6

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?

Travis Willse
  • 108,056
TAPLON
  • 1,956
  • 1
  • 19
  • 40
  • by the way, there are no non-abelian groups of order $2$ – J. W. Tanner Nov 13 '19 at 02:33
  • 1
    Ah yes thank-you I realize now that I merged the problem from before it. We do not assume the group is non-abelian – TAPLON Nov 13 '19 at 02:34
  • Perhaps this https://en.wikipedia.org/wiki/List_of_finite_simple_groups Otherwise I don't see any fast way of solving this. – WhatsUp Nov 13 '19 at 02:36
  • 2
    This is more like an advanced undergraduate project than a single exercise. Standard techniques can be used to rule out most orders, particularly if you are allowed to use Burnside's $p^aq^b$ theorem, but there still remain a handful of more difficult orders, such as 336 and 432. Also it is unclear whether you are expected to prove the uniqueness of the simple groups of orders 60, 168, 360, which is a difficult exercise in itself. – Derek Holt Nov 13 '19 at 08:33
  • @DerekHolt I basically agree that the question is something of a project, but NB some of the difficult cases are handled in earlier examples/exercises in the cited text (including $336$), so in OP's particular context the problem simplifies some. – Travis Willse Nov 13 '19 at 18:00
  • (And $432 = 2^4 3^3$, so Burnside's Theorem handles that case.) – Travis Willse Nov 13 '19 at 18:52
  • 1
    @TravisWillse Good point! I have some extensive notes on this problem, which for some reason avoid using Burnside's Theorem. The most difficult orders are 264, 288, 336, 420, 432, and 480. – Derek Holt Nov 13 '19 at 22:29
  • Perhaps because the standard proof of Burnside's Theorem is representation-theoretic? (There are more natively group-theoretic proofs that came much later, the 60s or 70s I think; I'm not sure whether they are harder.) If we invoke Burnside's Theorem, that leaves from your list just $264, 336, 420, 480$, none of which appear to have been asked about specifically on this site. – Travis Willse Nov 13 '19 at 22:42
  • (...and all of those but $480$ are resolved earlier in the text, thought $420$ is part (a) of the exercise of which OP's question is part (b).) – Travis Willse Nov 14 '19 at 17:59
  • @Aimingfor50points It looks to me as though you should be asking a new question rather than addressing questions specifically to me in comments to an old question. – Derek Holt Nov 30 '20 at 07:49
  • But you have already asked three previous questions. – Derek Holt Nov 30 '20 at 13:21
  • 1
    $420$ had in fact been previously resolve on this site: https://math.stackexchange.com/questions/773872/sylow-7-subgroup-of-a-group-of-order-4-cdot3-cdot5-cdot7-is-normal

    I recently written and self-answered questions addressing the $3$ erstwhile outstanding orders:

    $264$: https://math.stackexchange.com/questions/4723218/there-are-no-simple-groups-of-order-264/4723219

    $336$: https://math.stackexchange.com/questions/4723272/there-are-no-simple-groups-of-order-336/4723273

    $480$: https://math.stackexchange.com/questions/4723791/there-are-no-simple-groups-of-order-480/4723792

     – Travis Willse Jun 23 '23 at 00:07

1 Answers1

5

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.

Travis Willse
  • 108,056