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 $G\cong G'$?


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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?

  • $\begingroup$ @ Derek Holt. I would prefer to restrict our simple groups to $A_{p}$,for some prime $p$, $PSL(p,q)$ and $PSU(p,q)$ for some prime $p$ and some prime power $q$. What about nwo? $\endgroup$ Jan 6, 2018 at 17:56

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