Kernel of a Homomorphism.

Question

I am trying to figure out a rather simple and specific concept.

Let's say I have a group $G$ and a subgroup $H<G$.

Define $X=\{xH\mid x\in G\}$ – the set of all the left cosets of $H$

We have $G\curvearrowright X\;$ by $g\cdot xH = (gx)H$.

I know that this operation defines a homomorphism $\phi:G\rightarrow \operatorname{Per}(X)$ (the set of permutations on $X$), and I am really curious about the behaviour of $\operatorname{Ker}(\phi)$ and its relation to $H$.

Do we have $\operatorname{Ker}(\phi) = H$? If not, how do we get that? I'd love to get some insight! thanks

Answer 1

Note that kernel of the action is as same as the kernel of the corresponding homomorphism (can you prove this?). So, we can find kernel of the action in order to find $\text{Ker}(\phi)$.

Let $K$ be the kernel of the action you mentioned. Then, $$K = \{g\in G\ |\ (gx)H = xH, \forall\ xH \in X\} = \{g\in G\ |\ g \in xHx^{-1}, \forall x \in G\} = \bigcap_{x \in G}xHx^{-1} = \text{Ker}(\phi)$$

So, $\text{Ker}(\phi) \subseteq H$ since for $x = e_G$, we have $xHx^{-1} = H$. In order to have $\text{Ker}(\phi) = H$, we must have $H \subseteq xHx^{-1}, \forall x \in G$ since then $H$ will be the smallest set in the intersection $\bigcap_{x \in G}xHx^{-1}$. As a concrete example, if we have $H \unlhd G$, then $xHx^{-1} = H, \forall x \in G$ so we can say that $\text{Ker}(\phi) = H$.

Answer 2

$\ker(\theta)=\cap_{x\in G}xHx^{-1}$. This is called the core of $H$ in $G$, denoted by Core$_G(H)$. It is a subgroup of $H$ and a normal subgroup of $G$ (and thus a normal subgroup of $H$ as well). It has some other nice properties, which you can read about here if you want: https://groupprops.subwiki.org/wiki/Normal_core