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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 15solns.pdf

(1.) How do you envisage and envision to prove by contradiction? Why not direct proof?

(2.) What's the intuition?

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  • You would think to do it like this precisely due to the previous exercise. – Tobias Kildetoft Feb 24 '14 at 08:06
  • There is an intuition here that a non-abelian group has to have a non-abelian core (the centre can't be too large), and that the centre can't sit too simply in the group (the quotient by the centre can't be cyclic) – Mark Bennet Feb 24 '14 at 08:37

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Lemma: if $G/Z(G)$ is cyclic then $G$ is abelian.

So, only possible order for $G/Z(G)$ is $pq\implies$ $Z(G)=1$

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