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I want to show that

If $G$ non-abelian group of order $pq$ with $p,q$ primes then $Z(G)=\{e\}$.

It's immediate that if $Z(G) \neq \{e\}$ then $|Z(G)|=p$ or $|Z(G)|=q$ and using that any group of prime order is cyclic then $G / Z(G)$ is cyclic, which implies that $G$ is abelian and therefore we have a contradiction.

I wanted to know if it's possible to prove it without using the theorem that if $G/Z(G)$ is cyclic then $G$ is abelian because my professor hasn't mentioned it before. Any help would be greatly appreciated.

Shaun
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Juan Esteban Arevalo Gomez
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  • Yes, you should be able to prove the contrapositive—if $Z(G)\ne{e}$ then $G$ is abelian. Hint: if $x\in Z(g)\ne{e}$, what are the possible orders of $g$? Show that each possibility leads to an abelian group. – Greg Martin Apr 10 '21 at 17:59
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    The duplicate has an answer not using that if $G/Z(G)$ is cyclic then $G$ is abelian. – Dietrich Burde Apr 10 '21 at 18:12
  • If you know the Class Equation, I've posted an answer in the duplicate which doesn't use the quotient argument. –  Apr 10 '21 at 21:37

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