If $|G|=36$ then $G$ has either a normal $2$-Sylow or a normal $3$-Sylow

Question

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 $4$.

Answer 1

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.

Answer 2

$|G|=36=2^2·3^2$. For the number of Sylow 3-subgroups of $G$ we have: $n_3≡1\ (mod \ 3)$ and $n_3 \ | \ 2^2=4$. Suppose $G$ has no normal Sylow 3-subgroup. Then $n_3=4$. Our goal is to show that $G$ has a normal Sylow 2-subgroup.

Let $P∈Syl_3 (G)$. We have $n_3=|G∶N_G (P)|=4$. Let $G$ act on the set of distinct left cosets of $N_G (P)$ by left multiplication. This action affords a homomorphism $φ:G→S_4$ and, denoting $ker⁡ \ φ=K$, by the 1st Isomorphism Theorem we have: $G/K≅L≤S_4$. Then by the Lagrange’s Theorem, $|G/K| \ | \ |S_4 |=24$. Since $K≤N_G (P)$ and $|N_G (P)|=(2^2·3^2)⁄4=9$, the Lagrange’s Theorem enforces the following possible orders of $K$: $1,3,9$. There cannot be $|K|=1$ since then $|36/1|=36∤24=|S_4 |$. There cannot be $|K|=9$, since then there would be $K∈Syl_3 (G)$ and because $K⊲G$, $G$ would have a normal (hence unique) Sylow 3-subgroup, what contradicts our assumption. So it can only be that $|K|=3$. Then $|G/K|=36/3=12$. Since $G/K≅L≤S_4$ and $S_4$ has only one subgroup $L$ of order $12$, which is $A_4$, it follows that $G/K≅A_4$. In its turn, $A_4$ has one unique subgroup of order $4$, which is isomorphic to $V_4$ (Klein Viergruppe). Hence $G/K$ has one unique subgroup of order $4$, isomorphic to $V_4$, that because of order considerations (i.e., $|G/K|=12=2^2·3$ and $|V_4 |=4=2^2$) is a Sylow 2-subgroup of $G/K$.

Let $Q∈Syl_2 (G)$. Because $|K|=3$, $|Q|=2^2=4$, by the Lagrange’s Theorem $K ⋂ Q=1$, hence the image $Q/K=QK/K$ is of order $|QK/K|=(|Q||K|/|K ⋂ Q|)⁄|K| =|Q||K|⁄|K| =|Q|$ and so there is a bijection $Q→QK/K$ given as $q→qK$. Because of this we have $Q≅QK/K≅V_4$ (recall that $V_4$ is a unique subgroup of $G/K$ of order 4). Since $V_4⊲A_4$, it follows that $QK/K⊲G/K$ and so by the Correspondence Theorem $QK⊲G$.

Note that since $K⊲G$ and $K ⋂ Q=1$, the subgroup $QK=KQ$ can be written as a semidirect product $KQ≅K⋊Q$. And we already know the isomorphism types of $K≅Z_3$ and $Q≅V_4≅Z_2×Z_2$, therefore, $KQ≅Z_3⋊(Z_2×Z_2 )$.

Since $G/K≅A_4$, the part of the subgroup lattice of $G$ which contains $K$ reflects the subgroup lattice of $A_4$ by the Lattice Isomorphism Theorem. $A_4$ and hence $G/K$ contains four cyclic subgroups of order $3$ isomorphic to $Z_3$ (in $A_4$ these are the subgroups generated by $3$-cycles). Let's denote them as $P_1/K,P_2/K,P_3/K,P_4/K$. Their preimages in $G$ (each of which obviously contains $K$) are of order $9=|K|·|Z_3 |=3^2$ each, hence these preimages must be exactly the four Sylow 3-subgroups of $G$, that is, $P_1,P_2,P_3,P_4∈Syl_3 (G)$. So we have that $K$ lies in the intersection of all four Sylow 3-subgroups of $G$. Since $K$ is abelian, $K≤C_G (K)$. Furthermore, since $|P_i |=3^2$, the Sylow 3-subgroups of $G$ are abelian, hence each of them centralizes its subgroup $K$, that is, $P_1,P_2,P_3,P_4≤C_G (K)$. But only $G$ itself contains $K$ and $P_1,P_2,P_3,P_4$ together (recall that the part of the subgroup lattice of $G$ containing $K$ reflects the subgroup lattice of $A_4$). Thus, $C_G (K)=G$ and hence $K≤Z(G)$.

But this means that all elements of $K$ commute with all elements of $Q$ and hence the semidirect product $QK=KQ≅K⋊Q$ is just a direct product, i.e. $K⋊Q=K×Q$. Thus $QK≅K×Q$, therefore $Q⊲QK$. Since $Q∈Syl_2 (QK)$, this means that $Q \ char \ QK$. Because $QK⊲G$ as was shown before, $Q \ char \ QK⊲G$, therefore $Q⊲G$. Since $Q∈Syl_2 (G)$, this means that $G$ has a normal Sylow 2-subgroup $Q≅V_4≅Z_2×Z_2$.

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So, a group $G$ of order $36$ with no normal Sylow 3-subgroup always has a normal Sylow 2-subgroup. There doesn’t exist a group of order $36$ with no normal Sylow 3-subgroup and no normal Sylow 2-subgroup at the same time. And there exists a group $G$ of order 36 with a normal Sylow 3-subgroup, for example, the cyclic group $Z_{36}$ (this group also has a normal Sylow 2-subgroup at the same time).

⦁ Thus, a group $G$ of order $36$ has either a normal Sylow 2-subgroup or a normal Sylow 3-subgroup.