Question: Suppose that [itex]G[/itex] is a finite simple group of order greater than [itex]5[/itex] that contains no elements of order [itex]2[/itex]. Prove that [itex]G[/itex] contains no subgroup of index [itex]5[/itex]. Attempt: Suppose G has a subgroup H of index 5, i.e., [G]=5. Since 2 doesn't divide |G|, 2 doesn't divide |H|. So the order of H (and the order of G) is odd. If |H|=1, then |G|=5 x |H|= 5. This is a contradiction because |G|>5. So |H|> 2. Now if [itex]|G|=(5^a) m [/itex] where [itex]5 \not | m[/itex], then for [itex]n_5 = \#(Sylow\: 5\: subgroups)[/itex], [itex]n_5 | m [/itex] implies that [itex]n_5[/itex] is odd. If [itex]n_5 =1 [/itex], this contradicts that G is simple. So [itex]n_5 \geq 3[/itex]. From Sylow's theorem, we also know that [itex]n_5 \equiv 1 \mod 5[/itex]. So the last digit of [itex]n_5[/itex] must be 1. Okay, I think this is the correct approach but I don't see any contradiction. Please help... Thank you.