Exercise 3.2.4 (If $|G| = pq$ for some primes $p$ and $q$ then either $G$ is abelian or $Z(G) = 1$)

Question

Show that if $|G| = pq$ for some primes $p$ and $q$ (not necessarily distinct) then either $G$ is abelian or $Z(G) = 1$. [See Exercise 36 in Section 1.]

Richard Ganaye · Accepted Answer

\begin{proof} 
Suppose that $Z = Z(G) 
e \{1\}$. By Lagrange's Theorem $|Z|$ divides $|G| = pq$, where $p$ and $q$ are prime, therefore
$$|Z| \in \{p, q, pq\}.$$

\begin{itemize}
\item If $|Z| = pq$, then $G = Z$, so $G$ is abelian.

\item If $|Z| = p$, then (using $Z \unlhd G$), $|G/Z| = |G|/|Z| = q$. since $q$ is a prime number, $G/Z$ is cyclic. 

By Exercise 3.1.36, $G$ is abelian.
\item If $|Z| = q$, then similarly $|G/Z| = p$, so $G/Z$ is cyclic and $G$ is abelian
\end{itemize}
In every case, $G$ is abelian.

\bigskip
In conclusion, if $|G| = pq$ for some primes $p$ and $q$, then either $G$ is abelian or $Z(G) = 1$
\end{proof}

Exercise 3.2.4 (If $|G| = pq$ for some primes $p$ and $q$ then either $G$ is abelian or $Z(G) = 1$)

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Exercise 3.2.4 (If $|G| = pq$ for some primes $p$ and $q$ then either $G$ is abelian or $Z(G) = 1$)

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