Implying that a group homomorphism is Surjective

Question

Consider two groups $(G,\star)$ and $(H,\odot)$. Suppose $f:G\to H$ is a group homomorphism in the sense that for all $g_{1}, g_{2} \in G$ we have: $$ f( g_{1} \star g_{2} ) = f(g_{1}) \odot f(g_{2}) $$

What other conditions on $f$ would I need in order to infer that $f$ is surjective onto $H$?

I'm thinking along the lines of a "converse" to the first isomorphism theorem by looking at $\mathrm{Ker}(f)$, but I can't seem to find any literature on the matter.

EDIT: I'm thinking that if I can show that $G / \mathrm{Ker}(f) \approx H$ then I can conclude that $f$ is surjective? If there are other ways of deducing that $f$ is surjective I'd be really interested in that too.

Answer 1

If $H$ is finite, then $G/\ker(f) \cong H$ implies $f$ is surjective, as the first isomorphism theorem implies $$G/\ker(f) \cong f(G),$$ thus $f(G) \cong H$. Since $H$ is finite, this implies $f(G) = H$.

It's not true in general if $H$ is infinite. Consider $$f: \mathbb{Z} \to \mathbb{Z}$$ defined by $f(n) = 2n$.

Answer 2

There are two common ways to show that $f$ is surjective.

Method I:
By using definition of surjectivity, show that for all $y\in H$, there exists $x\in G$ such that $f(x)=y$.

Method II:
This method can be used if $G,H$ are finite and $|G|=|H|$.
In this case, $f$ is injective iff $f$ is surjective.
Hence, in order to show that $f$ is surjective, you may also try to prove that $f$ is injective, while is equivalent to prove that $\ker f=1$.