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):


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*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 $gH = Hg$.

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.
