Let be a Sylow p-subgroup of and assume that . Show that .
Could this be shown without giving out the cardinality of G first? I mean, for example without assuming first that G has a cardinality a power of a prime, or G has even cardinality, or anything like that. How do you start solving this?
Tell me if I got you right:
Well, is given to be a Sylow p-subgroup of . I also know that each conjugate of is also a Sylow p-subgroup. Now for all because . Since , then must be the only Sylow p-subgroup of . Since is a Sylow p-subgroup of (which is also the only Sylow p-subgroup of ), then must also be the only Sylow p-subgroup of . Therefore, .
With this proof, however, I used the fact that if then is unique. But the notes I have only guarantees the other way around: that if is the only Sylow p-subgroup of , then . Is this statement a really biconditional? Because if it is, I would have to show it first.