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The following question came up in the class I'm teaching right now. There definitely exist groups $G$ with subgroups $H$ such that there exists some $g \in G$ such that $g H g^{-1}$ is a proper subgroup of $H$. For instance, let $G$ be the (big) permutation group of $\mathbb{Z}$ (by the big permutation group, I mean that elements of $G$ can move infinitely many elements of $\mathbb{Z}$). Let $H \subset G$ be the big permutation group of $\mathbb{N}$ and let $g \in G$ be the permutation that takes $n \in \mathbb{Z}$ to $n+1 \in \mathbb{Z}$. Then $g H g^{-1}$ is a proper subgroup of $H$.

The same sort of trick produces many examples like this. However, a feature of all of them is that $G$ is "big" in some way -- for instance, $G$ is not finitely presentable. By Higman's embedding theorem, you can embed such a $G$ into a finitely presentable group, so there exist examples where $G$ is finitely presentable. However, in all the examples I can come up with, the subgroup $H$ is not finitely presentable. I'm pretty sure that there exist examples in which $H$ is finitely presentable. Does anyone know of one? Even better, are there examples in which both $G$ and $H$ are of finite type (ie have compact $K(\pi,1)$'s)?

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There are very simple examples with $H\cong\mathbb{Z}$. For instance let $G$ be the affine linear group over $\mathbb{Q}$ consisting of all maps $x\mapsto ax+b$ where $a\in\mathbb{Q}^*$ and $b\in\mathbb{Q}$. Let $H$ be the set of maps $x\mapsto x+b$ with $b\in\mathbb{Z}$. Then $x\mapsto 2x$ conjugates $H$ into the proper subgroup $2H$.

In this example we could restrict $a$ to be a power of $2$ and $b$ to be a dyadic rational. Then the new $G$ has generators $a:x\mapsto 2x$ and $b:x\mapsto x+1$ and is defined by the single relation $h^2=ghg^{-1}$.

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  • $\begingroup$ Thanks! I'm a little embarrassed that I missed this example! I'd like to point out the question mathoverflow.net/questions/20408/… I just asked, which asks how far this can be generalized. $\endgroup$ Commented Apr 5, 2010 at 18:50

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