Structures that are reciprocally homomorphic images but are not isomorphic

Question

Consider two algebras $\mathscr{A}=\langle A,O_i\rangle$ and $\mathscr{B}=\langle B,P_i\rangle$ with the same type. Suppose that there is a homomorphism from $A$ onto $B$, and a homomorphism from $B$ onto $A$. Clearly (edit: not really), $|A|=|B|$. It is easy to prove that if the algebras are finite, they must be isomorphic, but can we conclude the same thing for infinite algebras? I can find a counterexample for the same question about relational structures (sets equipped with relations) but my counterexample uses a one-many relation (the structures are $\big\langle\mathbb{N},\{\langle1,1\rangle,\langle1,2\rangle\}\big\rangle$ and $\big\langle\mathbb{N},\{\langle1,1\rangle\}\big\rangle$).

Answer 1

Consider the abelian groups $G = \mathbb{Z}/4\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}\oplus \dots$ (countably many copies) and $H = \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}\oplus \dots$ (countably many copies).

There is a surjective homomorphism $G\to H$ which is the quotient map $\mathbb{Z}/4\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ on the first component and the identity on all other components: $(a_1,a_2,a_3,\dots)\mapsto (\overline{a}_1,a_2,a_3,\dots)$ where $\overline{a}_1$ is the residue of $a_1$ mod $2$.

There is also a surjective homomorphism $H\to G$ which projects out the first component: $(a_1,a_2,a_3,\dots)\mapsto (a_2,a_3,a_4,\dots)$.

But these groups are not isomorphic. Note that the element $(1,0,0,\dots)$ in $H$ has order $2$ but is not divisible by $2$. On the other hand, every element of order $2$ in $G$ is divisible by $2$, since it has a $0$ or a $2$ in each component.

Answer 2

Let $G$ be an infinite group and let $H$ be a nontrivial proper subgroup. Let $A$ be the $G$-set that has countably infinitely many singleton orbits and countably infinitely many regular orbits (i.e., orbits isomorphic to ${}_GG$). Now let $B$ be constructed exactly like $A$, but include one more orbit isomorphic to ${}_G(G/H)$. $A$ is not isomorphic to $B$, since $B$ contains a point whose stabilizer is $H$ and $A$ does not. But there is a homomorphism of $A$ onto $B$ that collapses one regular orbit to ${}_G(G/H)$, and there is a homomorphism of $B$ onto $A$ that collapses the orbit ${}_G(G/H)$ onto a singleton.