There are no simple groups of order $264$

Question

Before this writing the smallest order for which the nonexistence of simple groups of that order is not explicitly demonstrated on this site is $264$; this self-answered question aims to fill that gap. (See Prove there are no simple groups of even order $<500$ except orders $2$, $60$, $168$, and $360$., the sole answer of which describes an outline for proving the titular statement but does not describe how to establish the claim for order $264$.) Other answers are, of course, welcome.

How does one show that there are no simple groups of order $264$?

Answer 1

This approach is essentially the one given in $\S$6.2 of Dummit & Foote's Abstract Algebra. A closely related approach is given in $\S$3 the cited article (as far as I know, the latter is the earliest proof of the claim for this order).

Suppose $G$ were simple of order $264 = 2^3 \cdot 3 \cdot 11$. Sylow's Theorems imply that $n_{11} = 1 \pmod {11}$ and $n_{11} \mid 264$; simplicity imposes $n_{11} > 1$ leaving $n_{11} = 12$ as the only possibility. So, we can identify $G$ with a subgroup of $S_{12}$, and since $G$ is simple, it contains no subgroup of index $2$, hence $G \leq A_{12}$. Now, let $P$ be a Sylow-$11$ subgroup; Frattini's Argument gives that $$|N_{A_{12}}(P)| = \frac{1}{2} |N_{S_{12}}(P)| = \frac{1}{2} (11)(11 - 1) = 55 .$$ But $|N_G(P)| = \frac{|G|}{n_{11}} = \frac{264}{12} = 22$, and $22 \not\mid 55$, which contradicts the fact that $N_G(P) \leq N_{A_{12}}(P)$.

Remark Cf. this answer, which uses a similar technique to resolve the case of order $336$.

F.N. Cole, "Simple Groups from Order $201$ to Order $500$," Amer. J. Math. 14(4) (October 1892), pp. 378$-$388.