I have got the following task: Prove that if $H$ is a subgroup of the group $G$, then $K:=\bigcap_{g \in G} g H g^{-1}$ is a normal subgroup in $G$, it lies inside $H$ and contains each normal subgroup of $G$ with lies in $H$.

For the first task, I have already found an answer here. However, I am still unsure about the two remaining ones.

If I want to prove that $K$ lies in $H$, is it enough to use that since $H$ is a subgroup of $G$, multiplying any element of $H$ with an arbitrary $g \in G$ will always remain in $H$?

For the last one, I should maybe use the fact that the intersection of normal subgroups is also a normal subgroup?

Any help appreciated.

  • 1
    $\begingroup$ The first one you're asking about has the easiest proof: $K=\bigcap_{g \in G} g H g^{-1}\subseteq eHe^{-1}=H$. No, it doesn't work to take an arbitrary $g\in G$, but a special one works (any $h\in H$ would work, but if $gHg^{-1}$ remains in $H$ for any $g\in G$, then $K=H$ and $H$ is normal itself). $\endgroup$ – Arthur Feb 17 '18 at 21:59
  • $\begingroup$ Oh yeah, it is really trivial, thanks for this. :) $\endgroup$ – Atvin Feb 17 '18 at 22:02
  • $\begingroup$ Note that the solution @Arthur gave also solves the last question. $\endgroup$ – TPace Feb 17 '18 at 22:03
  • 1
    $\begingroup$ Why does it solve the last question? I just don't see it :( $\endgroup$ – Atvin Feb 17 '18 at 22:15

I already answered the first question in the comments above. Here is an answer to the second question:

Let $N\subseteq H$ be a normal subgroup of $G$. We want to prove $N\subseteq K$.

We have for any $g\in G$ that $gNg^{-1}\subseteq gHg^{-1}$. But $gNg^{-1}=N$, since $N$ is normal. Therefore $N\subseteq gHg^{-1}$.

Since $N$ is contained in each of the $gHg^{-1}$, it must be contained in their intersection, which is $K$.

  • $\begingroup$ Thank you for this short and nice answer! $\endgroup$ – Atvin Feb 17 '18 at 22:45

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.