Problem. Let $p$ be an odd prime and $G$ be a non-abelian group of order $p^3$. Suppose there exists an element $x\in G$ of order $p^2$. Find an element $y\in G\setminus\langle x\rangle$ of order $p$.

It is true, by Cauchy's theorem or some other theorems, that $G$ has an element of order $p$. However, we are not sure whether it lies in $\langle x\rangle$ because $x^p,x^{2p},\ldots,x^{(p-1)p}$ all have order $p$.

I tried another approach: The number of element in $G$ of order $p^2$ must be a power of $$\varphi(p^2)=p(p-1),$$ where $\varphi$ is the Euler's totient function. There are $p^3-1$ non-identity elements in $G$. We have $$p^3-1=p(p-1)(p+1)+p-1=(p+1)\varphi(p^2)+p-1.$$ This also implies that $G$ has an element of order $p$, but the remainder $p-1$ is problematic as we already have $p-1$ elements of order $p$ above.

Does anyone have good ideas on this question?


1 Answer 1


In a nonabelian group of order $p^3$, we must have $[G,G] = Z(G)$ and $|Z(G)|=p$.

So, for any $a,b \in G$, using $ba= ab[b,a]$, and $[b,a] \in Z(G)$, we get $(ab)^p = a^pb^p[b,a]^{p(p-1)/2} = a^pb^p$, since $p$ is odd.

Let $y \in G \setminus \langle x \rangle$. Then, since $G$ is nonabelian, $y$ cannot have order $p^3$, so $y^p =x^{ap}$ for some $a$. Then $(x^{-a}y)^p = x^{-ap}x^{ap}=1$, so $x^{-a}y$ has order $p$.

(Note that this result is false for $p=2$ and the quaternion group $Q_8$ is a counterexample.)

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    $\begingroup$ Thanks. I also found this solution from K. Conrad's paper: kconrad.math.uconn.edu/blurbs/grouptheory/groupsp3.pdf just before. I was trying to understand the proof and planned to write it if nobody answered my question then. Anyway, thank you for your help :) $\endgroup$ Aug 9, 2021 at 20:42
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    $\begingroup$ Just one more comment, people can find the proof of $(ab)^n=a^nb^n[b,a]^{n(n-1)/2}$ from this post: math.stackexchange.com/questions/77149/…. $\endgroup$ Aug 9, 2021 at 20:51
  • $\begingroup$ Bernhard, you need not write it again here if it is already in one of Keith Conrad's notes. Of course, you can ask about details of the proof, which remain unclear. $\endgroup$ Aug 10, 2021 at 17:56
  • $\begingroup$ Do we really need lots of links for this? Isn't it preferable understand why this is true? To transform $(ab)^n$ to $a^nb^n[b,a]^{n(n-1)/2}$, you apply the rule $ba \to ab[b,a]$ $n(n-1)/2$ times and use the fact that $[b,a]$ commutes with $a$ and $b$. $\endgroup$
    – Derek Holt
    Aug 10, 2021 at 18:21

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