A question about generating set of groups and epimorphism

Question

Do there exist non-isomorphic finitely generated groups, $G$ and $H$, along with epimorphisms $\phi:G\rightarrow H$ and $\psi:H\rightarrow G$, such that every generating set of these groups is an image of a generating set, that's mean, for every generating sets $\{g_1,\dotsc,g_m\}$ and $\{h_1,\dotsc,h_n\}$ of $G$ and $H$ respectively, there are generating sets $\{y_1,\dotsc,y_m\}$ and $\{x_1,\dotsc,x_n\}$ such that $$g_i=\psi(y_i),\quad\text{and}\quad h_j=\phi(x_j)$$

Answer 1

In the paper,

S. Thomas, On the concept of "largeness'' in group theory J. Algebra 322 (2009), no. 12, 4181–4197,

it is shown that your "bi-surjectabilty relation" between finitely generated groups is strictly more complex (in the sense of Borel reducibility) than the isomorphism relation. The proof necessarily gives uncountably many examples of bi-surjective groups which are not isomorphic.

Answer 2

I only want to explain Yves's example in details.

Put $\mathbf Z:=\mathbb Z[t,t^{-1}]$, the ring of Laurent polynomials. Let $B$ be a group of matrices [\begin{pmatrix} 1&P&R\\ 0&D&Q\\ 0&0&1 \end{pmatrix} ] where $P$, $Q$ and $R$ belongs to $\mathbf{Z}$, and $D\in\langle t\rangle$. The group $B$ is easily checked to be finitely generated, and its center consists of unipotent matrices with a single possibly non-trivial element in the upper right corner. It is clearly isomorphic to $\mathbf{Z}$. Also, the map $$ \Phi:B\rightarrow B,\quad \begin{pmatrix} 1&P&R\\‎ 0&D&Q\\‎ 0&0&1 \end{pmatrix}‎\mapsto ‎‎ \begin{pmatrix} 1&P&tR\\‎ 0&D&tQ\\‎ 0&0&1 \end{pmatrix}‎$$ introduces an automorphism of $B$.

Set $N_k:=\bigoplus\{\mathbb Zt^m:m=k,k+1,\dots\}$, $k\in\mathbb Z$. The automorphism $\Phi$, implies that $B/N_k\cong B/\Phi(N_k)=B/N_{k+1}$, $k\in\mathbb Z$. Now, let $G:=B/N_0$ and $H:=B/(2\mathbb Z\oplus N_1)$. Then non-isomorphic groups $G$ and $H$ satisfy the conditions.