# Suppose $H<G$, let $N=\bigcap_{x\in G} xHx^{-1}$, show $N$ is the largest normal subgroup of $G$ contained in $H$.

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 $$N, since $$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.

## 2 Answers

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.

• Very concise, thanks! (Though there seems to be a small typo in the last equation.) – Yinfeng LU Jun 25 '20 at 6:51
• @Lu_Yinfeng Thanks – user803264 Jun 25 '20 at 6:52

HINT: show that any normal subgroup of $$G$$ contained in $$H$$ is contained on N