If $|G|=36$ then $G$ has either a normal $2$-Sylow or a normal $3$-Sylow
As many, I'm trying to classify all groups of order 36. I've seen many posts and in them, they claim this is true, but I can't find why. I know because of this, that $G$ is not simple. But I can't understand why the normal subgroup has to be a Sylow.
I know that the number of $2$-Sylows is either $1, 3$ or $9$, and the number of $3$-Sylows is either $1$ or $4$.
Assume contrariwise that the group $G$ has four Sylow $3$-subgroups. Let $X=\{P_1,P_2,P_3,P_4\}$ be the set of those. Assume further that $G$ also has more than one Sylow $2$-subgroups.
Conjugation action of $G$ on $X$ gives us a homomorphism $\phi:G\to Sym(X)\cong S_4$. Observe that as $[G:P_i]=4=|X|$, the groups $P_i$ are all equal to their own normalizers, $N_G(P_i)=P_i, i=1,2,3,4$. As groups of order $p^2$, $p$ a prime, they are abelian. All isomorphic to either $C_9$ or $C_3\times C_3$.
- If $z\in P_1\setminus P_2$ then $z$ normalizes $P_1$, but does not normalize any other group in $X$. This means that $\phi(z)$ is a 3-cycle with a unique fixed point $P_1$.
Similarly there are other $3$-cycles in $\mathrm{Im}(\phi)$ fixing other elements of $X$. It follows that all the $3$-cycles of $Sym(X)$ are in $\mathrm{Im}(\phi)$. The $3$-cycles of $S_4$ generate the subgroup $A_4$, so we can conclude that $Alt(X)\subseteq \mathrm{Im}(\phi)$.
But, the order of the image must be a factor of $36$. Therefore we can conclude that $|\mathrm{Im}(\phi)|=12$ and $\mathrm{Im}(\phi)\cong A_4$. Hence $|\mathrm{Ker}(\phi)|=3$. That kernel is the intersection $$N=\bigcap_{i=1}^4N_G(P_i)=\bigcap_{i=1}^4P_i,$$ a cyclic group of order three.
From elementary courses we know that $A_4$ has a unique Sylow $2$-subgroup isomorphic to the Klein Viergruppe. Hence $\mathrm{Im}(\phi)$ also has a unique Sylow $2$-subgroup $Q$. An immediate consequence of this is that all the Sylow $2$-subgroups of $G$ are also isomorphic to the Viergruppe, and they must also all be contained in $K:=\phi^{-1}(Q)\unlhd G$, a subgroup of order $12$.
Let $R\le K$ be a Sylow $2$-subgroup of $K$ (and hence also of $G$). As $N\unlhd K$, and $N$ intersects trivially with $R$, we have $K=N\rtimes R$. If $R\unlhd K$, then $K$ (and hence also $G$) has only a single Sylow $2$. So we are left with the possibility that $K$ has three Sylow $2$-subgroups, and that the semidirect product $N\rtimes R$ is not direct.
The automorphism group of $N\cong C_3$ is cyclic of order two, so up to isomorphism there is a single non-abelian semi-direct product $N\rtimes R=C_3\rtimes (C_2\times C_2)$, with exactly one factor $C_2$ commuting with $N$. As $C_3\times C_2\cong C_6$, it follows that $K\cong C_6\rtimes C_2\cong D_6$, the dihedral group of order twelve.
- Let $H=N_G(R)$ be the normalizer. As we work under the assumption that there are three Sylow $2$-subgroups, we have $|H|=36/3=12$. Therefore there exists an element $w\in H$ of order three. As $K$ also has three Sylow $2$-subgroups, $|N_K(R)|=4$. Therefore $w\notin K$.
- As an element of order three $w$ is contained in some Sylow $3$-subgroup, say (w.l.o.g.) $w\in P_1$. We saw that also $N\subset P_1$, so it follows that $P_1=\langle N\cup\{w\}\rangle$. Furthermore, there are more than two elements of order three in $P_1$, so $P_1\cong C_3\times C_3$.
- The dihedral group $K$ has exactly two elements of order six. Conjugation by $w$ thus must permute those. But $w$ has order three, so it must centralize both of them.
- Let us fix an element $r\in K$ of order six, so $N=\langle r^2\rangle$. The previous bullet implies that $r$ is centralized by the Sylow $3$-subgroup $P_1$. Therefore the subgroup $S=\langle r, w\rangle$ is abelian of order $18$.
- But, the previous result says that $S$ is contained in the normalizer of $P_1$. This is, at long last, a contradiction as we knew that $P_1$ equals its own normalizer.
So either the assumption that there are four Sylow $3$-subgroups or the assumption that there are more than one Sylow $2$-subgroup must be abandoned. This suffices to prove the claim.