Embeddings $A → B → A$, but $A \not\cong B$?

Solution 1:

Modules, rings: $A=\Bbb Q^{\oplus\omega}$, $B=A\oplus\Bbb Z$. To see $A\not\cong B$ consider additive divisibility.

Fields: For every char $p\ge0$ and cardinal $\kappa\ge{\frak c}$ there exists a unique algebraically closed field of characteristic $p$ and cardinality $\kappa$. If $F$ is an infinite field then $|\overline{F}|=|F|$. Let $F$ be an algebraically closed field of cardinality $|F|\ge{\frak c}$. Then $\overline{F(T)}\cong F$ which yields $F(T)\hookrightarrow F$. Thus we achieve a sequence $F(T)\hookrightarrow F\hookrightarrow F(T)$. To see why $F\not\cong F(T)$, note $F(T)$ is not algebraically closed.

Linear (hence lattice, partial) orders: $A=(0,1)$, $B=[0,1)$. To see $A\not\cong B$ consider minima.

The above is also an example for topological spaces: $B$ can be written as a disjoint union of a singleton and a connected subset, while $A$ cannot.

Solution 2:

For groups, you may consider $$\mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_4 \oplus \cdots \hookrightarrow \mathbb{Z}_4 \oplus \mathbb{Z}_4 \oplus \cdots \hookrightarrow \mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_4 \oplus \cdots.$$

They are not isomorphic: in the second group, any element of order two is divisible by 2.

Another example, but finitely-generated, is (where $\mathbb{F}_n$ denotes the free group of rank $n$): $$\mathbb{F}_2 \hookrightarrow \mathbb{F}_3 \hookrightarrow \mathbb{F}_2.$$

See for example here for the existence of the monomorphisms. To prove that $\mathbb{F}_2$ and $\mathbb{F}_3$ are not isomorphic, notice that their abelianizations are respectively $\mathbb{Z}^2$ and $\mathbb{Z}^3$.

Solution 3:

Shapes up to scale: a triangle embeds into a square embeds into a bigger triangle.