# Abstract Nonsense

## The Dual Group of the Product is the Product of the Dual Groups

Point of post: In this post we show that given a finite collection of finite groups the dual group of their product is the product of their dual groups.

Motivation

Often of course groups come naturally to us as the product of other groups. Of course it would be nice if would tell something about the dual group of the full group if we knew the dual group of each of the product factors. In this post we show, once again, that everything carries over just as we would hope and that the product of the dual groups is completely determined by the dual groups of each of its factors.

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The Dual Group of the Product of Groups

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We now show that in essence the dual group of the factors of a product of groups entirely determines the dual group for the full product. In fact, we show that the product of dual groups is equal to the dual of the product of the groups. Indeed,

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Theorem: Let $G$ and $H$ be finite groups, then $\widehat{\left(G\times H\right)_\mathfrak{L}}\cong\widehat{G_\mathfrak{L}}\times\widehat{H_\mathfrak{L}}$

Proof: Define $F:\widehat{G_\mathfrak{L}}\times\widehat{H_\mathfrak{L}}\to\widehat{\left(G\times H\right)_\mathfrak{L}}$ by $(\alpha,\beta)\mapsto \alpha\boxtimes\beta$ where $\alpha\boxtimes\beta$ is the tensor product of $\alpha$ and $\beta$. Clearly this map is well-defined in the sense that $F(\alpha,\beta)\in\widehat{\left(G\times H\right)_\mathfrak{L}}$ for every $(\alpha,\beta)$ since $\deg(\alpha\boxtimes\beta)=\chi^{(\alpha)\boxtimes(\beta)}(e_G,e_H)=\chi^{(\alpha)}(e_G)\chi^{(\beta)}(e_H)=1$. To see this is a homomorphism we first note that

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$F\left((\alpha,\beta)(\gamma,\delta)\right)=F\left((\alpha\otimes\gamma,\beta\otimes\delta)\right)=\left(\alpha\otimes \gamma\right)\boxtimes\left(\beta\otimes\delta\right)$

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That said, note that for every $(g,h)\in G\times H$ one has that

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\begin{aligned}\chi^{(\alpha\otimes\gamma)\boxtimes(\beta\otimes\delta)}(g,h) &=\chi^{(\alpha\otimes\gamma)}(g)\chi^{(\beta\otimes\delta)}(h)\\ &=\chi^{(\alpha)}(g)\chi^{(\gamma)}(g)\chi^{(\beta)}(h)\chi^{(\delta)}(h)\\ &= \left(\chi^{(\alpha)}(g)\chi^{(\beta)}(h)\right)\left(\chi^{(\gamma)}(g)\chi^{(\delta)}(h)\right)\\ &= \chi^{(\alpha)\boxtimes(\beta)}(g,h)\chi^{(\gamma)\boxtimes(\delta)}(g,h)\\ &= \chi^{(\alpha\boxtimes\beta)\otimes(\gamma\boxtimes\delta)}(g,h)\end{aligned}

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so that the characters of $(\alpha\boxtimes\beta)\otimes(\gamma\boxtimes\delta)$ and $(\alpha\otimes\gamma)\boxtimes(\beta\otimes\delta)$ are the same, and thus they are the same. So,

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$F\left((\alpha,\beta),(\gamma,\delta)\right)=(\alpha\otimes\gamma)\boxtimes(\beta\otimes\delta)=(\alpha\boxtimes\beta)\otimes(\gamma\boxtimes\delta)=F\left(\alpha,\beta\right)\otimes F\left(\gamma,\delta\right)$

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Thus, $F$ is a homomorphism.

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To see that $F$ is injective suppose that $(\alpha,\beta)\in\ker F$ then $F(\alpha,\beta)=\alpha\boxtimes\beta$ is the trivial character. Thus, let $g\in G$ be given then $\chi^{(\alpha)}(g)=\chi^{(\alpha)}(g)\chi^{(\beta)}(e)=\chi^{(\alpha)\boxtimes(\beta)}(g,h)=1$ and thus $\chi^{(\alpha)}=\chi^{\text{triv}}$ so that $\alpha=\alpha_\text{triv}$. Using the exact same techinque shows that $\beta=\beta_\text{triv}$ so that $(\alpha,\beta)=(\alpha_\text{triv},\beta_\text{triv})$ from where it follows that $\ker F$ is trivial and thus $F$ is injective.

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Finally, to see that $F$ is surjective we note by previous theorem that any $\gamma\in\widehat{\left(G\times H\right)_\mathfrak{L}}$ is of the form $\alpha\boxtimes\beta$ for $\alpha\in\widehat{G}$ and $\beta\in\widehat{H}$. But, we have that $1=\deg(\gamma)=\deg(\alpha)\deg(\beta)$ and since $\deg(\alpha),\deg(\beta)\geqslant 1$ we may conclude that $\deg(\alpha)=\deg(\beta)=1$ so that $\alpha\in\widehat{G_\mathfrak{L}}$ and $\beta\in\widehat{H_\mathfrak{L}}$ and so $(\alpha,\beta)\in\widehat{G_\mathfrak{L}}\times\widehat{H_\mathfrak{L}}$ and thus $\gamma=\alpha\boxtimes\beta=F(\alpha,\beta)\in \text{im}(F)$ and since $\gamma$ was arbitrary surjectivity follows. The conclusion follows. $\blacksquare$

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Clearly by induction it follows that:

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

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$\displaystyle \widehat{\left(\prod_{j=1}^{n}G_j\right)_\mathfrak{L}}\cong\prod_{j=1}^{n}\widehat{G_\mathfrak{L}}$

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References:

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