# $D_6$(12 elements) in terms of semidirect product of $Z_2\oplus Z_2$ and $Z_3$

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.

• Take the generator of $N$ to be rotation though $2\pi/3$ and generators of $H$ to be reflections in two orthogonal axes. – Lord Shark the Unknown Sep 24 '18 at 3:03

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 $$K$$?