I need to prove that a group of order 12 which doesn't have a normal subgroup of order 3 is isomorphic to A4. I know that the both groups has 4 3-sylow subgroups and 1 2-sylow subgroup. Does that means that they are isomorphic? if so, why?


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Let $G$ be a group of order $12$ with four $3$-Sylow subgroups (3SS).
$G$ acts on the set of 3SS (denoted $X$) by conjugation: $H^a=aHa^{-1} \: \:\forall H\in X$ .

This action induces a homomrphism $G\rightarrow S_4$. If $g$ is in the kernel then $H^g=H$ for all $H\in X$ and then $g\in Stab_G(H)$.
By Sylow theorems this action is transitive, and by the Orbit-Stabilizer theorem, $Stab_G(H)$ is of size $3$.
$H\subseteq Stab_G(H)$ because $H$ is abelian and therefore $Stab_G(H)=H$. We conclude that our $g$ is contained in all $4$ elements of $X$ and hence $g=e$, i.e the kernel is trivial.

Therefore out homomorphism is onto. So $G$ is isomorphic to a subgroup of $S_4$ of order 12, and the latter must be $A_4$.


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