# How to determine if a surjective homomorphism exists between two groups?

My question sheet asks about whether a surjective homomorphism exists between various symmetric groups and various $$Z_n$$ groups, for example between $$S_3$$ and $$Z_3$$, or $$A_4$$ and $$Z_3$$. To be honest, I don't really know where to start at all - I've been racking my brain for some property of homomorphisms to do with cardinality, or something like that, but I basically have no idea what I'm doing. Any pointers?

The biggest thing that's useful is knowing how many normal subgroups a group has. For example, $$S_n$$ has only itself, $$1$$, and $$A_n$$. By the first isomorphism theorem, the number of groups that can receive a surjective homomorphism from $$S_n$$ is severely limited.
A good place to start would be the first isomorphism theorem: if $$f: G\rightarrow H$$ is a homomorphism, then $$G/\ker f\simeq \operatorname{Im} f$$. So if $$f : G \rightarrow H$$ is a surjective homomorphism, then $$G / \ker f \simeq H$$ since surjectivity implies $$\operatorname{Im} f = H$$. So to determine whether such a homomorphism exists, you should determine whether or not there exists a normal subgroup $$G’ \leq G$$ such that $$G/ G’ \simeq H$$. If such a $$G’$$ exists, $$\varphi : G/G’ \rightarrow H$$ is an isomorphism, and $$\pi : G \rightarrow G/G’$$ is the quotient map, then $$\varphi \circ \pi$$ is a surjective homomorphism from $$G$$ to $$H$$.
As Randall pointed out in the comments below, we can use Lagrange's theorem to narrow our search even more. Since $$|H | = | G/G'|= |G| / |G'|$$, we see that $$|G| = |H| \cdot |G'|$$ (if we keep with the notation from above). Hence it is necessary (but not sufficient) for $$|H|$$ to divide $$|G|$$ and for $$G$$ to possess a subgroup of index $$|H|$$ in order for there to exist a surjective homomorphism from $$G$$ to $$H$$. If either of these conditions are not met, then such a homomorphism dots not exist.