bham10246
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Question: Suppose that G is a finite simple group of order greater than 5 that contains no elements of order 2. Prove that G contains no subgroup of index 5.
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 |G|=(5^a) m where 5 \not | m, then for n_5 = \#(Sylow\: 5\: subgroups), n_5 | m implies that n_5 is odd. If n_5 =1, this contradicts that G is simple. So n_5 \geq 3.
From Sylow's theorem, we also know that n_5 \equiv 1 \mod 5. So the last digit of n_5 must be 1.
Okay, I think this is the correct approach but I don't see any contradiction. Please help...
Thank you.
Attempt: Suppose G has a subgroup H of index 5, i.e., [G

If |H|=1, then |G|=5 x |H|= 5. This is a contradiction because |G|>5. So |H|> 2.
Now if |G|=(5^a) m where 5 \not | m, then for n_5 = \#(Sylow\: 5\: subgroups), n_5 | m implies that n_5 is odd. If n_5 =1, this contradicts that G is simple. So n_5 \geq 3.
From Sylow's theorem, we also know that n_5 \equiv 1 \mod 5. So the last digit of n_5 must be 1.
Okay, I think this is the correct approach but I don't see any contradiction. Please help...
Thank you.