The goal is to prove that any group of order $252 = 36 \cdot 7$ is solvable, and because I managed to confuse myself, I'm asking here.
Let $G$ be a group of order $252$. By Sylow's Theorems, the number of $7$-Sylow subgroups of $G$ is either $1$ or $36$. If it is $1$, we are done, because the quotient then has order $36$, and groups of order $7$ and $36$ are solvable.
Hence we are left with the much more interesting case in which the number of $7$-Sylow subgroups is $36$. One proof to show solvability is the following:
By the orbit-stabilizer theorem (since $G$ acts transitively on the set of its $7$-Sylow subgroups), the normalizer $N_G(P)$ of a $7$-Sylow $P$ of $G$ has order $7$, hence
$$N_G(P) = Z_G(P) = P,$$
where $Z_G(P)$ is the centralizer of $P$. By Burnside's Transfer Theorem, we obtain that $G$ contains a normal subgroup $N$ of order $36$. Since $|G/N| = 7$, we are done.
Questions to the second case (number of $7$-Sylows is $36$):
- I checked with GAP and saw that there is no group of order $252$, whose $7$-Sylow is not normal. Is there an easy way to see this without invoking a computer algebra system?
- Can one prove in a more elementary way that there is a normal subgroup of order $36$? Indeed, there are exactly $36 \cdot 6$ elements of order $7$, thus there are $36$ elements, whose order is coprime to $7$. How does one see that these $36$ elements form a subgroup? If we could see that in an elementary way, there is of course a unique subgroup of order $36$, hence a normal one, and there is no need to invoke Burnside's Transfer Theorem.
Best Answer
I suppose the more elementary way is to look at the Sylow $3$s next.
We know $n_3=1,4,7,28$. If $n_3=1$ we are done, and $n_3=4$ we have a homomorphism $G\to S_4$ with nontrivial kernel. So $n_3=7$ or $n_3=28$. But since we only have $36$ elements left, there must be two Sylow 3s, say $H_1,H_2$ that intersect nontrivially.
So $P=H_1\cap H_2$ has order $3$, whose centralizer (since $H_i$ are abelian) $C_GP$ contains at least the set $H_1H_2$ of $27$ elements. Therefore $\lvert C_GP\rvert$ has to be a factor of $252$ that is at least $27$ and divisible by $9,$ so must be $36$ (the other choices, $63$, would be an index-$4$ subgroup so again we have a nontrivial homomorphism to $S_4$, or $126$ which is index 2 hence normal). So $C_GP$ is every element with order prime to $7$.
But that is enough for contradiction. $C_GP$ contains all Sylow $3$s since we basically used up those elements, but the group generated by all Sylow $3$s is normal in $G$.