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I found this problem in a textbook of abstract algebra:

Let $H$ be a subgroup of $G$. Prove that $$\{x\in G:xHx^{-1}\subseteq H\}$$ is a subgroup of $G$.

It's easy to prove that the set is closed under multiplication, but I'm stuck on proving that it is closed under inverses.

If $H$ is finite, say $H=\{a_1,\ldots,a_n\}$, suppose $x$ is an element of the set. Then $xa_1x^{-1},\ldots,xa_nx^{-1}$ are all distinct, hence they are exactly $a_1,\ldots,a_n$, in some order. Therefore any element $b\in H$ can be written as $xcx^{-1}$ for some $c\in H$, and hence $x^{-1}bx=x^{-1}(xcx^{-1})x=c$ is also in $H$. So $x^{-1}$ is also an element of the set.

However, the above method does not work if $H$ is infinite. The main idea is to prove that $x^{-1}ax\in H$ for every $a\in H$, given that $xax^{-1}\in H$ for every $a\in H$. I was trying to do some substitutions of $a$ to get the required result, but I can't seem to get the $x^{-1}$ to the left.

Any help would be appreciated. It may be worth mentioning that I just started learning this group theory thing for a few days, so please adjust your explanation accordingly.

Thanks in advance.

1 Answers 1