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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.
Marktmeister
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    I’m rusty enough where I’m not able to help, but this is a very well-asked question. – Randall Aug 05 '20 at 13:14
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    user10354138 has answered the question, but I don't really why understand why you (or anyone) would want to avoid using Burnside's Transfer Theorem. It is really not that difficult a result for someone who understands Sylow theory. – Derek Holt Aug 05 '20 at 15:55
  • @Randall Thanks! – Marktmeister Aug 05 '20 at 17:45
  • @DerekHolt I agree and the "Burnside way" was the one I came up with. A colleague then claimed that a simple counting argument does the job, i.e., that the remaining $36$ elements form a unique subgroup of order $36$. However, they were not able to convince me. – Marktmeister Aug 05 '20 at 17:46
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    The remaining 36 elements form the Frobenius kernel don't they? The subgroup $H$ satisfies $H\cap H^g=1$ for all $g\in G\setminus H$ (as it's self-normalizing), so $G$ is a Frobenius group. The Frobenius kernel is all elements outside the conjugates of $H$, which is the $36$ ${2,3}$-elements left over. – David A. Craven Aug 05 '20 at 20:49
  • Good observation @DavidA.Craven. Thank you! – Marktmeister Aug 06 '20 at 10:05

2 Answers2

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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 $G$.

user10354138
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Another approach. Consider the following facts:

  1. $n_3\in\{7,2^2\cdot7\}$ ($2^2$ would violate the $n!$-theorem.)
  2. $n_7=2^2\cdot3^2$.
  3. Let $K=Q_1\cap Q_2$ be the largest possible intersection of two different Sylow $3$-subgroups.
  4. If $|K|=1$, then there would be $2^2\cdot3^2(7-1)$ elements of order $7$ plus no less than $7(9-1)$ of order $3$ or $9$, which gives a total of $272$ in a group of order $252$. Impossible. Then $|K|=3$.
  5. Define $J=N_G(K)$. Then $K\triangleleft Q_1,Q_2$ and we get $Q_1\cup Q_2\subseteq J$. Hence, $|J|\ge 15$. Since $9\mid|J|$, we get $|J|\ge18$. Then $|J|\ge2\cdot3^2$.
  6. If $|J|=2^2\cdot3^2\cdot7$, then $J=G$ and we are done.
  7. If $|J|=2\cdot3^2\cdot7$, then $|G:J|=2$ and we are also done.
  8. If $|J|=2\cdot3^2$, then $|J:Q_1|=2$ which is impossible because that would mean $Q_1\triangleleft J$.
  9. Therefore, we are left with the case $|J|=2^2\cdot3^2$.
  10. Let $S=\bigcup\text{Syl}_7(G)\setminus\{1\}$. Then $|S|=2^2\cdot3^2\cdot(7-1)$. Moreover, $S\cap J=\emptyset$, because elements in $S$ have order $7$ and elements in $J$ orders divisible by $2$ and/or $3$. Since $|S\cup J|=|S|+|J|=|G|$, we get a partition of $G=S\cup J$.
  11. Now take $Q\in\text{Syl}_3(G)$, since $Q\cap S=\emptyset$, it follows that $Q\subseteq J$. In other words, $J$ includes all Sylow $3$-subgroups of $G$, i.e., $n_3(J)=n_3$. But this is impossible because $7\mid n_3$ (fact 1) and $n_3(J)\mid2^2$ (fact 9).

Note. This is Problem 1E.8 from Isaacs' Finite Group Theory.

Leandro Caniglia
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