Let $G$ be finite and every nonidentity element have prime order. If $Z(G)\neq\{e\}$, prove that every nonidentity element of $G$ has the same order.

Question

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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.$\square$

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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.

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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 :)

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$\dagger$ 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.

Answer 1

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 $p=q$.

Answer 2

I am doing selfstudy and studying Group Theory, and using Gallian's book. Today I encounter this problem. I spend almost 2 hours on the problems and didn't find a clear path. So I thought to take some help (hint) from MSE, and so I reached at your question. After reading your question I found where my knowledge lack. Actually the part where you showed p and q are distinct by contradiction, here I was not getting contradiction.
Now I have wrote my proof without looking at the back of the book (solution). Please have a look.

Proof: Suppose $Z(G)$ is not trivial. Suppose $x \in Z(G)$ and $y \in G$ are both non-identity elements. Let $|x|=p$ and $|y|=q$. Assume for the sake of contradiction that $p$ and $q$ are distinct. Since $(xy)^{pq} =e$, it follows $|xy|$ divides $pq$. If $|xy|=pq$ then a contradiction because $pq$ is composite. WLOG assume $|xy|=p$. Then $e=(xy)^p=y^p$, so $q|p$, a contradiction. Thus $p$ and $q$ are equal. Since $p$ and $q$ were arbitrary, therefore every non-identity element of $G$ has same order. ∎

Edit: I just have looked at your accepted answer and solution in the book. I found I have missed a case in which order of $pq$ is equal to one.