We know $|PSL(2,p)|=p(p+1)(p-1)/2$. Let $p$ be Mersenne prime (that is $p+1=2^{n}$) and $r$ be prime divisor of $(p-1)/2$.
My question: What is the number of Sylow $r$-subgroups of $PSL(2,p)$?
We know $|PSL(2,p)|=p(p+1)(p-1)/2$. Let $p$ be Mersenne prime (that is $p+1=2^{n}$) and $r$ be prime divisor of $(p-1)/2$.
My question: What is the number of Sylow $r$-subgroups of $PSL(2,p)$?
Let $G=\operatorname{PSL}(2,q)$ for q an odd prime power. If r is an odd prime not dividing q but dividing $|G|$, then r divides exactly one of $\{q+1,q-1\}$ and a Sylow r-subgroup of G is cyclic with normalizer dihedral of order $q-\epsilon$ ($\epsilon = \pm1$). In particular, the number of Sylow r-subgroups is $q(q+\epsilon)/2$.
The normalizer is a at least this big because it obviously normalizes the subgroup, and it is no bigger, since it is a maximal subgroup.
This does not require q to be a Mersenne prime, or even prime (just odd), and does not require r divide $q-1$ (though in your case this means $\epsilon=1$).