# 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 (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.

• This question is related, but does not ask for $|H:gHg^{-1}|$ to be infinite. – user1729 Feb 21 '19 at 10:56
• @user1729 Does the matrix example work here? – YuiTo Cheng Feb 21 '19 at 10:58
• Nope; the subgroup $H$ there is cyclic. – user1729 Feb 21 '19 at 10:59

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 $$\,g.\,$$

• I believe this is the restricted wreath product of ${\mathbf Z}$ with itself. – tomasz Feb 24 '19 at 0:22

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.

• Any simpler example? – YuiTo Cheng Feb 21 '19 at 10:51
• Define "simpler". (Personally, I think HNN-extensions are very easy to work with, because of my background, whilst I shy away from matrix groups.) – user1729 Feb 21 '19 at 10:52
• Sorry, but I haven't taken combinatorial group theory... – YuiTo Cheng Feb 21 '19 at 10:53
• It will take me some time to read and understand your answer...Sorry for can't accept it now, but upvote it anyway. – YuiTo Cheng Feb 21 '19 at 11:02

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

• "Wreath product"? That sounds quite technical...Upvote it anyway. – YuiTo Cheng Feb 21 '19 at 12:10
• @YuiToCheng: It is not really. The first two paragraphs have a complete description, the wreath product is just a technical term. If you understand semidirect products, it should be clear. – tomasz Feb 21 '19 at 12:11
• Wreath product is one of the oldest group constructions, and occurs at many places. – YCor Feb 21 '19 at 13:20

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.