No simple group of order 756 : Burnside's proof

Question

I'm interested in a proof of the non-simplicity of groups of order 756. W.R. Scott, Group Theory, p. 392, exerc. 13.4.9, gives it as an easy exercise, but depending on rather advanced results.

I have a proof using classical methods (merging theorem, Sylow theorems, Burnside's transfer theorem) and also a theorem of Frobenius of which a particular case says that if $G$ is a finite group, if $d$ is a natural divisor of $|G|$, then the number of elements $x$ of $G$ such that $x^{d} = 1$ is divisible by $d$. (There is an English translation of Frobenius's paper by R. Andreev here : https://arxiv.org/abs/1608.08813 ) Unfortunately, my proof is very long.

Burnside gave a short proof in 1895 (Collected Papers, vol. I, p. 598-599 of the volume). He assumes that there exists a simple group $G$ of order 756 and here is the beginning of his proof (note that the subgroups of order 27 of $G$ are the Sylow $3$-subgroups of $G$) :

"If simple, the group must contain $3^{3}.7$ operations [= elements] of even order and $6.36$ operations of order $7$, leaving $350$ operations whose orders are powers of $3$. [Today, we would say : $351$ $3$-elements.] There are $28$ sub-groups of order $3^{3}$, and, if any of these have a common sub-group of order $3^{2}$, the group is certainly composite [= not simple]."

So far, no problem. (If anybody is interested, I can explain the details.) But then, Burnside says: "The sub-groups of order 3 which are common to two sub-groups of order $3^{3}$ form a single conjugate set." Since, as noted, the intersection of two different Sylow $3$-subgroups of $G$ always has order $1$ or $3$, this amounts to say that the subgroups of order $3$ of $G$ of the form $P \cap Q$, with $P$ and $Q$ two different Sylow $3$-subgroups of $G$, are all conjugate in $G$. (In any case, I understand it in this manner.) But how can we prove that ? Of course, it would be true if the conjugation action of $G$ on the set of its Sylow $3$-subgroups was 2-transitive. (And in this case, two different Sylow $3$-subgroups of $G$ should always have an intersection of order $3$, since it is easy to prove that if all Sylow 3-subgroups of $G$ intersected pairwise trivially, the group should have more than $756$ elements.) But is it the case that the conjugation action of $G$ on the set of its Sylow $3$-subgroups is 2-transitive?

As a sufficient condition for the conjugation action of a finite group $G$ on the set of its Sylow $p$-subgroups to be 2-transitive, I only find this (I can give a proof if anybody is interested :)

Theorem. Let $G$ be a finite group, let $p$ be a prime divisor of $\vert G \vert$, let $p^{m}$ denote the greatest power of $p$ dividing $\vert G \vert$. Assume that

1° $G$ has several Sylow $p$-subgroups;

2° $a$ is a natural number < $m-1$ such that the intersection of two different Sylow $p$-subgroups of $G$ is always of order $\leq p^{a}$; (we can take $a$ such that $p^{a}$ is the greatest possible order of $P \cap Q$, with $P$ and $Q$ two different Sylow $p$-subgroups of $G$)

3° for the same $a$, we have $\vert Syl_{p}(G)\vert < 2 p^{m-a}$.

Then the conjugation action of $G$ on the set of its Sylow $p$-subgroups is 2-transitive.

But, in Burnside's proof, we have $\vert Syl_{3} (G) \vert = 28$, $m = 3$ and the least possible value we can take for $a$ is $1$, not $0$ (as already said, it is easy to see that the $3$-Sylow subgroups of $G$ cannot intersect all pairwise trivially), so condition 3° is not satisfied.

Thus my question is : how can we prove Burnside's assertion : "The sub-groups of order 3 which are common to two sub-groups of order $3^{3}$ form a single conjugate set" ?

Thanks in advance.

$ \mathbf{Edit}$ (July 5, 2024)

Thanks to Derek Holt for his clear explanation. I think this explanation enables us to briefly finish the proof.

Let $G$ be, by contradiction, a simple group of order $756$.

Let us say that a subroup $R$ of order $3$ of $G$ is multiplist if it is contained in more than one Sylow $3$-subroup of $G$ and let us say that it is unicist in the other case.

Similarly, let us say that an element of order $3$ of $G$ is multiplist if it is contained in more than one Sylow $3$-subroup of $G$ and let us say that it is unicist in the other case.

