Using Exercise 37, show:
A nonabelian group G of order pq where p and q are primes has a trivial center.
Reference: http://users.humboldt.edu/pgoetz/Homework%20Solutions/Math%20343/hwi think 15
solns.pdf
(1.) How do you envisage and envision to prove by contradiction? Why not direct proof?
(2.) What's the intuition?
Best Answer
Lemma: if $G/Z(G)$ is cyclic then $G$ is abelian.
So, only possible order for $G/Z(G)$ is $pq\implies$ $Z(G)=1$