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.



A question about conjugacy in Higman's group
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 $u\_1=1$.


