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?
 A: Sounds like a fun problem. Basically, your strategy is going to be to write one down, or prove it doesn't exist.
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: 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. 
