Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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.)