“…A 5 has five copies of A 4 and any two A 4 in A 5 have non-trivial intersection, for otherwise |A 4 A 4 | = 144, which is not possible. Also H 1 := (12, 34) , H 2 := (12, 34), (13,24) , H 3 := (12, 34), (12354) , H 4 := (12, 34), (12453) , H 5 := (12, 34), (345) are proper subgroups of A 5 . Here H 1 is a subgroup of H i , for every i = 1, 2, 3, 4.…”