Let $G$ and $G'$ be two finite simple groups and $p$ be a prime divisor of $\vert G\vert$ and $\vert G'\vert$. Also suppose that every Sylow p-subgroup of $G$ and $G'$ is a prime order subgroup($C\_{p}$. If the number of Sylow p-subgroups of $G$ is equal to the number of Sylow p-subgroups of $G'$, then can we say that $G\cong G'$?



Recognition of finite simple groups by number of Sylow p-subgroups
A good place to look for counterexamples might be nonisomorphic simple groups of the same order.


And indeed we find that $A\_8$ and ${\rm PSL}(3,4)$ both have $960$ Sylow $7$-subgroups.


The next such pair is ${\rm PSp}(6,3)$ and ${\rm P}\Omega(7,3)$ and they both have the same numbers of Sylow $p$-subgroups for $p=5,7$ and $13$.


More generally, for odd $q$ and $2n \ge 6$, ${\rm PSp}(2n,q)$ has the same order as but is not isomorphic to ${\rm P}\Omega(2n+1,q)$, and I would expect there to be many more resulting examples.


Of course it is very likely that there are examples coming from simple groups that do no have the same order - why not?


