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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.

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  • $\begingroup$ Could you say something about how this condition arises, and why you suspect it might have this consequence? \\ $x$ and $x'$ are unrelated, except that they happen to have the same order? If so, then you might want to change the title slightly, or at least change the notation from $x'$ to something else; "commuting prime order elements" makes it easy to misread the question, though it is precisely stated. $\endgroup$
    – LSpice
    Commented Jan 30, 2023 at 15:16
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    $\begingroup$ For the rephrased version (which dropped the condition that $p, q, r$ divide $|G|$ to the first power only), I think $G=\Sigma_8$ with $p, q, r$ given by $2, 3,5$ is a counterexample. $\endgroup$
    – Achim Krause
    Commented Jan 30, 2023 at 16:22
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    $\begingroup$ (in the first version you restrict the primes to be odd, but $\Sigma_{12}$ with $3,5,7$ works just as well.) $\endgroup$
    – Achim Krause
    Commented Jan 30, 2023 at 16:28
  • $\begingroup$ Thanks for pointing out. I added back the assumption that $|G|$ is divisible by $p,q,r$ to the first power only. $\endgroup$
    – Nick
    Commented Jan 30, 2023 at 16:32
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    $\begingroup$ My apologies. I meant $x'$ and $y$ do not necessarily commute. $x'$ is just some random element of order $p$, not every element of order $p$. $\endgroup$
    – Nick
    Commented Jan 31, 2023 at 11:45

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