# If $h$ is element of a subgroup $H$ of the group $G$, all elements in the conjugacy class of $h$ in $G$ necessarily are members of that subgroup, too. [closed]

I started studying group theory a little while ago and now came across this statement that I am not sure why it is true?

Can someone provide me with a proof. I am sure it is not difficult, but I just do not see it.

## closed as off-topic by Namaste, Derek Holt, Alan Wang, Lord Shark the Unknown, ArsenBerkSep 21 '18 at 20:20

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• Normal subgroup? – Randall Sep 21 '18 at 14:07
• Do you mean the conjugacy class of $h$ in $H$? Or the conjugacy class of $h$ in $G$? – José Carlos Santos Sep 21 '18 at 14:07
• Let $G=S_3$ be the group of permutations of $\{1,2,3\}$, let $H=\{\operatorname{id},(1\,2)\}$. Note that $(1\,2\,3)^{-1}(1\,2)(1\,2\,3)=(1\,3)$ – Hagen von Eitzen Sep 21 '18 at 14:09
• Right, the example shows that it cant be true. – Marsl Sep 21 '18 at 14:26

For example, take $$G = S_n$$ to be the symmetric group on $$n\ge 3$$ letters, and set $$H=\langle(1~2)\rangle$$. The conjugates of $$(1~2)$$ are all the transpositions $$(i~j)$$, but $$H = \{e, (1~2)\}$$ does not contain all of these conjugates.
However, this is true for normal subgroups. If $$H\le G$$ is normal, then $$gHg^{-1}\subseteq H$$ for all $$g\in G$$. Given any $$h\in H$$, this means that $$ghg^{-1}\in H$$, so all conjugates of $$h$$ are in $$H$$.
• No wonder, I could not prove it, then. Thank you for clarifying. What is meant by H $\leq$ G though? – Marsl Sep 21 '18 at 14:28
• @Marsl For a group $G$, the notation $H\le G$ means that $H$ is a subgroup of $G$. – Santana Afton Sep 21 '18 at 14:29