[Hungerford, p99 Q8]

Let p be an odd prime. Prove that there are, at most, two nonabelian groups of order $\displaystyle p^{3}$.

One has generators a,b satisfying $\displaystyle |a|=p^{2}; |b|=p; b^{-1}ab=a^{1+p}$.

The other has generators a,b,c satisfying $\displaystyle |a|=|b|=|c|=p; c=a^{-1}b^{-1}ab; ca=ac; cb=bc$.