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For a finite group $G$, the prime graph of $G$ is an undirected graph such that its vertices are all prime divisors of $\vert G\vert$ and two distinct vertices $p$ and $q$ are adjacent when there is an element in $G$ of order $pq$.

Which finite groups does have a complete prime graph? i.e I am looking for finite groups in which for every two distinct prime divisors of $\vert G\vert$ say $p$ and $q$, there exists at least one element of order $pq$. For instance every Dedekind group has this property.

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    $\begingroup$ Certainly all nilpotent groups have this property, and this is a wider class that Dedekind groups. $\endgroup$ – verret Sep 14 '16 at 22:31
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    $\begingroup$ Victor Mazurov mailed to me this answer: This is a hard problem. For example, if G is any finite group then G\times G has this property. $\endgroup$ – H.Shahsavari Sep 15 '16 at 8:38
  • $\begingroup$ This paper should help: math.nsc.ru/~vdovin/Papers/adjeng.pdf The tables in the back imply, in particular, that you can't have $G$ being non-abelian simple. I'm guessing you also can't have $G$ being non-abelian almost simple. But, as you've mentioned, you can have $G$ being the direct product of 2 non-abelian simple groups, and that means that a full classification (using $F^*(G)$) is likely to be a bit tricky. $\endgroup$ – Nick Gill Sep 15 '16 at 8:59
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    $\begingroup$ Even the soluble case appears untractable to me. For the class of {2,3}-groups, for example, this is asking which ones have an element of order 6. I doubt this class has a simpler description than this. $\endgroup$ – verret Sep 15 '16 at 10:07
  • $\begingroup$ @verret, By $\{2,3\}$-groups, do you mean those with order $2^a3^b$? In this case, it seems possible that you get any group that isn't a Frobenius group. Here's the idea: if $G$ is such a group WITH NO ELEMENT OF ORDER 6, then $F(G)$ is either a 2-group or a 3-group. Let's suppose the latter for now. Then $P_2$ (a Sylow 2) acts fixed-point-freely on $F(G)$ and so $F(G)\rtimes P_2$ is a Frobenius group, and the structure of $P_2$ is rather restricted (quaternion or cyclic? Can't quite remember)... $\endgroup$ – Nick Gill Sep 15 '16 at 15:01
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To understand the "minimal troublemakers" in the case of solvable groups, I would start with a solvable group $G$ such that every proper section of $G$ has a complete prime graph but $G$ does not. Recall that a section of $G$ is a group $X/Y$ where $Y \lhd X$ and $X$ is subgroup of $G$.

Such a grooup $G$ must be a $\{p,q\}$-group for a pair of distinct primes $p$ and $q$ ( for otherwise $G$ has a Hall $\{p,q\}$-subgroup $H$ which is proper and $H$ contains an element of order $pq).$ Also, a Sylow $p$-subgroup of $G$ must be a maximal subgroup of $G$, and likewise a Sylow $q$-subgroup of $G$ must be maximal.

Now $G$ can't have both a normal Sylow $p$-subgroup and a normal Sylow $q$-subgroup. However, we have either $O_{p}(G) \neq 1$ or $O_{q}(G) \neq 1.$ Label so that $O_{p}(G) \neq 1.$ Then $G/O_{p}(G)$ must be a $q$-group ( otherwise it contains an element of order $pq$, and then so does $G$).

Hence $O_{p}(G)$ is a normal Sylow $p$-subgroup of $G$, and is hence a maximal subgroup of $G$, so that $[G:O_{p}(G)] = q.$ Hence $G$ is a Frobenius group with kernel $O_{p}(G)$ and cyclic complement of order $q$.

We can go a little further since a Sylow $q$-subgroup $Q$ of $G$ is a maximal subgroup. For then $G = QM$ for some minimal normal subgroup of $G$ contained in $O_{p}(G),$ and we deduce that $|G|$ has the form $qp^{e}$ where $e$ is the smallest positive integer such that $q$ divides $p^{e}-1.$

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This can be backed out from the referenced paper (here is the MathReview).

Maria Silvia Lucido and Ali Reza Moghaddamfar, MR 2063403 Groups with complete prime graph connected components, J. Group Theory 7 (2004), no. 3, 373--384.

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  • $\begingroup$ I have already read this paper. It does not answer my question. Please read its abstract again. $\endgroup$ – H.Shahsavari Sep 14 '16 at 22:04
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    $\begingroup$ The paper only classifies finite simple groups with complete components. The general problem is probably much harder, at least if you take the classification of finite simple groups as granted. $\endgroup$ – Jan-Christoph Schlage-Puchta Sep 15 '16 at 20:11

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