Let $p$, $q$, $r$ be three distinct odd primes, and $G$ a finite group with $|G|$ divisible by $p$, $q$, $r$ to the first power only. Let $x,y,z \in G$ be of order $p,q,r$ respectively. Assume (a.) $[x,y] = [y,z] = [z, x'] = 1$ where $x' \in G$ is some element of order $p$; (b.) $xy$, $yz$, $zx'$ are all all real elements in $G$; that is there exist $u,v, w \in G$ with $(xy)^u = (xy)^{-1}$, $(yz)^v = (yz)^{-1}$, $(zx')^{w} = (zx')^{-1}$.

Question: Do these assumptions guarantee the existence of an element of order $pqr$ in $G$, or if not is there a counter-example?

I should rephrase the question: if a finite group $G$ whose order is divisible by $p,q,r$ to the first power only has real elements of order $pq$, $qr$ and $rp$ where $p$, $q$, $r$ are distinct primes, then must $G$ have an element of order $pqr$?

Let $p$, $q$, $r$ be three distinct odd primes, and $G$ a finite group with $|G|$ divisible by $p$, $q$, $r$ to the first power only. Let $x,y,z \in G$ be of order $p,q,r$ respectively. Assume (a.) $[x,y] = [y,z] = [z, x'] = 1$ where $x' \in G$ is some element of order $p$; (b.) $xy$, $yz$, $zx'$ are all all real elements in $G$; that is there exist $u,v, w \in G$ with $(xy)^u = (xy)^{-1}$, $(yz)^v = (yz)^{-1}$, $(zx')^{w} = (zx')^{-1}$.

Question: Do these assumptions guarantee the existence of an element of order $pqr$ in $G$, or if not is there a counter-example?

I should rephrase the question: if a finite group $G$ whose order is divisible by $p,q,r$ to the first power only has real elements of order $pq$, $qr$ and $rp$ where $p$, $q$, $r$ are distinct odd primes, then must $G$ have an element of order $pqr$?

This question arises when I was looking into real conjugate classes and real representations of finite groups. I was hoping to establish that properties such as this (if true) will lead to dual statements on the real representations, or vice versa.