Cantor-Bernstein theorem for magmas

Question

Let $G$, $H$ be magmas.

$G_1 \subset G$ - submagma of $G$, $H_1 \subset H$ - submagma of $H$.

Let $G \simeq H_1$, $H \simeq G_1$.

Is true that $G \simeq H$?

Answer 1

No, here's a counterexample.

Let $H = \mathbb R^2$ under addition

$G=H_1 = \mathbb R \times [0,\infty)$

$G_1 = \mathbb R \times \{0\}$

It is well known that $H \cong G_1$ all we need to prove is that $G \not \cong H$.

To that end observe that each element of $H$ has an inverse but there are one or more elements of $G$ with no inverse. This prohibits there being ANY isomorphism between the two magmas.

Answer 2

No, not for magmas, not even for Abelian groups. Quoting this answer by Tom Leinster from Math Overflow:

[. . .] That is, there exists an abelian group $A$ isomorphic to $A^3$ but not $A^2$. This result is due to A.L.S. (Tony) Corner, and is the case $r = 2$ of the theorem described in the following Mathematical Review.

MR0169905 Corner, A.L.S., On a conjecture of Pierce concerning direct decomposition of Abelian groups. 1964 Proc. Colloq. Abelian Groups (Tihany, 1963) pp.43--48 Akademiai Kiado, Budapest.

It is shown that for any positive integer $r$ there exists a countable torsion-free abelian group $G$ such that the direct sum of $m$ copies of $G$ is isomorphic to the direct sum of $n$ copies of $G$ if and only if $m \equiv n (\mod r)$. This remarkable result is obtained from the author's theorem on the existence of torsion-free groups having a prescribed countable, reduced, torsion-free endomorphism ring by constructing a ring with suitable properties. It should be mentioned that the question of the existence of algebraic systems with the property stated above has been considered by several mathematicians. The author has been too generous in crediting this "conjecture" to the reviewer.

Reviewed by R.S. Pierce

Setting $r=3$ in Corner's theorem, we get a countable Abelian group $A$ which is isomorphic to $A^4$ but not to $A^2.$ Setting $B=A^2,$ we have countable Abelian groups $A$ and $B$ such that $A\cong B^2$ and $B\cong A^2$ but $A\not\cong B.$ (Of course, $A$ and $B$ are elementarily equivalent; it is a theorem of model theory that, whenever $A\times B\times C$ is elementarily equivalent to $A,$ then so is $A\times B.$)