A question about conjugacy in Higman's group

Question

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.

Answer 1

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 $u_1=1$.

Answer 2

Below is a geometric argument based on the action of Higman's group on its natural CAT(0) square complex. (Of course, Yves' argument is more elementary, but I find this alternative viewpoint enjoyable.)

The key observation is that $H$ can be described as the fundamental group of a square of groups: $$\begin{array}{ccc} \langle x,y \rangle & \leftarrow \langle y \rangle \rightarrow & \langle y,z \rangle \\ \uparrow & \uparrow & \uparrow \\ \langle x \rangle & \leftarrow \{1\} \rightarrow & \langle z \rangle \\ \downarrow & \downarrow & \downarrow \\ \langle w,x \rangle & \leftarrow \langle w \rangle \rightarrow & \langle z,w \rangle \end{array}$$ Now, this square of groups is nonpositively curved (as defined by Gersten and Stallings), so it comes from an action of $H$ on some CAT(0) square complex $X$. (Following the construction of the Bass-Serre tree associated to a graph of groups, $X$ can be described directly in terms of cosets of the vertex-, edge-, square-groups of the square of groups.) The action $H \curvearrowright X$ is studied in details in Alexandre Martin's paper On the cubical geometry of Higman's group, and there it is proved the following weak acylindricity (see Corollary 2.2, whose proof uses only elementary arguments):

Weak acylindricity: For any two vertices $x,y \in X$ at distance at least three in the $1$-skeleton $X^{(1)}$, the intersection $\mathrm{stab}(x) \cap \mathrm{stab}(y)$ is trivial.

Now, let $\alpha$ be an element of $H$ which belongs to $\langle x,y \rangle$ and which is conjugate to an element of $\langle z,w \rangle$, say $g^{-1}\alpha g \in \langle z,w \rangle$ for some $g \in H$. Consequently, if $A$ (resp. $B$) denotes the vertex of $X$ corresponding to the coset $\langle x,y \rangle$ (resp. $\langle z,w \rangle$), then $\alpha$ belongs to $\mathrm{stab}(A) \cap \mathrm{stab}(gB)$. Two cases may happen:

• Either $d(A,gB) \geq 3$, so that $\mathrm{stab}(A) \cap \mathrm{stab}(gB)=\{1\}$ as a consequence of the weak acylindricity.
• Or $d(A,gB) \leq 2$. In this case, it is not difficult to show that the only possible configuration is that $A$ and $gB$ are two diametrically opposed vertices of some square $C$, hence $$\mathrm{stab}(A) \cap \mathrm{stab}(gB) \subset \mathrm{stab}(C) = \{1\},$$ since square-stabilisers are trivial (this follows from the fact that the square-group of the square of groups defining $H$ is itself trivial).

The conclusion is that $\alpha$ is necessarily trivial.