# Find an element of order $p$ in a non-abelian group $G$ of order $p^3$ where $p$ is an odd prime

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?

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.)
• 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/…. Aug 9, 2021 at 20:51
• 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 $b$. Aug 10, 2021 at 18:21