If $H$ is a union of conjugacy classes, then $gH = Hg$? I'm going through some properties of normal subgroups. One of them is that if a subgroup $H$ is a union of conjugacy classes of a group $G$, then $H$ is normal in $G$.
I found a couple of proofs for this, one for example on the proof wiki goes roughly like this (summary of my understanding):


*

*Assume $H$ is the union of the conjugacy class of each $x \in H$. Call this conjugacy class $C_x$.

*Then, for each $x \in H$, $C_x \subseteq H$.

*Then, for each $x \in H$ and for each $g \in G$, $gxg^{-1} \in H$.

*So for each $g \in G$, $gHg^{-1} \subseteq H$.

*This means $H$ is normal in $G$.


At the moment the proof concludes with:


*

*So for each $g \in G$, $gHg^{-1} \subseteq H$.


Which is equivalent to $gH = Hg$ (as both show a subgroup is normal), but I was wondering if we can find a chain of implications that if $H$ is a union of conjugacy classes, then $gH = Hg$.
 A: Perhaps this is what you want: For every conjugacy class $C_x$ and every $g\in G$, $gC_x=\{gh^{-1}xh=((hg^{-1})^{-1}x(hg^{-1}))g\mid h\in G\}=C_xg$. So every union of conjugacy classes satisfies the same property.
A: Do you want something like this? If $g\in G$ and $h\in H$ then $gh=ghg^{-1}g=h'g$ for some $h'\in H$, as $h'=ghg^{-1}$ is conjugate to $h$, so lies in $H$. Thus $gH\subseteq Hg$. The converse is similar.
A: Prove that $gC=Cg$ where $C$ is any conjugacy class. Indeed, if $x\in C$, then also $y=gxg^{-1}\in C$ and so
$$
gx=gxg^{-1}g=yg\in Cg
$$
The converse follows in the same way.
A: This might be what you are looking for ($X\subseteq G$):
\begin{alignat}{1}
g^{-1}Hg &= g^{-1}\bigcup_{x\in X}\operatorname{Cl}(x)\space g \\
&= g^{-1}\bigcup_{x\in X}\{g'xg'^{-1}\mid g'\in G\}\space g \\
&= \bigcup_{x\in X}\{(g^{-1}g')x(g'^{-1}g)\mid g'\in G\} \\
&= \bigcup_{x\in X}\{(g^{-1}g')x(g^{-1}g')^{-1}\mid g'\in G\} \\
&= \bigcup_{x\in X}\{g''xg''^{-1}\mid g''\in G\} \\
&= \bigcup_{x\in X}\operatorname{Cl}(x) \\
&= H
\end{alignat}
