Let $G$ be a finitely generated group splitting non-trivially over
a infinite virtually cyclic subgroup $V$, namely an amalgamated free product 


$$
A\*\_V B,\; V \neq A,B.
$$
Question: can the group $G$ be simple? 



Finitely generated group splitting non-trivially over an infinite virtually cyclic subgroup
The answer is no: it's not simple.


It's an easy adaptation of the remarks done in Button's paper. 


The background: (a) Minasyan-Osin: let $G$ be an amalgam $G=A\ast\_C B$ of two groups over a proper subgroup $C$, such that $C$ is weakly malnormal in $B$. Then $A\ast\_C B$ is acylindrically hyperbolic. [In case of index (2,2), there's a homomorphism onto the infinite dihedral group.] (b) (Osin) Every finitely generated acylindrically hyperbolic group is SQ-universal (and hence has continuum many normal subgroups).


Given this it is not hard to conclude. Consider an amalgam $A\ast\_C B$ with $C$ virtually cyclic. It remains to consider the case when $C$ is weakly malnormal in neither $A$ nor $B$. Since $C$ is infinite virtually cyclic, it follows that it is a commensurated subgroup, in the sense that it is commensurate to all its conjugates, in both $A$ and $B$ (two subgroups of a group are commensurate if their intersection has finite index in both).


Given a group $H$, its abstract commensurator $\mathrm{Comm}(H)$ is the set of isomorphisms $f$ between two finite index subgroups $H\_1$, $H\_2$, modulo coincidence on a smaller finite index subgroup. If $L$ has finite index in $H$, then the inclusion $L\to H$ induces an isomorphism $\mathrm{Comm}(L)\simeq\mathrm{Comm}(H)$.


Hence for every amalgam $A\ast\_C B$ such that $C$ is commensurated in both $A$ and $B$, we have a homomorphism $A\ast\_C B\to\mathrm{Comm}(C)$. Here, $C$ being virtually (infinite cyclic), we have $\mathrm{Comm}(C)\simeq\mathbf{Q}^\*$.


Hence assuming the amalgam has a finite abelianization, this homomorphism is trivial, which means that deep enough cyclic subgroup $N$ of finite index in $C$ are normal in both $A$ and $B$. Hence $G$ is not simple, and even better, admits the quotient $(A/N)\ast\_{C/N}(B/N)$. In particular, unless $C$ has index 2 in both $A$ and $B$, $G$ is SQ-universal. (If the index is twice 2, we deduce that $G$ is virtually $\mathbf{Z}^2$.)


[Actually the latter conclusion (SQ-university except in index 2-2 case) holds if both homomorphisms $A\to\mathrm{Comm}(C)\leftarrow B$ have a finite image. If both have a nontrivial image and at least one has an infinite image, then since both vanish on $C$, we get a free product quotient and get SQ-universality. It remains the case where one has infinite image and the the other is trivial. In this case Button gets SQ-universality with a more complicated argument in the case $C$ is cyclic. I haven't tried to see if it adapts to $C$ virtually cyclic, but in all cases this is not asked by the OP.]


