alpha_1)}(g_1)\cdots\chi^{(\alpha_n)}(g_n)$

and that

$\displaystyle \left\langle \chi^{(\alpha_1\boxtimes\cdots\boxtimes\alpha_n)},\chi^{(\beta_1\boxtimes\cdots\boxtimes\beta_n)}\right\rangle=\prod_{j=1}^{n}\left\langle \chi^{(\alpha_j)},\chi^{(\beta_j)}\right\rangle$

The last thing we note which can often come in hand for practical computation (since, for instance it can be done on math world) is the following:

Theorem: Let $G$ be a finite group with irreducible characters $\chi^{\text{triv}},\chi^{(\alpha_1)},\cdots,\chi^{(\alpha_n)}$ and conjugacy classes $\mathcal{C}_1,\cdots,\mathcal{C}_n$. Moreover, let $H$ a finite group with irreducible characters $\chi^{\text{triv}},\chi^{(\beta_1)},\cdots,\chi^{(\beta_1)},\cdots,\chi^{(\beta_m)}$ and conjugacy classes $\mathcal{D}_1,\cdots,\mathcal{D}_m$. If $M_G$ is character table of $G$ thought of as a matrix with $(i,j)^{\text{th}}$ entry $\chi^{(\alpha_i)}\left(\mathcal{C}_j\right)$ and $M_H$ is the character table of $H$ thought of as the matrix with $(i,j)^{\text{th}}$ entry $\chi^{(\beta_i)}\left(\mathcal{D}_j\right)$. Then, if $M_{G\times H}$ is the character table of $G\times H$ where the rows are arranged left to right $\mathcal{C}_1\times\mathcal{D}_1,\cdots,\mathcal{C}_1\times\mathcal{D}_m,\cdots,\mathcal{C}_n\times\mathcal{D}_1,\cdots,\mathcal{C}_n\times\mathcal{D}_m$ and the columns arranged up to down $\chi^{\text{triv}\boxtimes(\beta_1)},\cdots,\chi^{\text{triv}\boxtimes(\beta_m)},\cdots,\chi^{(\alpha_n)\boxtimes(\beta_1)},\cdots,\chi^{(\alpha_n)\boxtimes(\beta_m)}$. Then, $M_{G\times H}=M_G\otimes M_H$ where $\otimes$ is the Kronecker product.

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Proof: This follows immediately from the definition $\blacksquare$

References:

1. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Math. Soc., 1996. Print.