Task is:
Let $p$ be a prime number and let $C$ be a cyclic subgroup of order $p$ in $G = S_p$. Compute the order of
the normalizer $N_G(C)$.
It's clear that $e$ is in the normalizer, as well at C itself(since it's abelian), so $|N_G(C)| \geq p+1$, but I can't find a way to make any conclusions for the rest of the group.
Best Answer
Let $C=\langle \alpha \rangle$, then $\alpha=(a_1\, a_2 \ldots a_p)$. There are $p!$ such expressions, but each cycle can be written $p$ ways as such an expression. This gives us $(p−1)! $ $ p-$cycles in $S_p$, and we know they are all conjugate. Each of these cycles generates a group of order $p$, and each such group has $p−1$ generators. Thus there are $(p−2)!$ cyclic subgroups of order $p$ in $S_p$, all conjugate. Hence the normalizer of anyone of them has index $(p−2)!$ and hence has order $p(p−1)$.