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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 $A_5$.

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I think the answer to the q