How can I prove that the Core of $H$ in $G$ is a normal subgroup?
By Sophia Vance
Given $G$ a group and $H$ a subgroup of $G$, how can I prove that ${\rm Core}_H G:=\bigcap_{g\in G} gHg^{-1}$ is a normal subgroup of $G$?
I've proved that is a subgroup, but how can I prove that it is a normal subgroup?
If $x \in{\rm Core}_H G,\ y \in G$, how can I prove that $yxy^{-1}$ is in ${\rm Core}_H G$?
As $x \in{\rm Core}_H G,\ \forall g\in G,\ \exists a\in H:x=gag^{-1} $Moreover, as ${\rm Core}_H G \subseteq H$, (since it is subgroup), $x\in H$
How can I continue and show that $yxy^{-1}\in {\rm Core}_H G$
$\endgroup$ 13 Answers
$\begingroup$For all $k \in G$, \begin{align*} k \mathrm{Core}_G(H)k^{-1} &= k \left(\bigcap_{g \in G} g H g^{-1} \right) k^{-1} \\ &= \bigcap_{g \in G} kg H (kg)^{-1} \end{align*}As $g$ ranges through $G$, $kg$ ranges through $G$. (That is, left-multiplication by $k$, $\varphi_k: G \rightarrow G: g \mapsto kg$, is a bijection.) So the expression on the right is the same intersection as in the definition of the core, with the sets permuted. Intersection is invariant under permutation of its arguments. Therefore,$$ \forall k \in G, k \mathrm{Core}_G(H) k^{-1} = \mathrm{Core}_G(H) \text{,} $$so $\mathrm{Core}_G(H)$ is normal in $G$.
$\endgroup$ 2 $\begingroup$Yet another way to get this result: $\operatorname{Core}_G(H)$ is the kernel of the $G$-action by left multiplication on left quotient set $G/H$ (see here for a proof). As such, it is normal in $G$.
$\endgroup$ $\begingroup$Another way to look at this: $core_G(H)$ is the largest normal subgroup of $G$ contained in $H$. For if $N \unlhd G$ and $N \subseteq H$, then for every $g \in G$, we have $g^{-1}Ng=N \subseteq g^{-1}Hg$, so $N \subseteq core_G(H)$.
Yet another observation: $core_G(H)$ is the subgroup of $H$ generated by all the $G$-conjugacy classes contained in $H$.
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