Edit on March 2, 2018: I just noticed that this is almost identical to a question asked on MO by David Harden in 2011, and that I had even given an (incomplete) answer to that one. I would delete the question, but I think this can't be done when an answer has been accepted ( I did try).


The title says it all really. I know that in some "small" simple groups such as ${\rm PSL}(2,p)$ it is known that if the order is divisible by $60,$ then $A\_{5}$ does occur as a subgroup ( this is because sufficient control of the character table is available to exhibit an involution and an element of order $3$ whose product has order $5$). However, at the moment, I don't see a general strategy for attacking this question ( or similar ones), though of course it seems likely that some use of the classification of finite simple groups would be needed.


Later edit: It is a little late for motivation, and I did not realise it when I asked the question, but it seems to follow from the affirmative answer outlined by Derek Holt that a finite group $G$ is $5$-solvable (ie has all composition factors of order $5$ or of order prime to $5$) if and only if every subgroup of $G$ has a Hall $\{2,5\}$-subgroup.
Recall that a Hall subgroup of $G$ is one whose order and index are relatively prime.


Here is an outline of the difficult half of the proof: suppose then that every subgroup of $G$ has a Hall $\{2,5\}$-subgroup, but that $G$ is not $5$-solvable, and $G$ is chosen to be of minimal order subject to this. Note that every proper subgroup of $G$ is $5$-solvable. I claim that $G/F(G)$ is non-Abelian simple. For let $M$ be a maximal normal subgroup of $G$, and suppose that $M \neq 1$. Let $p$ be a prime divisor of $|M|$ and let $P$ be a Sylow $p$-subgroup of $M$. Then $G = MN\_{G}(P)$ by a Frattini argument. If $N\_{G}(P) < G$ then it easily follows that both $M$ and $G/M$ are $5$-solvable, and then $G$ is, contrary to assumption. Hence $P \lhd G$. Since $p$ as arbitrary, $M$ is nilpotent. Hence $M = F(G)$ is the unique maximal normal subgroup of $G$, and $G/M$ is non-Abelian simple. Now it easy to check that every subgroup of $G/M$ has a Hall $\{2,5\}$-subgroup, since $G$ has that property.


Now, however, $G/M$ can't have order divisible by $60$, using the answer to this question (otherwise, it has a subgroup isomorphic to $A\_{5}$, but $A\_{5}$ has no subgroup of order $20$). However, $G/M$ must have order divisible by $5$, otherwise $G$ would be $5$-solvable. Also, by Feit-Thompson, $G/M$ has order divisible by $4$. Hence $G/M$ has order prime to $3$, so that $G/M$ is a Suzuki group. However, no simple Suzuki group has a Hall $\{2,5\}$-subgroup, a contradiction.



Does a finite simple group of order divisible by $60$ have $A_{5}$ as a subgroup?
Let me make my comment into an answer just get things off the ground. I claim that ${\rm PSL}(n,q)$ contains $A\_5$ whenever its order is divisible by $60$.


Clearly ${\rm SL}(n,q)$ contains ${\rm SL}(m,q)$ for all $m \le n$, and hence ${\rm PSL}(n,q)$ contains some central quotient of ${\rm SL}(m,q)$ (which is something between ${\rm SL}(m,q)$ and ${\rm PSL}(m,q)$). In particular, ${\rm PSL}(n,q)$ contains a central quotient of ${\rm SL}(5,q)$ for all $n \ge 5$. Since the centre of ${\rm SL}(5,q)$ has order $1$ or $5$, and $A\_5$ has no perfect $5$-fold central extension, if ${\rm PSL}(5,q)$ contains $A\_5$, then so does ${\rm SL}(5,q)$. Also $|{\rm PSL}(5,q)|$ is divisible by $60$ for all $q$. So it suffices to prove the claim for $n \le 5$.


The result is well-known for $n=2$. 


For $n=3$, ${\rm PSL}(3,q)$ contains ${\rm PSL}(2,q)$ as the subgroup ${\rm P}\Omega(3,q)$, and $|{\rm PSL}(3,q)|$ is divisible by $60$ if and only if $|{\rm PSL}(2,q)|$ is, so the result holds.


For $n=4$, ${\rm PSL}(4,q)$ contains the subgroup ${\rm P}\Omega^-(4,q) \cong {\rm PSL}(2,q^2)$ which has order divisible by $60$ for all $q$ and hence contains $A\_5$.


For $n=5$, ${\rm PSL}(5,q)$ contains $A\_5$ as the image of the natural permutation representation.


Let me carry on with the classical groups. This is not too hard.


Virtually the identical argument works for the unitary groups. In dimensions $3$ and $4$ they contain the same orthogonal groups, and the permutation representation of $A\_5$ clearly preserves the identity matrix as unitary form.


For the symplectic case, the obvious containment ${\rm Sp}(m,q) \le {\rm Sp}(n,q)$ for $m \le n$, $m,n$ even projects onto ${\rm PSp}(m,q) \le {\rm PSp}(n,q)$, and ${\rm PSp}(4,q)|$ is divisible by $60$ for all $q$, so it is enough to do it for $n=4$. From the ATLAS, the $4$-dimensional irreducible representation of ${\rm SL}(2,5)$ has indicator $-$ and integer character values, so its image reduces to a subgroup ${\rm SL}(2,5) < {\rm Sp}(4,q)$ for all $q = p^e$ with $p>5$. In fact it works also for $p=5$, and the cases ${\rm PSp}(4,2) \cong S\_6$ and ${\rm PSp}(4,3)$ (which contains $2^4:A\_5$) can be done separately.


Finally, all of the simple orthogonal groups in dimension at least $5$ contain ${\rm P}\Omega(5,q) \cong {\rm PSp}(4,q)$, so there is nothing new to do there.


Now for some of the smaller rank exceptional groups. The Suzuki groups $^2B(q)$ have order not divisible by $3$. There are containments ${\rm PSL}(2,q) < {\rm SL}(3,q) < G\_2(q) < {^3D}\_4(q)$, and their orders are all either divisible by $60$ or not, so that covers them. Similarly ${\rm PSL}(2,q) < {^2G}\_2(q)$.


That leaves $F\_4$, $^2F\_4$, $E\_6$ $^2E\_6$, $E\_7$ and $E\_8$. For $^2F\_4(2^{2n+1})$, I suspect that they all contain the smallest one in the class, the Tits group $^2F\_4(2)'$, which (according to the ATLAS) contains $A\_5$.


This is probably not the best reference, but I think Table 2 in


<http://seis.bristol.ac.uk/~tb13602/docs/large_4.pdf>


is enough to deal with the remaining exceptional groups. For example $F\_4(q)$ contains $D\_4(q)$ = $\Omega^+(8,q)$, which contains ${\rm P}\Omega(5,q)$, which we considered above. There are lots of papers around on subgroups of the exceptional groups, but it's hard to extract exactly what we want.


