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

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.

Answer 1

Let $\,G\,$ be the free group generated by $\,g,x_1,x_2,\dots\,$ modded by the equations $\,gx_ng^{-1} = x_{n+1}\,$ for all $n>0.$ Let $\,H\,$ be the subgroup generated by all the $\,x_n.\,$ The index of $\,gHg^{-1}\,$ in $\,H\,$ is infinite because $\,x_1\,$ has infinite order. Note that there are more concrete ways to represent the groups and specializations where all the $\,x_n\,$ commute with each other but not with $\,g.\,$

Answer 2

Yes.

For example, take $G$ to be the HNN-extension* $$\langle a, b, t\mid t^{-1}at=[a, b], t^{-1}bt=b\rangle$$ and $H=\langle a, b\rangle$. By the theory of HNN-extensions, $H$ embeds into $G$ in the natural way. Clearly $t^{-1}Ht\leq H$. To see that $|H:t^{-1}Ht|$, note that the presentation $\langle a, b\mid [a, b], b\rangle$ defines an infinite group. Hence the smallest normal subgroup of $H$ containing $[a, b]$ and $b$ has infinite index in $H$; hence the smallest subgroup of $H$ containing $[a, b]$ and $b$, which is $\langle [a, b], b\rangle$, has infinite index in $H$. As $t^{-1}Ht=\langle [a, b], b\rangle$, the result follows.

*For background on HNN-extensions see Wikipedia or the book Combinatorial group theory by Lyndon and Schupp.

Answer 3

$\newcommand{\bZ}{\mathbf Z}\newcommand{\bN}{\mathbf N}$ 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.

Answer 4

This is a very good question, and the answer is yes.

An example is given by $G= \mathfrak{S}\mathbb{Z}$, the group of permutations of $\mathbb{Z}$, and $H$ is the subgroup of permutations that fix $\mathbb{Z}_{-}$ pointwise (so the image of $\mathfrak{S}\mathbb{N}$ under the obvious morphism)

Now for $g\in G$, $gHg^{-1}$ is the group of permutations that fix $g\mathbb{Z}_{-}$ pointwise, so whenever $\mathbb{Z}_{-}\subset g\mathbb{Z}_{-}$, we have $gHg^{-1}\subset H$.

In particular if $g\mathbb{Z}_{-}\setminus\mathbb{Z}_{-}$ is infinite, then $gHg^{-1}$ has infinite index in $H$, and of course this can happen if you choose $g$ well enough.