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.


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.

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.