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.


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

This is not true in general.

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 $H$.

  • $\begingroup$ No wonder, I could not prove it, then. Thank you for clarifying. What is meant by H $\leq$ G though? $\endgroup$ – Marsl Sep 21 '18 at 14:28
  • $\begingroup$ @Marsl For a group $G$, the notation $H\le G$ means that $H$ is a subgroup of $G$. $\endgroup$ – Santana Afton Sep 21 '18 at 14:29
  • $\begingroup$ @Marsl where did you "come across this statement"? $\endgroup$ – Randall Sep 21 '18 at 14:56

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