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Let $H=\mathbb Z_2\oplus \mathbb Z_2$ and $N=\mathbb Z_3$. Let $\varphi: H \to \text{Aut}(N)$ be determined by $(1,0) \mapsto -1$, $(0,1) \mapsto -1$. I wish to show the semidirect product $N\rtimes_{\varphi}H \cong D_6$.

This is different from the standard semidirect product $C_6\rtimes C_2 \cong D_6$ and thus I don't know how to find the generators corresponding to the rotation of reflection.

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  • $\begingroup$ Take the generator of $N$ to be rotation though $2\pi/3$ and generators of $H$ to be reflections in two orthogonal axes. $\endgroup$ – Lord Shark the Unknown Sep 24 '18 at 3:03
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Let $G = N \rtimes_\varphi H$. If we find a normal cyclic subgroup $C_6 \simeq N' \unlhd G$ whose quotient $G/C_6$ acts on $C_6$ by inversion, and a complement $K \leq G$ satisfying $G = N'K$ and $N' \cap K = \{1\}$, we are done.

We must find a candidate for $N'$. It must have order $6$ so it contains $N \rtimes_\varphi \{1\}$. It must be abelian, so any element of $H$ it contains must keep $N$ fixed. The only elements of $H$ fixing $N$ are $(0,0)$ and $(1,1)$. Now, who is the subgroup $N \rtimes_\varphi \langle (1,1)\rangle$? Who can be a cadidate for $K$?

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