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StatusX said:I guess, but I just laid out the whole argument for you.
Ah, so you have! I replied to you before I read your edit.
Since I still have time I think I'm going to try to do it the way my professor suggested as well. One of his favorite sayings throughout the entire semester has been that it's good to be able to prove things more than one way.
OK, here we go. We know that [itex]G[/itex] has 1 element of order 1, 24 elements of order 5, and 48 elements of order 15. That accounts for 73 elements of [itex]G[/itex].
The number [itex]\nu_2[/itex] of 2-Sylow subgroups are:
[itex]\nu_2\in\{1,3,5,9,15,45\}[/itex]
Because the 2-Sylow subgroups intersect only in the identity, there is a 1-to-1 correspondence between 2-Sylow subgroups and elements of order 2. S we can rule out [itex]\nu_2=45[/itex] immediately.
Next up: [itex]\nu_2=15[/itex].