I am trying to answer this question:
Show that we also have: $$D_{6} \cong S_{3} \times \mathbb{Z}_{2}.$$ Hint: Label the vertices of a regular hexagon $1,2,3,4,5,6$ consecutively, and consider the stabilizer of the set $\{ 2,4,6 \}$ by the action of $D_{6}$ on the set of vertices.
I got this hint as a part of the solution:
$D_{6}$ has a two element subgroup, isomorphic to $\Bbb Z_2$, whose intersection with $S_3$ is trivial (consider a reflection that doesn't stabilise $\{2,4,6\}$.)
Here is the link for the hint Understanding why proving a hint will prove that $D_{6} \cong S_{3} \times \mathbb{Z}_{2}.$
My questions are:
1- I know that there are 3 reflections that do not stabilize $\{2,4,6\},$ which are reflections through the 3 axes of symmetry connecting the midpoints of opposite sides but I do not know how those give us a two element subgroup? and why this group is isomorphic to $\mathbb{Z}_{2}$ specifically? and why this group is a normal subgroup in $D_{6}$
Could anyone explain this to me, please?
2- I think @Chris (the author of the hint ) meant to say to me the intersection of the 2 element subgroup (that is isomorphic with $\mathbb{Z}_{2}$) with the stabilizer of $\{2,4,6\}$ is trivial not with $S_{3},$ am I correct?
the stabilizer of $\{2,4,6\}$ in $D_{6}$ is defined here Understanding how reflection acts on rotation in $D_{6}.$ which is the following set $\{id. , \rho^{2}, \rho^{4}$, reflection through the axis of symmetry passing through the vertex 2 and between vertices 4 & 6, reflection through the axis of symmetry passing through the vertex 4 and between vertices 2 & 6, reflection through the axis of symmetry passing through the vertex 6 and between vertices 2 & 4 } where $\rho$ means rotation.
My definition of $D_{6}$ is that it is the dihedral group of order 12 defined on Wikipedia.