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Is there solvable group $G$ such that prime graph $G$ equal to prime graph $PGL(2,p)$ and $|G|=|PSL(2,p)|$?

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Why not remind us what the prime graph is. –  Derek Holt Apr 22 '12 at 13:01
    
I think the prime graph is defined as follows: Vertex: all the primes dividing $|G|$. The vertices $p, q$ have a edge if and only if there exists $g \in G$ such that $|g|=pq$. –  Wei Zhou Apr 22 '12 at 13:24
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3 Answers

You should consult the papers of Akhlaghi, Khosravi and Khatam - they have two that are relevant. I don't have subscription access to the full text of the articles but I can access enough to say the following.

With regard to the group $PGL(2,q)$, the situation depends dramatically on whether or not $q$ is prime.

Case 1: $q=p$, a prime. Let me quote from the mathscinet review of this paper:

There are infinitely many nonisomorphic finite groups with the same prime graph as $PGL(2,p)$. In this paper, the authors determine the structure of finite groups $G$ such that $\Gamma(G)=\Gamma(PGL(2,p))$, where $11\neq p \neq19$ and $p$ is not a Mersenne or Fermat prime. In particular, if $p\neq 13$ then $G$ has a unique nonabelian composition factor which is isomorphic to $PSL(2,p)$ and if $p=13$ then G has a unique nonabelian composition factor which is isomorphic to $PSL(2,13)$ or $PSL(2,27)$.

Here I'm writing $\Gamma(G)$ to mean the prime graph of a group $G$. So, to answer your question, this result means that if a solvable group $G$ is to satisfy $\Gamma(G)=\Gamma(PGL(2,p))$ for some prime $p$, then $p$ is a Mersenne or Fermat prime.

Case 2: $q$ is not prime. Then this paper proves that the group $PGL(2,q)$ is characterized by its prime graph, i.e. there are no other groups sharing the same prime graph.

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Though "sara" has long since disappeared, your answer sheds better light on this kind of old problem, which is easy to formulate after a basic course in group theory. It's not so easy to resolve, but of course there is a long paper trail by now, including numerous multi-author papers (including the ones you cite) on the recognition problem relative to prime graphs or sets of element orders. Probably a more useful question than the one asked here would be whether any single source gives a complete survey of what's known and unknown at this point along with basic methods. –  Jim Humphreys Dec 19 '12 at 20:33
    
Jim, good comment. This is potentially a useful community-wiki type question for MO. Something like "What is the current status and priority of group-recognition questions?" Experts may then be able to explain (a) why a particular group-recognition question is useful, and (b) how much is known on the given question... –  Nick Gill Dec 20 '12 at 10:25
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I have to say that I have no work about this problem. But I know something relate to this problem. Let $\pi_i$ ($i=1, \cdots, t$) be the connected components of the prime graph. Then $|G|=m_1\cdots m_t$, where $\pi(m_i)$ is the vertext set of $\pi_i$. The integer $m_i$ are called the order components of $G$. Then $PSL(2,q)$ ($q$ is odd prime power) is uniquely determined by its components. (see G.Y. Chen, A new characterization of PSL(2,q), Southeast Asian Bull. Math. 22 (1998), 257-263). In your problem, $G$ and $PSL(2,q)$ have the same order and prime graph, then their order components are same, and then $G \cong PSL(2,q)$. (I am sorry that I have not read this paper.)

By the way, the problem are relate to Thompson conjecture, and G. Y. Chen had some good work about this conjecture. (see G.Y. Chen, On Thompson's conjecture, J. Algebra 15 (1996), 184-193.)

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I am very sorry for misunderstanding your question. –  Wei Zhou Apr 23 '12 at 13:17
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There is such an example for $p=7$, namely the group ${\rm A \Gamma L}(1,8)$ of order 168. Its prime graph has vertices 2,3 and 7, and a single edge joining 2 and 3. The group ${\rm PGL}(2,7)$ has the same prime graph. (This is different from the prime graph of ${\rm PSL}(2,7)$, which has no edges.) This might be the only example.

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