# Dihedral group of order 8

Writing Dihedral group $D_{8}$ (order 8)as a semidirect product $% V\left\langle \alpha \right\rangle$, where $V$ is a four-group and $\alpha$ an involution, and so we have $D_{8}=\left\langle \alpha ,\beta \right\rangle$ and $V=\left\langle \beta ,\beta ^{\alpha }\right\rangle .$

I want to show that: Let $D$ be isomorphic with $D_{8}=\left\langle \alpha ,\beta \right\rangle$ and suppose that $D$ acts on a group $G$ in such a manner that $C_{G}(V)$ is finte, where $V=\left\langle \beta ,\beta ^{\alpha }\right\rangle$ and $\alpha$ sends every element of $G$ to the inverse, then $C_{G}\left( \beta \right)$ is finite.

My solution:

Let $\phi \in AutG,$ then $g^{\phi \alpha }=\left( g^{-1}\right) ^{\phi }$ and $g^{\alpha \phi }=\left( g^{-1}\right) ^{\phi }$. Therefore $\alpha \in Z\left( AutG\right)$ and so $C_{G}(\left\langle \beta ,\beta ^{\alpha }\right\rangle )=C_{G}\left( \beta \right)$ is finite. ($V=\left\langle \beta \right\rangle$???)

I think there's a problem in my solution for the Klein group is not cyclic.

Help!

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Your solution looks basically OK to me. Since the action of $\alpha$ is central in ${\rm Aut} G$, the actions of $\beta$ and $\beta^\alpha$ on $G$ are the same, so $C_G(\beta) = C_G(\beta^\alpha) = C_G(V)$. Why should this imply that $V = \langle \beta \rangle$? –  Derek Holt Dec 17 '12 at 16:57
Because $V=\left\langle \beta ,\beta ^{\alpha }\right\rangle$ and $\beta ^{\alpha }=\beta .$ Would not it be so? –  User2040 Dec 17 '12 at 17:26
It is not true that $\beta^\alpha = \beta$. They induce the same automorphism of $G$, but that does not make them equal. It implies that $\beta^{-1}\beta^\alpha$ induces the identity automorphism of $G$, which means that the action of $D_8$ on $G$ is not faithful. –  Derek Holt Dec 17 '12 at 22:18
Lima:I understood. Thank you! –  User2040 Jan 2 '13 at 12:39