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I am not sure how to do the following question:

Let $G$ be a group and $a \in G$, let $[g,a]=gag^{-1}a^{-1}$, and let $M(a) = \{g \in G | [g,a] \in Z(G) \}$. It is easy to see that the map $\psi : M(a) \rightarrow Z(G)$ defined by $\psi(g) = [g,a]$ is a homomorphism. Does this imply that $M \cong C_G(a) \rtimes Z(G)$ ?

I am not really sure in general how to show that something implies a semidirect product and hence I do not have an idea of how to approach this.

Thank you very much.

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1 Answer 1

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A useful property related to the semidirect product construction is that if $G \cong H \rtimes K,$ with $H, K \le G,$ then we must have $H \cap K =1.$ Now, if $x\in Z(G),$ then $xax^{-1} = a$ so that $x\in C_G (a).$ Thus any group with a nontrivial center cannot be factored in this form.

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It can be if we are using an external semidirect product; and the use of $\cong$ rather than $=$ would allow this. For example, all even dihedral groups are such a semidirect product: let $a$ be a generator of the maximal normal (cyclic) subgroup, so that $C_G(a)=\langle a \rangle$. We have $Z(G)\cong\mathbb{Z}_2$. Then $G\cong C_G(a)\rtimes Z(G)$ as an external semidirect product; but not an internal one. –  zibadawa timmy Jan 12 at 15:17

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