Suppose $G$ is a finite simple group and $G$ is a multiple of $60$. Does it follow that $G$ has a subgroup isomorphic to $A_{5}$? If so, can this be proven without using the Classification?

1$\begingroup$ The "converse" in the header is overstated, since the question concerns only a very special type of divisibility. Aside from that minor point, it's hard to avoid quoting a big chunk of the classification work such as what's involved in determining all possible minimal nonabelian simple groups and their orders. $\endgroup$– Jim HumphreysApr 16, 2011 at 17:07
3 Answers
I think the answer to the question is yes, but it is very unlikely that it can be proved without using the classification of finite simple groups.
Note that $A_5$ of order 60 is the only simple group order for which this statement is true, because for all higher order simple groups $G$, there will be groups $L_2(p)$ with order divisible by $G$ that do not contain $G$ as subgroup.
Let's quickly look at all families if finite simple groups. I hope someone will correct any mistakes I make!
The Suzuki groups (Lie type $^2B_2$) are not divisible by 3, so we can forget them. All other finite simple groups have order divisible by 12, so their order is divisible by 60 if and only if it's divisible by 5.
The claim is clearly true for $A_n$, $n \ge 5$.
It is wellknown that $L_2(q)$ contains $A_5$ if and only if $q \equiv \pm{1} \bmod 5$, which is equivalent to $G$ divisible by 5.
$U_3(q)$, $L_3(q)$, $G_2(q)$, $^3D_4(q)$ all contain $L_2(q)$ and also have order divisible by 5 if and only $q \equiv \pm{1} \bmod 5$.
$^2F_4(2^{2e+1})$ contains $^2F_4(2)$, which contains $A_5$.
$^2G_2(3^{2e+1})$ never has order divisible by 5.
$S_4(q)$ contains $L_2(q^2)$, which always contains $A_5$ for all $q$.
All remaining groups of Lie type contain $S_4(q)$ and hence contain $A_5$.
It is easily checked, for example by looking at their lists of maximal subgroups in the ATLAS or on
http://brauer.maths.qmul.ac.uk/Atlas/v3/
that the sporadic groups contain $A_5$.

1$\begingroup$ Hey! Don't leave us hanging like that! Gerhard "Ask Me About System Design" Paseman, 2011.04.17 $\endgroup$ Apr 17, 2011 at 20:28

$\begingroup$ Good. I can breathe easier now. Gerhard "Really Likes To Breate Easy" Paseman, 2011.04.17 $\endgroup$ Apr 17, 2011 at 22:16

$\begingroup$ For a layman (nay, this is already too generous, an ignoramus) as myself, this is a quite impressive answer. $\endgroup$– OlivierApr 18, 2011 at 8:48

$\begingroup$ Thanks, but I am hoping somebody will read it critically, and highlight any dubious claims or possible inaccuracies! $\endgroup$ Apr 18, 2011 at 14:01
I am not sure what the answer to this question is offhand, but once you know the character table of a finite group G, it is, in principle, straightforward to determine whether G has a subgroup isomorphic to $A_5$. For G has such a subgroup if and only if G contains elements $x,y,z$ of respective orders 2, 3 and 5, with $xyz = 1$. This can be done from the character table, using "class algebra constant" calculations, using a formula which dates back at least as far as W.Burnside, and can be found in most texts on character theory of finite groups. The trick to checking this efficiently for a group whose character table you know is to try to choose the elements $x,y,$ and $z$ so that lots of irreducible characters vanish one at least one of $x,y,z.$ For example, all nontrivial irreducible characters of $A_5$ vanish on one of $x,y,z,$ when $x,y,z$ have those orders. It is hard to believe that this problem could be resolved without the classification of finite simple groups. A related question is a theorem of Graham Higman, who proved that $A_5$ is the only finite simple group which has a maximal subgroup which is dihedral of order 10. This did not use the classification of all finite simple groups, but did use the fact that Suzuki groups were the only simple groups of order prime to 3.
See http://en.wikipedia.org/wiki/Ngroup_%28finite_group_theory%29 where it talks about minimal simple groups. You'd need to check in particular that the example of 2x2 projective special linear groups can't have order divisible by 60. But I don't believe that.
Edit: I was however using the wrong formula for the order, which can indeed avoid divisibility by 5.

1$\begingroup$ $PSL_2(2^p)$, $PSL_2(3^p)$, $PSL_2(p)$ with $p=2,3$ mod 5, $Sz(2^p)$ all have order not divisible by 60. $\endgroup$– JunkieApr 16, 2011 at 7:50

1$\begingroup$ It's not that easy. If the minimal simple groups are checked, and that says that the only one that has order equal to a multiple of 60 is A_5 itself, then it is still possible that some group larger than A_5 is not minimal among simple groups, but minimal among those whose order is a multiple of 60 (and hence a counterexample). Also, the list of minimal simple groups is incomplete as given here, since it doesn't include $PSL_{3}(3)$. $\endgroup$ Apr 16, 2011 at 16:56

$\begingroup$ If my arithmetic is correct (not guaranteed), 5 doesn't divide the order of
$PSL_3(3)$
; the comment by Junkie should dispose of the other minimal simple groups. Still, as David points out there are further possibilities. It seems that 60 does divide the order of all but one of the 26 sporadic simple groups, so they need more scrutiny. Even assuming CFSG is in hand, what is the strategy for deciding whether or not$A_5$
occurs as a subgroup when 60 divides? $\endgroup$ Apr 16, 2011 at 19:48 
$\begingroup$ My strategy for disposing of the claim having failed, it seems I'm out of my depth. $\endgroup$ Apr 16, 2011 at 20:48

$\begingroup$ My strategy was essentially to look at (known) maximal subgroups, as Derek Holt did. I am not as facile with the subject as he. $\endgroup$– JunkieApr 18, 2011 at 4:57