From Derek Holt's explanation, it results that if $R$ is a multiplist subgroup of order $3$ of $G$, then $\vert C_{G}(R) \vert = 36$. Since every proper subgroup of $G$ has order $\leq 36$, it follows that $\vert N_{G}(R) \vert = 36$. As stated by Burnside and explained by Derek Holt, all multiplist subgroups of order 3 of $G$ form a conjugacy class, so they are in number $756/36 = 21$.

Thus the multiplist elements of order $3$ of $G$ are in number $42$.

By definition, two distinct Sylow $3$-subroups of $G$ never have a common unicist element of order $3$. Moreover, since all Sylow $3$-subroups of $G$ are conjugate, they all have the same number of unicist elements of order $3$. Thus (since, as noted by Burnside, $G$ has exactly $28$ Sylow $3$-subroups) the number of unicist elements of order $3$ in $G$ is divisible by $28 \times \varphi (3)$ and thus by $8$.

As noted by Burnside, two distinct Sylow $3$-subroups of $G$ never have a common subgroup of order $9$ and thus never have a common element of order $9$. In the same manner as precedingly, we conclude that the number of elements of order $9$ in $G$ is divisible by $28 \times \varphi (9)$ and thus by $8$.

From the fact that every Sylow $3$-subroup of $G$ is its own normalizer in $G$ and from Burnside's transfer theorem, we could conclude that the Sylow $3$-subroups of $G$ are not abelian and thus not cyclic, so $G$ has no element of order $27$, but without using Burnside's transfer theorem, we can say that clearly no element of order $27$ can be in two distinct Sylow $3$-subroups of $G$, so, as above, the number of elements of order $27$ in $G$ is divisible by $28 \times \varphi (27)$ and thus by $8$.

From what precedes, the $3$-elements of $G$ different from $1$ are in number congruent to $2$ modulo $8$. But Burnside has shown that the $3$-elements of $G$ different from $1$ are in number $350$, not congruent to $2$ modulo $8$, so we have done.

Here is Burnside's complete proof :

"If simple, the group must contain $3^{3}.7$ operations [= elements] of even order and $6.36$ operations of order $7$, leaving $350$ operations whose orders are powers of $3$. [Today, we would say : $351$ $3$-elements.] There are $28$ sub-groups of order $3^{3}$, and, if any of these have a common sub-group of order $3^{2}$, the group is certainly composite [= not simple]. The sub-groups of order 3 which are common to two sub-groups of order $3^{3}$ form a single conjugate set; and when the group is expressed in $28$ symbols a simple calculation will show that each sub-group of order $3^{3}$ must contain $8$ operations keeping $1$ symbol unchanged and $18$ keeping $4$ symbols unchanged. There are therefore $63$ sub-groups of order $3$ which occur in more than one sub-group of order $3^{3}$. On the other hand, such a sub-group of order $3$ must (Note VI) be permutable in a sub-group of order $2^{2} /dot 3^{3}$ at least; and must therefore be one of a set of $21$ conjugate sub-groups at most. The supposition that the group is simple thus leads to a contradiction."

I don't understand the part of the proof beginning with "and when the group is expressed in $28$ symbols...". Isn't my reasoning sufficient and simpler?

Answer 1

Let $R = P \cap Q$ with $P,Q \in {\rm Syl}_3(G)$ and $|R|=3$. Then $C_G(R)$ contains subgroups of order at least $9$ from both $P$ and $Q$, so it has at least $4$ Sylow $3$-subgroups, and hence has order at least $36$. It is easy to see that it cannot have order greater than $36$: it contains no element of order $7$ and if it had order $108$ then it would have index $7$ in $G$, which is impossible because $3^3$ does not divide $|A_7|$.

So $|C_G(R)|=36$ and has $4$ Sylow $3$-subgroups, so $C_G(R)/R \cong A_4$ and hence $C_G(R)$ has a normal Sylow $2$-subgroup $T$, which is also a Sylow $2$-subgroup of $G$. Again $N_G(T)$ cannot have order greater than $36$, so $C_G(R) = N_G(T)$, and $R$ is a Sylow $3$-subgroup of $C_G(T)$, and hence all such subgroups are conjugate in $G$, because the Sylow $2$-subgroups are all conjugate.

Your approach will not work, because the action by conjugation on the Sylow $3$-subgroups cannot be $2$-transitive. If it was, then the action would be as a Frobenius group, and the Sylow $3$-subgroups would have trivial intersection.