Okay, this one is easy but I really like it: ( here is the Euler totient function)
Let be a group of order Prove that if then is abelian.
by Counting by order of elements theorem since is a divisor of number of elements in G of order n is a multiple of for some positive integer.
gcd( )=gcd( )=1 for some positive integer.
or
if ord(G)=1 then G={e} G is abelian
if ord(G)=2 then G is abelian since if a,b in G then a*b=b*a because (a=b) or (either a or b equal e).
is this correct?
here's a solution: excluding the trivial case we must have where are distinct primes, since in this case now
for any let be the number of Sylow subgroups of suppose then since for some we must have but we also have that which
is impossible becasue then and thus for all i.e. every Sylow subgroup of is normal. let be the Sylow subgroup of then:
so we actually proved that is more than abelian, it's cyclic!