Why aren't all homomorphic abelian groups isomorphic as well?

Question

I don't think I understand isomorphism well enough.

Consider the albelian groups $G = (G_1, e_1, o_1 )$ and $H = (H_1, e_1, o_1 )$. Consider a homomorphism $f:G\rightarrow H$ that exists in case of these two groups.

Then consider the homomorphism $g:H\rightarrow G$ that is the inverse of the function $f$, i.e: $g=f^{-1}$.

Then $f \circ g$ and $g \circ f$ are both identity mappings.

Does this not satisfy the conditions for isomorphism?

Answer 1

As a commentator already indicated, there does not always exist an inverse homomorphism.

For instance, consider the two groups $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z} $ under addition. Clearly the former group can be embedded (Via a homomorphism) inside the second group, but there is not an onto inverse of that embedding.

An isomorphism is, by definition, a homomorphism with an inverse.