Let $G$ be a group, $H < G$, and $A_H=\{g\mid\bar{g}\subset H\}$ where $\bar{g}$ is the conjugacy class of $g$. Prove $A_H\lhd G$. Suppose $G$ is a group and $H < G$ and $A_H=\{g \mid \bar{g} \subset H \}$ where $\bar{g}$ is the conjugacy class of $g$. I want to prove $A_H\lhd G$.

First approach:
Define $X$ as the set of left cosets of $H$, then define a homomorphism $ \varphi(g) = f_g$ from $G$ to $S_X$ where $f_g(aH)=gaH$.
$g_1=g_2$ implies $f_{g_1}=f_{g_2}$, so $\varphi$ is well defined. Also $\varphi(g_1 g_2)=f_{g_1 g_2}=f_{g_1}(f_{g_2})=\varphi(g_1) \circ \varphi(g_2)$ shows $\varphi$ is a homomorphism.
Now, $\varphi(g)=e_{S_X}$ implies $f_g(aH)=gaH=aH$ for all $a \in G$. Equivalently, it implies  $a^{-1}ga \in H$ for all $a \in G$. Therefore, $\ker(\varphi)=A_H$, and so $A_H \lhd G$.

Now, I want to directly prove, first, $A_H$ is a subgroup of $G$, then it is normal in $G$.
I cannot even prove $A_H$ is a subgroup because I cannot prove $A_H$ is closed under inverse.
Can anyone prove $A_H \lhd G$ directly?
 A: Let $B$ be the subgroup generated by $A_H$. Since $B$ is generated by a set closed under taking conjugates, it is a normal subgroup of $G$. Also, since $A_H\subseteq H$, $B\leq H$. But now $B$ is a union of conjugacy classes since it is normal, and thus every conjugacy class in $B$ lies in $A_H$ by definition of $A_H$. Thus $A_H=B$.
A: For inverses: the fact that ${a \in A_H}$ we know means ${[a]\subseteq H}$. In particular, ${\forall\ g \in G: g^{-1}ag\in H}$. Since $H$ is closed under inverses we must then also have ${g^{-1}a^{-1}g \in H}$ for every ${g \in G}$, and so then ${[a^{-1}]\subseteq H}$. Hence ${a^{-1} \in A_H}$.
If you want to see the rest let me know and I'll add more
A: Despite all the answers above, I have not seen that $$A_H=core_G(H):=\bigcap_{g \in G} H^g$$
the intersection of all conjugates $H^g=g^{-1}Hg$ of $H$ in $G$. This is the largest normal subgroup of $G$ contained in $H$. It also appears as the kernel of the action of $G$ via right (left) multiplication on the right (left) cosets of $H$.
A: I will use the one-step subgroup test.
By definition, $A_H\subseteq G$.
For all $g\in G$, we have $geg^{-1}=e\in H$, so that $\overline{e}=\{e\}\subseteq H$. Thus $e\in A_H$. Thus $A_H\neq\varnothing$.
Let $a,b\in A_H$. Then $\overline{a}, \overline{b}\subseteq H$. Consider $\overline{ab^{-1}}$: it is the set of all $k\in G$ such that $k=g(ab^{-1})g^{-1}$ for some $g\in G$; but then
$$k=(gag^{-1})(gb^{-1}g^{-1})\in \overline{a}(\overline{b^{-1}}),$$ where $\overline{b^{-1}}$ is a subset or equal to $H$, since $H$ is closed under inverses and $\overline{b}\subseteq H$ (specifically: $gb^{-1}g^{-1}=(gbg^{-1})^{-1}\in H$). But the set product $AB$ of any $A,B\subseteq H$ is in $H$ as $H$ is itself a group. Therefore, $\overline{ab^{-1}}\subseteq \overline{a}(\overline{b^{-1}})^{-1}\subseteq H$. Hence $ab^{-1}\in A_H$.
Hence $A_H\le G$.

For normality, let $g\in G$ and $a\in A_H$. Then $\overline{a}\subseteq H$ and so
$$\begin{align}
\overline{gag^{-1}}&=\{d(gag^{-1})d^{-1}\mid d\in G\}\\
&=\{(dg)a(dg)^{-1}\mid d\in G\}
\end{align}$$
is simply $\overline{a}$ because $\rho_g:G\to G$, given by $q\mapsto qg$, is a bijection. Therefore, $\overline{gag^{-1}}\subseteq H$, so that $gag^{-1}\in A_H$.
Hence $A_H\unlhd G$.
A: The only things (slightly)  non-trivial here are closure under multiplication and inverses of $A_H$.
($e$ is its own conjugacy class and in $A_H$, and a subgroup that is a union of conjugacy classes is always normal.)
For closure under multiplication,  let $a,b\in A_H$.  Then for all $g\in G$, we have $gag^{-1}\in H$ and $gbg^{-1}\in H$.  Thus $gabg^{-1}=gag^{-1}gbg^{-1}\in H$, since $H$ is a subgroup.   Thus $\bar{ab}\subset H$.
For closure under inverses, suppose $a\in A_H$.  Then $\bar{a}\subset H$, so $gag^{-1}\in H,\,\forall g\in G$.  Since $H$ is a subgroup,  $H\ni (gag^{-1})^{-1}=(g^{-1})^{-1}(ga)^{-1}=ga^{-1}g^{-1}$.  Since this is true for any $g\in G$, we have $\bar{a^{-1}}\subset H$.

@Nicky Hekster found us out.   It was in a bit of a disguise.   But this is just the normal core:   $$\bigcap_{g\in G}gHg^{-1}$$.
