I am working on the following problem:
Prove that if $\lvert G \rvert = 12$ and $G$ has $4$ Sylow 3-subgroups, then $G \equiv A_4$.
If you let $G$ acts by conjugation on the set containing the $4$ Sylow 3-subgroups, then this action induces a homomorphism between $G$ and $S_4$. The issue is that I am having a hard time trying to prove that this homomorphism is injective. Note that this is important as the order of $G$ is $12$. Any thoughts?
The action is transitive, so the image has order divisible by $4$. If your homomorphism is not injective, then its kernel has order $3$. But that would mean the kernel was a normal Sylow $3$-subgroup, so unique.