Image and kernel of the action of the group $D_{12}$ on the $D_{12}/H$

Question

Let $D_{12}=\langle(1,2,3,4,5,6),(2,6)(3,5)\rangle$ and $H=\langle(1,4)(2,5)(3,6)\rangle$. Then $|H|=2$. I want to determine kernel and image of the action of $D_{12}$ on $D_{12}/H$ (set of right cosets of $H$). I know that this group action is defined by $(Hx)^g=Hxg$ for every $x, g\in D_{12}$

So $kernel=\{ g\in D_{12}: (Hx)^g=Hx \text{ for every } x\in D_{12} \}= \{ g\in D_{12}: Hxgx^{-1}=H \text{ for every } x\in D_{12} \}$
So actually I need to determine for which elements $x,g\in D_{12}$ hold $xgx^{-1}\in H$? If so, how to do that? It doesn't seem like an easy question.

Next, $image=\{ (Hx)^g: x,g\in D_{12} \}=\{ Hxg: x,g\in D_{12} \}=\{ Hx': x'\in D_{12} \}=D_{12}/H$ ?

Did I understand these definitions correctly and how to proceed?

Answer 1

Hint: For all $\sigma, \tau\in S_n$ with $\tau=(t_{a_1}\dots t_{a_j})\dots(t_{z_1}\dots t_{z_k})$ being the cyclic decomposition of $\tau$, we have

$$\sigma\tau\sigma^{-1}=(\sigma(t_{a_1})\dots \sigma(t_{a_j}))\dots(\sigma(t_{z_1})\dots\sigma(t_{z_k})),$$

where $\sigma(x)$ is $\sigma$ evaluated at $x$.