(A very limited instance of) Lagrange's Theorem's converse and A_5

Question

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?

Answer 1

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 well-known 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 $A_5$.

Answer 2

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 non-trivial 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.

Answer 3

See http://en.wikipedia.org/wiki/N-group_%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.