On groups of order $p^aq^br^c$ containing elements of order $pq$, $qr$, and $pr$, but not $pqr$

Question

Let $G$ be a solvable group of order $p^aq^br^c$ (for distinct prime $p,q,r$) containing elements of orders $pq$, $qr$, and $pr$, but no element of order $pqr$. Furthermore, assume that $G$ is minimal with respect to this property, i.e. any proper subgroup of $G$ lacks a element of order $pq$, $qr$, or $pr$.

1. What are some examples of groups like this? $^\star$

2. Can they be characterized? Are there additional properties that come from these conditions?

$^\star$ Note: trivially, it suffices to search for groups containing elements of orders $pq$, $qr$, and $pr$, but no element of order $pqr$, as then we can just take subgroups until we find one satisfying the minimality property. So, I would also enjoy to hear about any non-minimal examples of this type, as they contain minimal examples.

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Things I can tell:

If any proper subgroups of $G$ have order $p^xq^yr^z$ ($x,y,z>0$), they need to come in one of three type:

1. Frobenius group whose Frobenius kernel is a $p$-group and Frobenius complement is the product of a $q$-group and a $r$-group. This means the $q$ and $r$ groups are either cyclic or generalized quaternion, and that $q^yr^z\mid p^x-1$.

2. Frobenius group whose Frobenius kernel is the direct product of a $q$-group and a $r$-group and Frobenius complement a $p$-group. So, $p^x\mid q^yr^z-1$, and the $p$-group is either cyclic or generalized quaternion.

3. If there are no elements of order $pq$ or $qr$, there is still a element of order $pr$, and the group has the form $G=PQR$ (where $P$, $Q$, and $R$ are subgroups of corresponding prime power order) with $Q$ a cyclic $q$-group of odd order, $R$ a cyclic $r$-group, where $PQ$ is a Frobenius group whose kernel is $P$ and $QR$ is also a Frobenius group whose kernel is $Q$. (The corresponding prime congruences hold similar to #1 and #2.)

I think if we make group $P$ and group $Q$ elementary abelian of dimension $r$ and then let $R$ be a cyclic group of prime order with faithful action, $$(a_1,a_2,\ldots, a_{r-1},a_r)\mapsto (a_2,a_3,\ldots, a_r,a_1)$$ (same on both $P$ and $Q$,) then it works, is it true?

Answer 1

Since all $\{p,q \}$-subgroups are contained in a Hall $\{p,q\}$-subgroup up to conjugacy, we may suppose (by symmetry in p,q,r) that a minimal such $G$ (of order divisible by $pqr$) has a Hall$\{p,q \}$-subgroup $H$ which contains no element of order $pq$ and whose Fitting subgroup is a $p$-group. Beyond that, no more can really be said (apart from describing the possible structure of $H$ itself, as you have now done yourself). For a solvable group $G$ with such a Hall $\{p,q\}$-subgroup $H$ certainly contains no element of order $pqr,$ as it already contains no element of order $pq.$ It can be deduced that $H$ must be a maximal subgroup of $G$ by the minimality of $G.$ Also, no proper subgroup of $G$ can have order divisible by $pqr,$ as every proper subgroup of $G$ of order divisible by $pqr$ would have a Hall $\{p,q\}$-subgroup of the same kind. This implies that $F(G)$ is Abelian of squarefree exponent. If $|F(G)|$ is divisible by two primes, then $F(G)$ has prime index. In any case, $F(G)$ is Abelian of squarefree exponent.