# 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 $$4$$.

• More generally (try to prove this if you want), if $|G| = p^2q^2$, where $p<q$ are primes, then $G$ has a normal Sylow $q$-subgroup, except when $|G|=36$. – the_fox Jul 12 at 13:59
• It is not that trivial to show that. You can for example find that statement in kconrad.math.uconn.edu/blurbs/grouptheory/sylowapp.pdf Theorem 5.14. I think the proof by Keith Conrad should be written well enough for you to understand it. – ThorWittich Jul 12 at 15:48
• See also this, this and perhaps also this. – Jyrki Lahtonen Jul 12 at 19:58
• The question now is though: can you use this information to describe the list of groups of order $36$ exhaustively? – the_fox Jul 13 at 11:12

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 $$C_3\times C_3$$.

1. 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 $$P_1$$.
2. 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 $$Alt(X)\subseteq \mathrm{Im}(\phi)$$.

3. 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.

4. 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 $$12$$.

5. 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.

6. 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.

7. 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 $$w\notin K$$.
8. 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 $$P_1\cong C_3\times C_3$$.
9. 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.
10. 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 $$18$$.
11. 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.

• Posting this admittedly kludgy piece of reasoning. The starting idea was somewhat developed in the comment exchange under quasi's deleted answer. I think special facts about low order groups that made an appearance are needed. I did not check Keith Conrad's blurb. – Jyrki Lahtonen Jul 12 at 18:50
• Perhaps it's worth noting (I had missed that before) that if $G$ has $4$ Sylow $3$-subgroups and if $P \in \operatorname{Syl}_3(G)$ then $\mathbf{N}_G(P) = P$. Since $P$ is abelian, it follows that $P$ centralises its normaliser in $G$ and thus by Burnside's theorem $G$ has a normal $3$-complement which is just a Sylow $2$-subgroup of $G$. Of course, it is unreasonable to expect that someone taking a first course in group theory would have met Burnside's theorem. – the_fox Jul 12 at 20:08