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In his paper A finitely generated infinite simple group (J. London Math. Soc., 1951), Higman introduced the following finitely presented group:

$$ H = \langle x,y,z,w \mid [x,y]=y, \, [y,z]=z, \, [z,w]=w, \, [w,x]=x \rangle . $$

This group has many remarkable properties, including being acyclic.

Let $H_{x,y} \le H$ be the subgroup generated by $x$ and $y$, and let $H_{z,w}\le H$ be the subgroup generated by $z$ and $w$. Both $H_{x,y}$ and $H_{z,w}$ are isomorphic to the Baumslag--Solitar group $BS(1,2)=\langle a,b \mid aba^{-1}=b^2\rangle$ (although this is not particularly relevant to my question).

Does there exist a non-trivial element of $H_{x,y}$ which is conjugate in $H$ to an element of $H_{z,w}$?

It would seem to me that the answer is no, but I'm having difficulty proving or disproving it. Probably I am missing some well-known technique for deciding this type of question, which I would be very happy to hear about.

Motivation: My co-authors and I are writing about the topological complexity (in the sense of Farber) of aspherical spaces, and the above fact (if it were true) would give a nice example illustrating our techniques.

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I'd say that the usual technique for deciding this sort of question is to find a nice quotient in which the answer is 'obvious'. Since Higman's group has no finite quotients, and hence very few 'nice' quotients, this technique isn't going to work here. –  HJRW Aug 7 '13 at 12:28
    
I vaguely remember there was some nice lemma in the book of Higman, Scott "Existentially closed groups" that two elements of a group are conjugate if and only if some system of group equations has a solution, and, I can't remember which one, but some of Higman's simple groups was used for that heavily. Sorry I don't have this book -- this is in Chapter 1 –  Victor Aug 7 '13 at 12:42
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1 Answer

up vote 10 down vote accepted

No, there are no such element. Indeed $H$ is amalgam of the subgroups $\langle x,y,z\rangle$ and $\langle z,w,x\rangle$ over the intersection $\langle x,z\rangle$, which is free (as we see by viewing $\langle x,y,z\rangle$ itself as amalgam of the Baumslag-Solitar $\langle x,y\rangle$ and $\langle y,z\rangle$ over $\langle y\rangle$). [EDIT: my previous argument was mistaken] In an amalgam $A*_CB$, an element of $A$ conjugate to an element of $B$ is actually conjugate to an element of $C$.

Thus an element $u_1$ of $\langle x,y\rangle$ conjugated to an element $u_2$ in $\langle z,w\rangle$ is conjugated to an element in the amalgamated subgroup defining $H$, namely $\langle x,z\rangle$. But we can also view $H$ as an amalgam of $\langle w,x,y\rangle$ and $\langle y,z,w\rangle$ and the argument now implies that $u_1$ is conjugated to an element in $\langle y,w\rangle$.

So some element of $\langle x,z\rangle$ (conjugate to $u_1$) is conjugated to some element in $\langle y,w\rangle$. Using the first amalgam decomposition of $H$ and its Bass-Serre tree, $\langle x,z\rangle$ acts with a fixed vertex. On the other hand, an element of $\langle y,w\rangle$ not conjugated into the cyclic subgroups $\langle y\rangle$ or $\langle w\rangle$ acts hyperbolic. So every element in $\langle y,w\rangle$ conjugate to $u_1$ is conjugate in $\langle y,w\rangle$ to an element in $\langle y\rangle\cup\langle w\rangle$. Again using this argument in the other decomposition of $H$, we get that every element in $\langle x,z\rangle$ conjugate to $u_1$ is conjugated in $\langle x,z\rangle$ to an element in $\langle x\rangle\cup\langle z\rangle$.

Since these element exist, we deduce that that some element in $\langle x\rangle\cup\langle z\rangle$ (conjugate to $u_1$) is conjugated to an element in $\langle y\rangle\cup\langle w\rangle$.

By symmetry, we can suppose that some element in $\langle x\rangle$ (conjugate to $u_1$) is conjugate to an element in $\langle y\rangle$.

Now I use that if in an amalgam $A*_CB$ an element of $A$ is conjugate to an element of $C$, then it's conjugate to an element of $C$ by an element of $A$.

Using the first decomposition of $H$, we deduce that some element of $\langle y\rangle$ (conjugate to $u_1$) is conjugate to an element of $\langle x,z\rangle$ by an element of $\langle x,y,z\rangle$, and by the "every" statement above, we can choose the latter element to belong to $\langle x\rangle\cup\langle z\rangle$.

Now in the group $\langle x,y,z\rangle$ (with its explicit presentation with 2 relators), nontrivial powers of $y$ cannot be conjugate to nontrivial powers of $x$ or $z$. Indeed the powers of $x$ survive in the abelianization but not those of $y$ or $z$; on the other hand, nontrivial powers of $y$ are at most exponentially distorted, while $z^{k2^{2^n}}=x^nyx^{-n}z^kx^ny^{-1}x^{-n}$ and thus $z^k$ is partially more than exponentially distorted. So finally the only possibility is $u_1=1$.

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Wonderful, thanks! Do you know of a reference for the fact that an element of $A-C$ can't be conjugate to an element of $B-C$ in $A\ast_C B$? –  Mark Grant Aug 7 '13 at 13:38
    
@Mark it's false actually ($aca^{-1}$ being conjugate to $bcb^{-1}$). What's true is that an element of $A$ conjugate to an element of $B$ is conjugate to an element of $C$. Let me try to fix the answer. –  Yves Cornulier Aug 7 '13 at 14:20
    
... and it's now much more complicated; I hope it's OK... –  Yves Cornulier Aug 7 '13 at 16:03
    
Thanks so much for your time. I will try to check the details (and hopefully learn some Bass--Serre theory in the process). –  Mark Grant Aug 7 '13 at 16:07
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