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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 $gH = Hg$.

BMBM
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4 Answers4

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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.

markvs
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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.

David A. Craven
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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.

egreg
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  • The converse: Let $x \in C$. And let $y \in C$ for $y = g^{-1} x g$. Then $xg \in Cg$, we rewrite $xg = g g^{-1} x g = g y$. Thus $Cg \subseteq gC$. We conclude $gC = Cg$ for all conjugacy classes of a group $G$. – BMBM Jul 09 '20 at 04:19
  • But then what is the chain of implications for the question I raised? $H$ is a union of conjugacy classes of $G$, and since for each conjugacy class $gC = Cg$, $H$ is normal in $G$? – BMBM Jul 09 '20 at 04:29
  • @Max If $H=\bigcup C_x$, then $gH=\bigcup gC_x=\bigcup C_xg=Hg$. – egreg Jul 09 '20 at 07:13
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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}