This is a general question about homomorphisms on groups, rings, and fields.
If we are given a surjective homomorphism $f:A \rightarrow B$ and an injective homomorphism $g:A \rightarrow B$, can we say that A and B are isomorphic? Does the answer change depending on the structure of A and B, i.e. whether they are groups, rings, or fields?
Best Answer
This is false for groups and rings. For example, let $A=\prod_{i=1}^\infty \mathbb Z$ and $B=\mathbb Z/2\mathbb Z\times A$. Here, $A$ and $B$ can be viewed either as groups or as rings.
There is an obvious injection $A\to B$. There is also a surjection given by $$(a_1,a_2,\ldots)\mapsto (a_1\pmod 2, a_2,a_3,\ldots).$$ But certainly $A\not\cong B$, since the latter has an element of order $2$.
However, this is true for fields: if $g:A\to B$ is a surjective homomorphism of fields, then its kernel must be an ideal of $A$. Since $g$ is surjective, the kernel cannot be the whole of $A$. Hence $\ker g$ is trivial, so $g$ is also injective.