# Is Higman's group a free product?

Higman proved the existence of a finitely generated simple group here:

Higman, Graham A finitely generated infinite simple group. J. London Math. Soc. 26 (1951), 61–64.

It is a quotient of what is called Higman's group, which is an amalgamated free product.

Question: is either group isomorphic to a (non-trivial) free product? (i.e. amalgamated over the trivial group)

• Simple groups are not non-trivial free products. – Benjamin Steinberg Jan 13 at 1:14
• Higman never "constructed" such a simple quotient. He constructed the so-called Higman group, and said it admits simple quotient (finitely generated, maybe not finitely presented by the way). It's not a construction. – YCor Jan 13 at 8:09
• Thanks! Corrected these points in the question. – Guest Jan 15 at 16:26

As pointed out by @BenjaminSteinberg in the comments, a simple group cannot be a non-trivial free product.

It's also true that the Higman group cannot be a non-trivial free product. It has a presentation

$$G=\langle a,b,c,d | a^{-1}ba=b^2, b^{-1}cb=c^2, c^{-1}dc=d^2, d^{-1}ad=a^2\rangle.$$

Suppose that $$G \cong A\ast B$$, a free product of non-trivial groups $$A, B$$. Consider the subgroup $$H=\langle a,b\rangle$$. Clearly the relation $$a^{-1}ba=b^2$$ holds in the subgroup $$H$$, and in fact one can show that this is the only relation needed, so $$H\cong BS(1,2)$$, a solvable Baumslag-Solitar group. $$H$$ is non-trivial and not a non-trivial free product or free group since it is solvable and non-cyclic. Then by the Kurosh subgroup theorem, $$H$$ is conjugate into $$A$$ or $$B$$, let's say into $$A$$. Now we may take a non-trivial quotient of $$G\cong A\ast B \twoheadrightarrow B$$ by killing $$A$$. In turn, this kills the subgroup $$H$$ which is conjugate into $$A$$, so kills $$a, b$$. But one sees directly from the presentation that killing $$a$$ and $$b$$, one kills $$c$$ and $$d$$ as well, giving the trivial group, a contradiction.

• Why must the subgroup generated by a be conjugate to a subgroup of a factor? Its an infinite cyclic group if I am not mistaken and so the Kurosh theorem says nothing about it as far as being conjugate into a factor. If you take a product is a nontrivial element of A with a nontrivial element of B you get an element of infinite order which is hyperbolic in its action on the free product Bass Serre tree – Benjamin Steinberg Jan 13 at 2:42
• But the subgroup generated by a,b is a solvable Baumslag Solitar group and so one-ended and you can apply your argument to that. – Benjamin Steinberg Jan 13 at 2:50
• @BenjaminSteinberg Of course you're right. In fact, that was the argument that I started writing up, then thought that this gave a slicker proof (forgetting about the statement of Kurosh). I'll fix, that, thanks. – Ian Agol Jan 13 at 2:56
• This applies more generally as follows: if $G$ is a group, $H$ a subgroup with $H$ freely indecomposable and not infinite cyclic and $H$ generates $G$ normally (i.e. killing $H$ kills $G$): then $G$ is freely indecomposable. – YCor Jan 13 at 8:15
• Btw this promotes to a proof that Higman's group $G$ has no quotient $Q$ that is a nontrivial free product. Indeed, let $R$ be the image of the Baumslag-Solitar group $H$ in $Q$. It is part of the original argument that $G$ has no nontrivial finite quotient, that the only quotient of $G$ in which some of the four generators is torsion, is trivial. Since any proper quotient of the Baumslag-Solitar group $H$ makes the second generator torsion, it follows that $R$ is a copy of $H$, hence my previous comment applies. – YCor Jan 15 at 16:35

Here are some details about constructing normal subgroups and some properties of negative curvature of groups (such as splitting as a free product).

The connection between these two subjects comes from small cancellation theory. Initially, the goal was to solve the word problem in some groups admitting specific presentations $$\langle X \mid R \rangle$$, which amounts to determining when an element of the free group $$F(X)$$ belongs to the normal subgroup $$\langle \langle R \rangle \rangle$$. Such methods have been generalised to many groups, including free products. And, as an application, it allows us to construct "many" non-trivial normal subgroups. One of the main ideas is that, if you take a group $$G$$ and an element $$g \in G$$ that satisfies a good small cancellation property, then $$\langle \langle g^k \rangle \rangle$$ is a normal subgroup of $$G$$ that does not contain $$g$$ if $$k$$ is large enough. The most recent and general theorem is this direction is:

Theorem: (Dahmani-Guirardel-Osin) A group admitting a non-elementary acylindrical action on a Gromov-hyperbolic space is SQ-universal, i.e. every countable group embeds in some quotient of $$G$$.

Notice that being SQ-universal implies having uncountably many normal subgroups, so it is a very strong negation of being simple. The theorem applies to (non-elementary) free products as they act acylindrically on their Bass-Serre trees (but some classical small cancellation can be also used instead, allowing more explicit constructions).

So simple groups are very far away from free products in the world of groups. For Higman's group, the situation is less extreme. In fact, the group is acylindrically hyperbolic, and in particular SQ-univeral like (non-elementary) free products. However, there is a classical hierarchy among negatively curved groups: $$\text{hyperbolic groups \subset relatively hyperbolic groups \subset acylindrically hyperbolic groups}.$$ (Here, I consider only non-virtually cyclic groups for simplicity.) Free products turn out to hyperbolic relative to their factors. Again, this can be seen from the actions on their Bass-Serre trees. However:

Fact: Higman's group is not relatively hyperbolic.

The justification is close to Ian Agol's argument:

Higman's group $$H$$ contains four subgroups $$H_1,H_2,H_3,H_4$$ such that:

1. they are all isomorphic to the Baumslag-Solitar group $$BS(1,2)$$;
2. they generate $$H$$;
3. the intersections $$H_1 \cap H_2$$, $$H_2 \cap H_3$$, $$H_3 \cap H_4$$, $$H_4 \cap H_1$$ are all infinite.

If $$H$$ is relatively hyperbolic, then we know by (1) that each $$H_i$$ must lie in a peripheral subgroup $$P_i$$. But the collection of peripheral subgroups is almost malnormal in a relatively hyperbolic group, so (3) implies that the $$P_i$$'s must be all the same. We conclude by (2) that our peripheral subgroup must be the whole group.

So, loosely speaking, free products are "more negatively curved" than Higman's group.

• I think the argument shows that whenever $H$ acts isometrically on a (nonempty) Gromov-hyperbolic space, either the action is horocyclic (there's a unique fixed point at infinity and no loxodromic), or each generator, actually each $H_i$, acts with bounded orbits. In particular every proper such action is horocyclic (while an infinite relatively hyperbolic group always admits a proper non-horocyclic action). – YCor Jan 13 at 9:46
• Thanks! This is very informative. – Guest Jan 15 at 16:28

The abelianization of the Higman group $$G$$ is trivial, because in the presence of commutativity the relations become $$a = 1$$, $$b =1$$, $$c =1$$, $$d=1$$. Therefore $$G$$ is not free.

• This doesn't answer the question. – YCor Jan 15 at 16:37
• @YCor I don’t understand? The abelianization of a nontrivial free group is not zero. Since the abelianization of $G$ is zero, it’s not free? – Jeff Strom Jan 15 at 18:19
• Your argument is correct: precisely it proves that it's not (nontrivial free); it's actually not a trivial group, which is not that obvious, so it's not free at all. But the question is not whether the Higman group is free, but whether it's a nontrivial free product. – YCor Jan 15 at 18:41