I've been working on this same math problem, and think I have it... see if you can poke holes in my answer:
If |G| = pq, where p and q are primes that are not necessarily distinct, prove that |Z(G)| = 1 or pq.
- We know Z(G) is a subgroup of G, and G is a finite group (since its order is finite). By Lagrange's theorem, then, Z(G) must be 1, p, q, or pq.
- There are two cases for G: G is Abelian, or G is non-Abelian.
- If G is Abelian, Z(G) = G, so |Z(G)| = |G| = pq.
- If G is non-Abelian, |Z(G)| = 1, p, or q.
- Assume |Z(G)| = p. By Lagrange, |G/Z(G)| = |G| / |Z(G)| = q, a prime. Groups of prime order are cyclic, so G/Z(G) is cyclic. By G/Z theorem, then, G is Abelian, a contradiction. Therefore, |Z(G)| is not p. Similarly, there is a contradiction for |Z(G)| = q. Therefore, if G is non-Abelian, |Z(G)| = 1. (This is alluded to in the Gallian text in the paragraphs following the proof of the G/Z theorem-- "it follows immediately from this statement and Lagrange's Theorem that a non-Abelian group of order pq, where p and q are primes, must have a trivial cener.")
- Therefore, |Z(G)| = 1 or pq.
Edited by Albeda, 31 October 2007 - 07:05 PM.