TITLE: Let $G$ be a group, if $H=\{b\in G\ |\ bab^{-1} \in \langle a \rangle\} $ is H a subgroup of G?
QUESTION [5 upvotes]: I've seen this question around when $G\wedge\langle a \rangle$ are finite, but what if $G$ is infinite and $\langle a \rangle $ is finite?
My approach:


Show $H$ is closed, which just follows from the cyclicality of $\langle a \rangle$. 
In the case that $G$ is infinite I also have to show that $\forall b \in H,\ b^{-1} \in H$.


I can't figure out how to show condition 2. I was hoping somone could help me out

REPLY [2 votes]: For $b\in H$, since $c=bab^{-1}$ has the same order as $a$, it is a generator of $\langle a\rangle$. That is, $a=c^k$ for some $k$. Thus
$$b^{-1}ab=b^{-1}c^kb=b^{-1}(bab^{-1})^kb=a^k.$$
So $b^{-1}\in H$.

Note: the argument can be reversed. If $b,b^{-1}\in H$, then 
$$b^{-1}ab=a^k=b^{-1}c^kb.$$
So $a=c^k$, and $c$ is a generator of $\langle a\rangle$.
In the case $a$ has infinite order, it implies that $c=a$ or $c=a^{-1}$. So in general, $H$ is not a subgroup if $a$ has infinite order.