# Abstract Nonsense

## Representation Theory of Semidirect Products: The Preliminaries (Pt. II)

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

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What we next do is ‘dualize’  the action $\varphi:K\to\text{Aut}\left(A\right)$ to an action $\widehat{\varphi}:K\to\text{Aut}\left(\text{irr}(A)\right)$ (where we, for notational convenience, have denoted $\text{Hom}(A,\mathbb{T})$ by $\text{irr}(A)$. Since $A$ is abelian we know the two coincide) by $\left(\widehat{\varphi}_k(\chi)\right)(a)=\chi\left(\varphi_k^{-1}(a)\right)$. It is not at all clear that 1) $\widehat{\varphi}_k(\chi)$ is really a homomorphism, 2) $\widehat{\varphi}_k$ is an automorphism, and 3) $\widehat{\varphi}$ is a homomorphism. Indeed:

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Theorem: Let $\widehat{\varphi}:K\to\text{Aut}\left(\text{irr}(A)\right)$ be defined as above. Then:

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\begin{aligned}&\mathbf{(1)}\quad \widehat{\varphi}_k(\chi)\textit{ is a homomorphism }G\to\mathbb{T}\textit{ for all }k\in K,\chi\in\text{irr}(A)\\ &\mathbf{(2)}\quad \widehat{\varphi}_k:\text{irr}(A)\to\text{irr}(A)\textit{ is an automorphism}\\ &\mathbf{(3)}\quad \widehat{\varphi}\textit{ is a homomorphism }K\to\text{Aut}\left(A\right)\end{aligned}

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Proof:

1) We note that for any $k\in K$, $a,a'\in A$, and $\chi\in\text{irr}(A)$ one has that

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\displaystyle \begin{aligned}\left(\widetilde{\varphi}_k(\chi)\right)(aa') &= \chi\left(\varphi_k^{-1}\left(aa'\right)\right)\\ &= \chi\left(\varphi_k^{-1}(a)\varphi_k^{-1}(a')\right)\\ &= \chi\left(\varphi_k^{-1}(a)\right)\chi\left(\varphi_k^{-1}(a')\right)\\ &= \left(\widehat{\varphi}_k(\chi)\right)(a)\left(\widehat{\varphi}_k(\chi)\right)(a')\end{aligned}

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(note the use of the abelianess of $A$).

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2) We note that for any $k\in K$, $a\in A$, and any $\chi,\chi'\in\text{irr}(A)$ one has

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\displaystyle \begin{aligned}\left(\widehat{\varphi}_k\left(\chi\chi'\right)\right)(a) &= \left(\chi\chi'\right)\left(\varphi_k^{-1}(a)\right)\\ &= \chi\left(\varphi_k^{-1}(a)\right)\chi'\left(\varphi_k^{-1}(a)\right)\\ &= \left(\widehat{\varphi}_k(\chi)\right)(a)\left(\widehat{\varphi}_k\left(\chi'\right)\right)(a)\end{aligned}

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so that $\widehat{\varphi}_k$ is an endomorphism. To prove it is an automorphism it suffices to prove (since $A$ is finite) to prove it is injective. To do this we note that if $\chi\in\ker\widehat{\varphi}_k$ and $a\in A$ then $\varphi_k(a)\in A$ and so

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$1=\left(\widehat{\varphi}_k(\chi)\right)(\varphi_k(a))=\chi\left(\varphi_k^{-1}\left(\varphi_k(a)\right)\right)=\chi(a)$

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Thus, $\chi(a)=1$ for every $a\in A$, and thus $\chi=\chi^{\text{triv}}$.

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3) We note that for any $\chi\in\text{irr}(G)$ and $k,k'\in K$ one has

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\begin{aligned}\left(\widehat{\varphi}_{kk'}(\chi)\right)(a) &= \chi\left(\varphi_{kk'}^{-1}(a)\right)\\ &= \chi\left(\varphi_{k'}^{-1}\left(\varphi_k^{-1}(a)\right)\right)\\ &= \left(\widehat{\varphi}_{k'}(\chi)\right)\left(\varphi_k^{-1}(a)\right)\\ &= \left(\widehat{\varphi_k}\left(\widehat{\varphi}_{k'}(\chi)\right)\right)(a)\end{aligned}

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and so $\widehat{\varphi}_{kk'}=\widehat{\varphi}_k\circ\widehat{\varphi}_{k'}$. $\blacksquare$

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References:

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