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Given the data of a triple $(G,h,k)$ where $G$ is a finite group, and $h,k\in G$ of the same order which together generate $G$, I'm interested in understanding the possible pairs $(i,\alpha)$, where $i : G\hookrightarrow \tilde{G}$ is an injection, and $\alpha\in\tilde{G}$ satisfying

  1. $\tilde{G}$ is a finite group.

  2. There is an $\alpha\in \tilde{G}$ such that $\tilde{G} = \langle G,\alpha\rangle$, and

  3. $\alpha h\alpha^{-1} = k$.

There is a natural way to do this, which is to embed $G$ inside the symmetric group $S_G$ on $G$ via the left regular representation. I'll translate their proof into my situation: Let $H := \langle h\rangle$, $K := \langle k\rangle$. Now let $\sigma$ denote a pair of transversals $x_1,\ldots, x_n$ for $G/H$, and $y_1,\ldots,y_n$ for $G/K$, such that $y_1 = x_1 = 1$. Now define $$\alpha_\sigma : G\rightarrow G \qquad\text{by}\qquad \alpha_\sigma(h^jx_i) = k^j y_i$$ In particular, we have $\alpha_\sigma(h) = k$.

Then, if $\ell_t$ for $t\in G$ denotes the permutation given by left-multiplication by $t$, we wish to check that $\ell_k = \alpha_\sigma\circ\ell_h\circ\alpha_\sigma^{-1}$. To check this, for any $g\in G$, write $g = k^j y_i$, then note: $$\alpha(h^jx_i) = k^jy_i = g\quad\text{so}\quad h^jx_i = \alpha^{-1}(k^jy_i) = \alpha^{-1}(g)$$ Thus, $$(\alpha_\sigma\circ\ell_h\circ\alpha_\sigma^{-1})(g) = \alpha_\sigma(h\cdot h^jx_i)= k^{j+1}y_i = kg = \ell_k(g)$$

In particular, letting $L_G\subset S_G$ be the image of the left regular representation of $G$, then for any choice $\sigma$ as above, letting $G_\sigma := \langle L_G,\alpha_\sigma\rangle\subset S_G$, the pair $(G\subset G_\sigma,\alpha_\sigma)$ satisfies our desired properties.

Unfortunately, the link above doesn't provide any references for this proof.

Some questions: What is the relation between such $G_\sigma$ and the infinite HNN extension $\langle G,\alpha | \alpha h\alpha^{-1} = k\rangle$?

Does this construction yield all pairs $(i,\alpha)$ satisfying (1),(2),(3)? (E.g., is every finite quotient of $\langle G,\alpha | \alpha h\alpha^{-1} = k\rangle$ isomorphic to some $G_\sigma$? If so, given a pair $(i,\alpha)$, how does one recover the transversals $\sigma$ and the permutation $\alpha_\sigma\in S_G$? If not, is there a more general technique for constructing the ones we're missing?

Does this construction satisfy some kind of universal property?

Does there exist a group $G$ for which any pair of generators of $G$ are conjugate (ie, the two generators are conjugate elements) in $G$?

Are any of these questions answered in the literature? (references would be appreciated).

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  • $\begingroup$ Could you clarify what you mean exactly by "any two generating pairs are conjugate"? $\endgroup$ – verret Oct 3 '17 at 1:00
  • $\begingroup$ @verret Sorry that was a typo/badly phrased. Corrected now. $\endgroup$ – Will Chen Oct 3 '17 at 2:29
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This is probably an extended comment rather than an answer. An HNN extension of a finite group is residually finite, in fact, virtually free. The element $\alpha$ has infinite order in the HNN extension so you can find finite images where it maps to elements of arbitrarily large order. Your construction only produces conjugators of bounded order (in fact you create only finitely many images of the HNN extension). So you are missing many quotients.

Added. Let $H$ be the HNN extension. Then your construction shows that there is a normal subgroup $N$ of finite index such that $G\cap N=\{1\}$ (take the quotient map $H\to G_{\sigma}$). It follows that $N$ contains no conjugate of an element of $G$. Therefore, the action of $N$ on the Bass-Serre tree of $H$ is free and so $N$ is a free group. Therefore, $H$ has a free subgroup of finite index and hence $H$ is residually finite. Therefore, any finite set of elements of $H$ can be separated by a finite quotient of $H$. In particular, $G$ and any power of $\alpha$ can be separated in a finite image of $H$.

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  • $\begingroup$ Is it obvious that the infinite HNN extension surjects onto groups $\tilde{G}$ where $\alpha$ has arbitrary large order and yet $\tilde{G}$ is finite? For example, it's clear you can impose the relation $\alpha^m = 1$ for any $m\ge 1$, but I believe this would still give you an infinite group. Is it clear that you can consistently add more relations to make the group finite yet preserving the order of $\alpha$ (and the injectivity of $G\hookrightarrow\tilde{G}$)? $\endgroup$ – Will Chen Oct 3 '17 at 16:12
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    $\begingroup$ As I said, it is well known that an HNN extension of a finite group is residually finite (and in fact has a free subgroup of finite index). You can find this in Serre's book on trees I believe or in the book of Dunwoody and Dicks on groups acting on trees. Since the group is residually finite and $\alpha$ has infinite order, it is possible to find a finite homomorphic image in which $G$ maps invectively and $\alpha^m$ is not $1$. $\endgroup$ – Benjamin Steinberg Oct 3 '17 at 17:37

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