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Let $G = HN$ be a Frobenius group with complement $H$ and kernel $N$. The following two results are well known (for example, [Huppert, Endliche Gruppen V.8.18]):

  • If $|H|$ is even, then $H$ has a unique subgroup of order $2$;
  • If $|H|$ is odd, then $H$ has a unique subgroup of order $p$, for all primes $p$ dividing $|H|$.

My question is: what about when $|H|$ is even, but $p$ is odd? I assume it will not be true in general that $H$ will only have a single subgroup of order $p$, but is there a known count of subgroups of order $p$ in $H$ (or in $G$) in this case?

Thanks in advance!

Gauss
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  • Cauchy's Theorem seems relevant here: for prime $p\mid |G|$, there exists $g\in G$ with $|g|=p$. – Shaun Jul 23 '25 at 18:04
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    Sure, but I was hoping for something a bit more substantial than just saying this number had to be at least $1$ (per conjugate of $H$ in $G$) – Gauss Jul 23 '25 at 18:28
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    ${\rm SL}(2,3)$ and ${\rm SL}(2,5)$ are examples of Frobenius complements with more than one subgroup of order $3$, or respectively $3$ and $5$. Looking at the results on the structure of Frobenius complements here, I wonder whether they are essentially the only such examples. The result that all subgroups of order $pq$ for $p,q$ prime are cyclic implies that there is a unique subgroup of order $p$ in metacyclic Frobenius complements. – Derek Holt Jul 23 '25 at 18:37

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