1. Let G be a simple group with PI > 60. Show that G has no subgroups of index less or equal to 5. 2. Let G be a subgroup of Sq that acts transitively on {1,2,..., a}. (The action is the restriction...

Solution 1: Suppose that G has a subgroup H of index 5.

The group G acts by left multiplication on the set of left cosets of H in G. Since G is a simple group,
the kernel of this action is trivial.

Therefore G embeds in S5 and we may assume that |S5 : G| = 2. Then G S5 and G  A5 .

Now suppose that G has no subgroup of index 5.

Let np denote the number of Sylow p-subgroups of G.

If np ≤ 4, then by considering the action of G, by conjugation, on the set of its Sylow p-subgroups, we
see that G embeds in S4, which is impossible.

Therefore, np > 5 i.e. G has no subgroups of index less than or equal to 5.

Solution 2a: If G is transitive, then there exists element

g12, g13, . . . ., g1n ϵ G

that map 1 to 2, 3, . . . ., n respectively. Consider the sets

G1g12,G1g13, . . . ., G1g1n....