How do I prove that $S_3 \times \mathbb{Z}_2$ is isomorphic to $D_6$? I was able to find an element of order 6 within $S_3 \times \mathbb{Z}_2$. I was also able to define a function mapping all 6 elements of $S_3 \times \mathbb{Z}_2$ to $D_6$ I the homomorphism portion would be too exhaustive to prove all 36 cases. How do I prove this.
 A: The best way to find an isomorphism is to picture the sets that these groups are naturally acting on. $D_6$ consists of the symmetries of a rgular hexagon, while $S_3\cong D_3$ is the symmetries of an equilateral triangle. This suggests that the $S_3$ component of $S_3\times Z_2$ should be acting on some triangle embedded in the hexagon. 
There are only two such triangles which are subsets of the hexagon, and these are $180^\circ$ rotations of each other. Furthermore, $180^\circ$ roation commutes with every symmetry of $D_6$, and $180^\circ$ rotation behaves like $Z_2$. This gives you your isomorphism: $S_3$ acts on the hexagon by rotations/reflections which preserve the two equilateral subtriangles, namely, rotation by multiples of $120^\circ$ and reflection through these triangles' axes of symmetry, while $Z_2$ is $180^\circ$ rotation.
As for the explicit mapping, note that $S_3$ is generated by $(1\,2)$ and $(1\,2\,3)$, $Z_2$ is generated by $z$, and $D_3$ is generated by $r$ and $s$, corresponding to rotation by $60^\circ$ and reflection through some axis. The isomorphism I described is given by
$$
(1\,2)\mapsto s,\qquad (1\,2\,3)\mapsto r^2,\qquad z\mapsto r^3
$$
The same idea works for any even polygon to show that $D_{2n}\cong D_n\times Z_2$.
