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I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.51.$^\dagger$
Suppose that $G$ is a finite group with the property that every nonidentity element has prime order (e.g., $D_3$). If $Z(G)$ is not trivial, prove that every nonidentity element of $G$ has the same order.
Thoughts:
Lemma: If $G$ is abelian with the property that every nonidentity element has prime order, then every nonidentity element has the same prime order.
Proof: If $G$ is abelian (i.e., $G=Z(G)$), then consider $g,h\in G$ such that $\lvert g\rvert=p$ and $\lvert h\rvert=q$ for distinct primes $p$ and $q$. We have
\begin{align}
(gh)^{pq}&=(g^p)^q(h^q)^p\\
&=e,
\end{align}
so that $\lvert gh\rvert$ divides $pq$.
If $\lvert gh\rvert=pq$, then it is composite, a contradiction; thus without loss of generality $\lvert gh\rvert=p$. Now we have
\begin{align}
e&=(gh)^p\\
&=g^ph^p\\
&=eh^p \\
&=h^p,
\end{align}
but now $q\mid p$, which is a contradiction since $p\neq q$ and $p$ is prime.
Thus all nonidentity elements of $G$ have the same prime order.$\square$
That's all I have so far.
I've considered proving some version of the contrapositive but nothing springs to mind other than, "yeah . . . contrapositive might work" followed by a shrug.
Edit:
This comment gives me some idea of how to finish; however, I'm not sure where the finiteness of $G$ comes into play.
Please help 🙂
$\dagger$ I've just noticed that this exercise has a solution in the book. It makes sense to me. If anyone would like to answer it here anyway, go ahead! I might post an answer later summarising the proof in the text.
Best Answer
The following paraphrases the proof in the solutions section of the book the exercise is from.
Let $z\in Z(G)$ such that $\lvert z\rvert=p$ for a prime $p$. Consider $g\in G$. We have $\lvert g\rvert=q$ is prime. Then
\begin{align} (zg)^{pq}&=(z^p)^q(g^q)^p \\ &=e^qe^p \\ &=e \end{align}
and thus $\lvert zg\rvert\in\{p, q\}$ (since it has to be prime; and if $\lvert zg\rvert=1$, then $z=g^{-1}$, so $p=q$). Assume without loss of generality that $\lvert zg\rvert=p$. Then
\begin{align} e&=(zg)^p \\ &=z^pg^p \\ &=eg^p \\ &=g^p, \end{align}
so $q\mid p$. Hence $p=q$.