This post comes directly from a talk of Elisa Covato at the seminar of the the PhD students of Bristol.

Definition of the prime graph of a group
Let G be a finite group. Let $\pi (G)$ be the set of prime factors of $\lvert G\rvert$. The prime graph $\Gamma(G)$ has $\Pi(G)$ as vertex set and two primes $r,s$ are neighbours if there exists $g\in G$ whose order is multiple of $rs$.

Note that Lagrange’s theorem implies that every order of an element of $G$ is a divisor of $\lvert G\rvert$, hence is a product of elements of $\Pi(G)$ ; and that thanks to Cauchy, we know that every element of $\Pi(G)$ is the order of somebody. But a product $rs$ of elements of this set is not always the order of somebody, not even a divisor of the order of somebody.

In the case of the unique abelian group of order $p_1\cdots p_k$ where the $p_i$ are pairwise distinct primes, the prime graph is the complete graph.

In the case of $\mathcal{A}_{10}$, we have $\Pi(G)=\{2,3,5,7\}$. And there’s somebody of order 6, for instance $(123)(45)(67)$ ; of order 10, for instance $(12345)(67)(89)$. There’s also somebody of order 15, for example $(12345)(678)$. There’s somebody of order 21, $(1234567)(89A)$ (où A=10). But no one has order 14 nor 35. Indeed, recall that the order of a permutation is the ppcm of the orders of the cycles in its decomposition in a product of disjoint cycles ; and if 14 is the ppcm of some numbers, then 2 and 7 ; and in order to make an even permutation with at least one cycle of length 2 and at least one of length 7, you need 11 elements. For $5\times 7$, you need 12 elements.

A lot of information about $G$ is encoded in the prime graph but not every thing.

Recognition problem
Can a group $G$ be uniquely determined by its prime graph (at least among a more or less restricted class of groups) ? So if $\Gamma (G_1)=\Gamma(G_2)$ are the groups isomorphic ? Be careful, when I say that the graphs are equal, I mean it ! Not only isomorphic ! Otherwise for instance all the $\mathbb{Z}_P$ have the same graph, a single vertex.

The answer is rather no in general.

Theorem (Lucchini, Morigi, Shumyatski)
Let $G$ be a finit group. Then there exists $H\leq G$ which is 3-generated and has the same prime graph.

Of course, the theorem doesn’t claim that $H\neq G$ so it’s only interesting when $G$ is not himself 3-generated (generated by a set of at most 3 elements). Moreover, if $G$ is simple, it is even 2-generated (if you believe in the classification of finite simple groups).

Problem : let G be finite and simple. Determine $(H,H)$ with $H\leq G,\Gamma(H)=\Gamma(G)$.

Theorem (Burness, Covato)
Let $G$ be finite, simple and $H$ be a maximal subgroup. Then $\Gamma (H)=\Gamma (G)$ implies that one of the following holds:
a) $(G,H)$ belongs to a quite short table of exceptions.
b) $G=\mathcal{A}_n,H=(\mathcal{S}_k\times\mathcal{S}_{n-k})\cap\mathcal{A}_n$
Moreover, $\Gamma(H)=\Gamma (G)$ whenever they are in the table.

I didn’t really get how the proof goes but apparently it uses a former result of Liebeck, Prager & Saxl who determined in which case the set of prime factors is the same. And then it uses an old result of Zsygmondi (1892) : for all prime power q, e integer, $q^e$ has a prime divisor which divides no other $q^i-1$ with $i\leq e$.

When b) holds, it is not always true that the prime graphs are equal.

Conjecture : Suppose $G=\mathcal{A}_n,H=(\mathcal{S}_k\times\mathcal{S}_{n-k})\cap\mathcal{A}_n$ with $1< k< n$ and $p\leq n\implies p\leq k$ for every prime p.
Then $\Gamma(H)=\Gamma (G)\Longleftrightarrow$ either
i)$n\geq 25$ is odd,k=n-1, n-4 is composite
ii)(n,k) is (6,5) or (10,7)

Goldbach's conjecture implies this conjecture. In fact
Lemma : suppose $G=\mathcal{A}_{p+1}, H=\mathcal{A}_{p},p\geq 7$. Then
$\Gamma(H)=\Gamma (G)\Longleftrightarrow p+1=r+s$.

We have restricted ourselves in all this to maximal subgroups H. Can we go further?
Theorem : let G be finite simple and H a proper subgroup. Let’s even suppose that $G=\mathcal{A}_n$ and H is transitive (so we forbid the situation of b)). Then they have same prime graphs if and only if either
i) H is maximal and (G,H) are in the table.
ii) H is a second maximal group (maximal subgroup of maximal subgroup M).