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Let G be a finite solvable group. If p,q,r are distinct primes dividing |G|, then G contains an element of order the product of two of these three primes. This is lucido's three prime lemma. I encountered it in a paper. Could not solve it yet, but also had a thought that will there be a bigger class of finite groups for which it holds, a class which contains all solvable groups.

Can somebody give a proof of original, and tell me has there been some research in the direction of classes for which it is true?

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  • $\begingroup$ Maybe also this characterization of the prime graphs of solvable groups by Gruber, Keller, Lewis, Naughton and Strasser is of interest to you: arxiv.org/pdf/1305.2368.pdf $\endgroup$ – Stefan Kohl Nov 4 '14 at 19:44
  • $\begingroup$ @StefanKohl this is exactly tha paper I want to do, but before that I want to do Lucido's three lemma as it is the beginning step in grubers paper. Gruber himself suggested this paper to me on mathstack in concern of a question. I have emailed Gruber, lets see may be he will send me a copy. Thanks anyways. $\endgroup$ – Bhaskar Vashishth Nov 4 '14 at 19:59
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    $\begingroup$ Also on M.SE: math.stackexchange.com/questions/1005985/… $\endgroup$ – user26857 Nov 4 '14 at 19:59
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    $\begingroup$ Usually, cross-posting a question simultaneously on M.SE and MO is frowned upon. $\endgroup$ – Stefan Kohl Nov 4 '14 at 20:01
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    $\begingroup$ The point, Bhaskar, is that when you post to two sites, the least you can do is leave a link at each site to the question at the other one. $\endgroup$ – Gerry Myerson Nov 4 '14 at 22:01
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I do not know the published proof, but here is an outline proof of the solvable case (which may be the same as the published one): we may take a finite solvable group $G$ of order divisible by $pqr$, and assume by induction that no proper subgroup of $G,$ and no proper homomorphic image of $G,$ has order divisible by $pqr.$ Hence we may suppose that $G$ has a minimal normal subgroup $M$ which is a Sylow $p$-subgroup. Arguing similarly, we may suppose that $G/M$ has a minimal normal Sylow $q$-subgroup $N/M,$ and that $G/N$ is cyclic of order $r.$ Let $Q$ be a Sylow $q$-subgroup of $G.$ Now $C_{G}(m) = M$ for each non-identity element of $M$ (or we are done). Hence $G$ is a Frobenius group with kernel $M$ (and complement a Hall $\{q,r\}$-subgroup of $G$ containing $Q$). Now it is well-known that (as $Q$ is elementary Abelian), we even have $Q$ of order $q$. Hence $G$ has a subgroup of order $qr$, which must be cyclic (as it is contained in a Frobenius complement), as was known to Burnside.

I am not sure what sort of generalization you had in mind to non-solvable groups. For example ${\rm PSL}(2,5)$ and ${\rm PSL}(2,7)$ are groups whose order has three prime divisors in which every element has prime power order.

Further to Frieder Ladisch's answer, I believe that it is possible to describe the structure of all finite groups in which every element has prime power order. This is a question of collecting together several characterization theorems- I may well be duplicating known arguments, though I am not sure whether these results have been collected together in one place in the literature, or in some group theory text. In particular, results of M. Suzuki are the main tools.

Let $G$ be a finite group which is not a $p$-group for any prime $p,$ but in which all elements have prime power order. Suppose first that $F(G) \neq 1.$ Then $G$ has an elementary abelian minimal normal $q$-subgroup $Q$ for some prime $q.$ Also, $C_{G}(Q) = F(G) = O_{q}(G).$ Now for any other prime divisor $p$ of $|G|$, the Sylow $p$-subgroups of $G$ are cyclic or generalized quaternion. Suppose first that $q$ is odd. If $[G:O_{q}(G)]$ is even, then by the Brauer-Suzuki theorem (if the Sylow $2$-subgroup of $G$ is generalized quaternion) we see that $G = O_{2^{\prime}}(G)C_{G}(t)$ for some involution $t.$ Since $C_{G}(t)$ is a $2$-group, we see that $O_{2^{\prime}}(G)$ is an Abelian $q$-group for the given prime $q$. On the other hand, if $q$ and $[G:O_{q}(G)]$ are both odd, then $G/O_{q}(G)$ is metacyclic, so has a normal Sylow $p$-subgroup its largest prime divisor $p.$ Then $G/O_{q}(G)$ must be a cyclic $p$-group by the earlier argument.

If we have $F(G) = 1,$ then $G$ has a unique component $L,$ which is simple. Furthermore, $L \leq G \leq {\rm Aut}(L),$ and $L$ is a simple CIT-group (the centralizer of every involution is a $2$-group). These were classified by M. Suzuki. Not all these have the property that every element has prime power order, but this reduces the problem to checking which possibilities for $L$ have this additional properties, and which subgroups of ${\rm Aut}(L)$ do.

Consider the case that $q =2$ and $F(G)$ a $2$-group. If there is a solvable normal subgroup $N$ of $G$ with $N > F(G)$, then $G/O_{2}(G)$ is easily checked to be metacyclic. If there is no such normal subgroup $N,$ then $G/O_{2}(G)$ has the same structure as described in the single component case, though it may be that not all possibilities for the component occur, since we also require that non-identity elements of odd order act fixed point freely on $O_{2}(G).$ Note that at least some cases occur here though: the semidirect product of ${\rm SL}(2,2^{n})$ with its natural module is one such example.

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  • $\begingroup$ Also, it is possible that a nonsolvable group has the property in question. For example $\operatorname{PSL}(3,3)$ has order $2^4 \cdot 3^3 \cdot 13$ and contains an element of order $6$. $\endgroup$ – Mikko Korhonen Nov 5 '14 at 19:10
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The original source is Lucido: The diameter of a prime graph of a finite group, J. Group Theory 2 (1999), 157−172.

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  • $\begingroup$ I do not have access to de gruyter. Can you send it to me on email. I know I should not ask, but I m really interested in this question. $\endgroup$ – Bhaskar Vashishth Nov 4 '14 at 18:53
  • $\begingroup$ I don't have online access either. I am sure you can find it in a library. If not, then perhaps someone else can help you. $\endgroup$ – GH from MO Nov 4 '14 at 19:32
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What you call "Lucido's three prime lemma" is actually part of Theorem 1 of Graham Higman's 1957 paper Finite groups in which every element has prime power order (J. London Math. Soc. 32 (1957), 335–342). This paper contains already some results about non-solvable groups, and some of the papers citing Higman's paper may contain more. I wouldn't be surprised if now there is some classification of groups having the property in the title of Higman's paper.

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Lucido's proof goes like this:

Let $G$ be a counterexample of minimal order and let $N$ be a minimal normal subgroup of $G$. If $pqr$ divides $[G:N]$, then there exists an element in $G/N$ whose order is the product of two of those primes, and thus an element of this order in $G$. We can therefore assume without loss of generality that $N$ is a $p$-group.

If $H$ is a $\{q,r\}$-Hall subgroup of $G$, there cannot exist an element of order $qr$ in $H$. Thus $H$ must have a trivial center. Now $H$ acts by conjugation on $N$ and, by the choice of $G$, $H$ must act fixed-point-freely on $N$. By theorem 8.5 of [6], $NH$ is a Frobenius group with Frobenius complement $H$; by Theorem 8.18 of [6], $H$ has a non-trivial center. But this is a contradiction.

[6] is Endliche Gruppen by Huppert.

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