Suppose there exists an surjective homomorphism from group $G$ to group $H$ and a second surjectibehomomorphism from $H$ to $G$. Does this imply $G$ and $H$ are isomorphic?
I know there exists an invertible function from the set in $G$ to the set in $H$ by the Schröder–Bernstein theorem, but I am unsure if this invertible function will also be a homomorphism.
Note: I initially did not follow etiquette on edits, so I am reverting to the question that was answered in the top response. I will not do so again.
Note: Original question was about surjections, not injections. It was changed by the OP long after this answer was posted.
No. There exists a countable torsionfree abelian subgroup $A$ which is isomorphic to $A\times A\times A$, but not to $A\times A$. This is a famous result of A.L.S. Corner..
Let $G=A$ and $H=A\times A$. Then each has the other as a quotient (use the isomorphism of $A$ with $A\times A\times A$ to get $H$ as a quotient of $G$), but they are not isomorphic.