Prove there are no simple groups of even order $<500$ except orders $2$, $60$, $168$, and $360$.

Question

In Dummit & Foote, Abstract Algebra, $\S6.2$, Exercise 17(b) is:

Prove there are no simple groups of even order $<500$ except orders $2$, $60$, $168$, and $360$.

The fact that the we have to check all groups of less $<500$ makes me think there is a faster way of solving this rather than brute force. Even using various formulas to wipe out entire families of orders still seems like it would take an unreasonable amount of effort for an exercise.

Is there something I'm missing with this problem? Is there a faster way to reduce the work that I am not seeing?

Answer 1

Hint

There are $246$ orders to eliminate.

1. Recall that Burnside's Theorem (Theorem 11(1) in $\S$6 of Dummit & Foote) implies that the order of any non-Abelian, finite, simple group has at least three distinct prime factors. Eliminating the other groups leaves $98$ orders.

2. Recall that if $2$ divides the order $n$ of a group $G$ exactly once, then $G$ has a subgroup of index $2$ ($\S$4.2, Exercise 12), but any such group is normal ($\S$3.2, Example (2)), so unless $n = 2$, we have $2^2 \mid n$, leaving just $35$ orders.

3. Recall that Sylow's Third Theorem implies that the number $n_p$ of Sylow $p$-subgroups of a finite group $G$ satisfies $n_p \mid |G|$ and $n_p \equiv 1 \pmod p$, and recall that a Sylow $p$-subgroup is normal in $G$ (and hence $G$ is not simple) iff $n_p = 1$. Checking the $35$ orders shows that $n_p = 1$ for at least one prime factor $p$ of $G$ for all but $13$ orders:

$$120, \quad 132, \quad 180, \quad 240, \quad 252, \quad 264, \quad 280, \quad 300, \quad 336, \quad 380, \quad 396, \quad 420, \quad 480 .$$

A problem in a previous section ($\S$4.5, Exercise 22) resolves order $132$. Orders $264$ and $396$ are given as examples in the problem's section ($\S$6.2), and orders $336$, $380$, and $420$ are resolved in earlier (sub)problems in that section (Exercises 9, 3, and 17(a), respectively).

Now only $7$ orders remain, all of which have been addressed elsewhere on this site:

$\qquad\qquad\qquad\qquad\quad$$120$, $\quad$$180$, $\quad$$240$, $\quad$$252$, $\quad$$280$, $\quad$$300$, $\quad$$480$.

Remark It's only a little more work to show that there also no simple group of odd order $< 500$ other than the $94$ odd prime cyclic groups. Burnside's Theorem reduces the list from $155$ to just $18$. Sylow's Third Theorem eliminates all of these orders but $105 = 3 \cdot 5 \cdot 7$, $315 = 3^2 \cdot 5 \cdot 7$, and $495 = 3^2 \cdot 5 \cdot 11$. Counting elements of Sylow $p$-subgroups under the assumption that $n_p \neq 1$ for all $p$ quickly dispenses of orders $105$ and $495$. The remaining case, $315$, is slightly trickier.