I need to prove that group G of order 36 is solvable. This is what I came with so far:
There is either one 3-sylow subgroup or 4 3-sylow subgroups.
-first case: if there is only one 3-sylow subgroup than P- a 3-sylow subgroup of G is normal. it is also abelian and therefor solvable. [G:P] = 4 and therefor G/P is abelian so it is solvable and we are done.
-second case: if there are four 3-sylow subgroups it means by sylow theory that there is a subgroup N of index 4 so G/N is abelian and therefor solvable and N is of order 9 so it's also abelian and solvable.
the only thing I'm missing is that in the second case I don't know how to prove that N is also normal in G.