Let $G$ be a group of order $12.$ Prove that either $G$ has a normal $ 3$-Sylow subgroup or $ G$ is isomorphic to $A_4$.
I know that $|G|=12=2^23$ and that either $n_3=1$ and there is a $3$-sylow subgroup, or either $n_3=4$. If $n_3=4$ then there are $4$ $3$-sylow subgroups whose intersection is $\{e\}.$ That means there are $8$ elements of order $3,$ leaving $4$ element of order that is not $3$. There is a $2$-sylow subgroup of order $4.$ Its elements are of order dividing $4.$ This subgroup is normal. There is the identity, $8$ elements of order $3$ and $3$ of order $2$ or $4.$ So there is one element of order $6$, I guess. But I don't know how to continue and show isomorphism. Any help?
Best Answer
$G$ acts by conjugation on the 3-Sylow subgroups. So we have a homomorphism $\phi: G \rightarrow S_4.$ Let $K$ = ker$\phi.$ Then $K \leq N_G(P),$ where $P$ is a 3-Sylow subgroup of $G$. From this conclude that $K = \{e\}.$ So the map $\phi$ is injective. Now $G$ contains eight element of order 3. The number of elements of order 3 in $S_4$ is also eight and all are contained in $A_4.$ This shows that $G \cong A_4.$