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?
Best Answer
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 $A_H=B$.