TITLE: Is there a group $G$ and subgroup $H$, such that there exists $g\in G$ with $gHg^{-1} \subset H$ and $|H:gHg^{-1}|$ is infinite?
QUESTION [16 upvotes]: Question (asking on behalf of my friend who studies abstract algebra):

Is there a group $G$ and subgroup $H$, such that there exists $g\in G$ with $gHg^{-1} \subset H$ and  $|H:gHg^{-1}|$ is infinite? ( I incline to think this is true.)

For such an example to exist, $H$ (and hence $G$) must be infinite and a non-normal subgroup of $G$. At first, it seems easy. However, I really don't know many types of infinite nonabelian groups (perhaps only the general linear group ${GL}_n(F)$ and the group of bijections). Thanks for your slightest effort.

REPLY [4 votes]: $\newcommand{\bZ}{\mathbf Z}\newcommand{\bN}{\mathbf N}$
Let $\overline{H}$ be the group of functions $\bZ\to \bZ$ (with pointwise addition), while $H$ is the group of functions which vanish on negative integers.
Now, consider the semidirect product $G:=\overline{H}\rtimes \bZ$, where $\bZ$ acts on $\overline{H}$ by shifting to the right, so $(1\cdot f)(k)=f(k+1)$. Then if you take $g=(0,1)\in G$, then $gHg^{-1}\unlhd H$ (it is just the set of functions $\bZ\to \bZ$ which vanish on non-positive integers) and $H/gHg^{-1}\cong \bZ$.
If you replace functions $\bZ\to \bZ$ with functions into any other group $K$, then you will have $H/gHg^{-1}\cong K$.
Note that this construction is the (unrestricted) wreath product of $\bZ$ with itself.