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I read in some text the following statement:

Let $H$ be a subgroup of $G$. Denote $N=\bigcap_\limits{x\in G} xHx^{-1}$, then $N$ is the largest normal subgroup of $G$ contained in $H$.

It's easy to show $N<G$, since $H$ is a subgroup of $G$ any conjugate $xHx^{-1}~(x\in G)$ of $H$ is also a subgroup of $G$, and the intersection of subgroups is also a subgroup. $N\lhd G$ is also easily shown, if $n\in N$ then for any $g\in G$ there exists $h\in H$ such that $n=ghg^{-1}$, and for any $x\in G$ we have $xnx^{-1}=x(ghg^{-1})x^{-1}=(xg)h(xg)^{-1}$. Since for any $g$ such an $h$ always exists and $x\mapsto xg$ is surjective, clearly $xnx^{-1}\in N$. $N\subseteq H$ because $1H1^{-1}=H$ is one of the intersecting subgroups.

It's left to show $N$ is the largest normal subgroup contained in $H$, which I do not know how to achieve. I appreciate any help or hint, thanks.

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Suppose there is a larger normal subgroup of $H$, i.e. some subgroup $N'$ exists such that $$N \subseteq N' \subseteq H.$$ We need to show that $N' = N$. We only need to show that $N' \subseteq N$.

Take any $g \in N'$ and $x \in G$. Then $x^{-1} g x \in N' \subseteq H$, thus $g \in xHx^{-1}$. This is true for any $x \in G$, hence $$g \in \bigcap_{x \in G} xHx^{-1} = N,$$ completing the proof.

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  • $\begingroup$ Very concise, thanks! (Though there seems to be a small typo in the last equation.) $\endgroup$ – Yinfeng LU Jun 25 '20 at 6:51
  • $\begingroup$ @Lu_Yinfeng Thanks $\endgroup$ – user803264 Jun 25 '20 at 6:52
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HINT: show that any normal subgroup of $G$ contained in $H$ is contained on N

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