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)



Is Higman's group a free product?
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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Baumslag%E2%80%93Solitar_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](https://en.wikipedia.org/wiki/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.


