Mathematical Reflections 2013, Issue 1 - Problem U257

Problem:
a) Let $p$ and $q$ be distinct primes and let $G$ be a non-commutative group with $pq$ elements. Prove that the center of $G$ is trivial.

b) Let $p, q, r$ be pairwise distinct primes and let $G$ be a non-commutative group with
$pqr$ elements. Prove that the number of elements of the center of $G$ is either $1$ or a prime number.

Proposed by Mihai Piticari.

Solution:
We use the following

Lemma
If $G$ is a non-abelian group, then $G/Z(G)$ is not a cyclic group.

Proof
Homework!

a) Since $Z(G)$ is a subgroup of $G$, $|Z(G)|$ divides $|G|$, which means  $|Z(G)| \in \{1,p,q,pq\}$. Therefore, $|G/Z(G)| \in \{pq,q,p,1\}$ and since $G/Z(G)$ cannot be cyclic, it must be $|G/Z(G)|=pq$, i.e. $Z(G)$ is trivial.

b) As above, $|G/Z(G)|$ is a divisor of $G$, so, $|G/Z(G)| \in \{1,p,q,r,pq,qr,rp,pqr\}$. Since $G/Z(G)$ cannot be cyclic, the only possibilities are $|G/Z(G)| \in \{pq,qr,rp,pqr\}$, which means that $|Z(G)|$ is either $1$ or a prime number.