This is a follow up to my question last night Groups of order 36 where I was confused about the first step of Lemma 5.4 of http://matwbn.icm.edu.pl/ksiazki/fm/fm92/fm9211.pdf. I am now confused about one of the last steps - that R cannot be isomorphic to $A_4$.
Definitions and relevant parts of the lemma: The goal is to prove that any group $G$ of order 36 has a normal Sylow subgroup. After supposing that $n_3,n_2 \neq 1$, he deduces the existence of a normal subgroup of order 3, by noting that $N_G(P)=P$ for any $P \in Syl_3{G}$. Thus if $\pi$ is the action by conjugation on $Syl_3{G}$, then $Ker \pi \subset P$. Also it must be nontrivial since $Ran \pi ||S_4|$, so $H:=Ker \pi$ is of order 3.
$im \pi$ must be $A_4$ so that $G/H$ has a normal subgroup of order 4,$S$ so that $G$ has a normal subgroup of order 12, $M$ that contains $H$. Using the fact that the group of order 4 is not normal in G, he shows that $R:=N_G(S)$ is a subgroup of order 12 that is not normal satisfying $H \rtimes_s R \cong G$ for some nontrivial action $s$.
My question: He then claims this implies R cannot be isomorphic to $A_4$. Why is this true?
I understand the steps after this on why this implies a contradiction. Maybe its true that $A_4 \triangleleft G$. This would imply his claim because the homomorphism into $Aut H$ would then need to be trivial. Maybe since there would be at least 5 groups of order 3 there would be too many for some reason. I have not been coming to a contradiction so far. Can I have hint?
