Is there a reason the prime factors of $|M_{24}|$ are all one less than the factors of $24$? Wikipedia says of the Mathieu group $M_{24}$, a $5$-transitive permutation group acting on $24$ points,
$$ |M_{24}|= 2^{10}\cdot3^3\cdot5\cdot7\cdot11\cdot23. $$
The prime factors $2,3,5,7,11,23$ are all one less than one of the factors $3,4,6,8,12,24$ of $24$.
Am I crazy? Is this a coincidence, or does it admit an explanation?
(I suppose it could be a combination of both: maybe the factors $11$ and $23$ for some $24$-related reason and $2,3,5,7$ because of the law of small numbers, for example.)

The fact that $M_{24}$ is 5-transitive forces its order to be
$$
|M_{24}| = 24 \cdot (24-1) \cdot (24-2) \cdot (24-3) \cdot (24-4) \cdot |H|,
$$
where $H$ is the subgroup fixing any ordered list of five distinct points.  (In fact, we have $|H| = 48$.)  It follows that $|M_{24}|$ is divisible by
\begin{align*}
24 \cdot \frac{24-1}{1} \cdot \frac{24-2}{2} \cdot \frac{24-3}{3} \cdot \frac{24-4}{4} & = 24 \cdot \left(\frac{24}{1}-1\right) \cdot \left(\frac{24}{2}-1\right) \cdot \left(\frac{24}{3}-1\right) \cdot \left(\frac{24}{4}-1\right) \\
& = 24 \cdot 23 \cdot 11 \cdot 7 \cdot 5.
\end{align*}
So the primes $5, 7, 11$, and $23$ arise "because" they're one less than divisors of $24$.  The remaining primes, $2$ and $3$, appear in several places:  as divisors of $24$ itself, as divisors of $|H|$, and in the denominators we introduced.
