For find and .
We directly computed the subgroup lattice of , so we know that and .
Note that . By Sylow’s theorem, we have the following.
We know that is simple, so . Now the Sylow 3-subgroups of have order 3; by this previous exercise, has 20 elements of order 3, all of which are in . Each of these is in a Sylow 3-subgroup of , and each Sylow 3-subgroup is generated by two elements. Thus . Note that the elements of order 4 in are precisely the 4-cycles, and none of these are contained in . Thus the nonidentity elements of any Sylow 2-subgroup of have order 2; these elements are precisely the products of two 2-cycles. By this previous exercise, contains 15 products of two disjoint 2-cycles, all of which are contained in and each of which is contained in some Sylow 2-subgroup. Note that the Sylow 2-subgroups of are abelian, and that every element of order 2 in fixes exactly one element in . Let such that fixes and does not fix . If , then ; becuase fixes only , we have , a contradiction. So . Thus, if are elements of order 2 that do not fix the same element, then no Sylow 2-subgroup contains both and . For each , there are precisely three products of two 2-cycles in which fix , corresponding to the image of (in ) under a renaming injection , and each such image is a Sylow 2-subgroup. Thus .
Now consider ; note that . By Sylow’s Theorem, we have the following.
Every Sylow 5-subgroup of is a Sylow 5-subgroup of , so . Similarly, every Sylow 3-subgroup of is a Sylow 3-subgroup of , so that . Now the nonidentity elements in Sylow 3-subgroups of are precisely the 3-cycles, and all of these are contained in . So has no additional Sylow 3-subgroups; hence . Note that , and that every Sylow 2-subgroup of is a Sylow 2-subgroup of . In this previous exercise, we saw that , , and are the Sylow 2-subgroups of . Then , , and are also distinct Sylow 2-subgroups of , and clearly , , and are pairwise distinct from , , and . (In particular, the former fix 5 and the latter do not.) So ; thus .