Prove that $|Z(G)|=1$ or $pq$.

Question

If $|G|=pq$ where $p$ and $q$ are prime that are not necessarily distinct, prove that $|Z(G)|=1$ or $pq$.

I understand everything up to $G/Z(G)$ is cyclic of prime order....

is $G/Z(G)$ being cyclic a general concept? and why would this imply that $q=1$ and therefore form a contradiction?

The phrasing is certainly weird. It's obviously not true that $G/Z(G)$ is generally cyclic, just take any $G$ with trivial center. Furthermore, being cyclic and of prime order are very compatible, even though the phrasing suggests that they're contradictory. Of course, since $G/Z(G)$ is of prime order, it therefore follows that it is cyclic. Then you can derive a contradiction with the fact that $G \neq Z(G)$. — Mees de Vries, Dec 07 '17 at 15:28
@stressed-out In general it's perfectly possible for $G/Z(G)$ to be a nontrivial abelian group. — anon, Dec 07 '17 at 15:31

score 0 · Accepted Answer · answered Dec 07 '17 at 15:26

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A well-known theorem says:

If $G/Z(G)$ is cyclic then $G$ is abelian

in which case, $G=Z(G)$ and so the only possibility is really $G/Z(G)=1$.

The quoted proof uses this result twice to eliminate $|Z(G)|=p$ and $|Z(G)|=q$ because in both cases $G/Z(G)$ would have prime order and so be cyclic. However, the result above shows that this cannot happen.

answered Dec 07 '17 at 15:26

lhf

216,483

See also https://math.stackexchange.com/questions/999247/if-g-zg-is-cyclic-then-g-is-abelian-what-is-the-point. – lhf Dec 07 '17 at 15:28
this brings in my next question, since we derive using this fact that q=1...why does this form a contradiction by definition of q? – rover2 Dec 07 '17 at 15:44
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@rover2 $q$ is a prime. Can $q=1$ be a prime? – Stefan4024 Dec 07 '17 at 16:05
oh no! forgot the formal definition of a prime! got it! thank you! – rover2 Dec 11 '17 at 13:30

Prove that $|Z(G)|=1$ or $pq$.

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