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I am not sure what the answer to this question is offhand, but one 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. $A_5$. For G has such a subgroup if and only if G contains elements x,y,z $x,y,z$ of respective orders 2, 3 and 5, with $xyz = 11$. 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, $x,y,$ and z $z$ so that lots of irreducible characters vanish one at least one of x,y,z. $x,y,z.$ For example, all non-trivial irreducible characters of A_5 $A_5$ vanish on one of x,y,z, $x,y,z,$ when x,y,z $x,y,z$ 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 $A_5$ 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.