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Problem on sets (own)
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Re: Problem on sets (own)
I just figured that the existence of a subgroup of order $2^n$ can be proved...without refering to Sylow.
It can be shoowed that, following my earlier arguments, that if any subgroup of order $2^k$ exists and it generates more than $1$ cosets, then the union of that subgroup and any other of its cosets is a subgroup, of order $2^{k+1}$.
It can be shoowed that, following my earlier arguments, that if any subgroup of order $2^k$ exists and it generates more than $1$ cosets, then the union of that subgroup and any other of its cosets is a subgroup, of order $2^{k+1}$.
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor