Find all the homomorphisms from $C_3$ to $A_4$.
I know that $C_3=\{1,c,c^2\}$ and $A_4$ is the group of even permutations on four elements.
So let $\lambda:C_3\rightarrow A_4$ be a homomorphism. Then $\lambda$ has to have this property $\lambda(c_1c_2)=\lambda(c_1)\lambda(c_2)$ for $c_1,c_2\in C_3$.
I'm guessing that I have to first find how many there are, then trial and error until I find them all?
How do I find how many there are and is there a method to finding them?
Every homomorphisms $C_3 \to A_4$ is either trivial or injective (because $C_3$ is simple), and the non-trivial homomorphisms correspond bijectively to the elements of order $3$, which we can easily find by using the cycle decomposition; the elements of order $3$ are $$ a_2 = (1,2,3), \; a_3 = (1,3,2), \; a_4 = (1,2,4), \; a_5 = (1,4,2), \\ a_6 = (1,3,4), \; a_7 = (1,4,3), \; a_8 = (2,3,4), \; a_9 = (2,4,3). $$ Together with $a_1 = 1$ we get nine homomorphisms $C_3 \to A_4$, each uniquely described by $1 \mapsto a_i$.
(Instead of looking at the cycle decomposition of each element we could also use the Sylow theorems to find that $A_4$ has four subgroups of order $3$, each of which gives us two elements of order $3$, resulting in the above eight elements.)