Be $H$ Sylow p-subgroup of $S_{2p}$ with $p > 2$ prime number. Be $N(H)$ the normalizer of H. Find the cardinality of $N(H)$ I've got some trouble doing this exercise

Be $H$ Sylow p-subgroup of $S_{2p}$ with $p > 2$ prime number. Be $N(H)$ the normalizer of H. Find the cardinality of $N(H)$

I know that H is generated by two disjoint p-cycle, so H is isomorphic to $\mathbb{Z}_{p}\times \mathbb{Z}_{p}$ but I'm stuck.

We note that $N(H) = \text{Stab}(H)$ under the conjugation action of $S_{2p}$ on $H$. By the Orbit-Stabilizer Theorem, we have that $n_{p} \cdot N(H) = |S_{2p}|$, where $n_{p}$ is the number of Sylow $p$-subgroups.
Start by determining $n_{p}$. We have two disjoint $p$-cycles. We select the elements for the first $p$-cycle in $2p \cdot (2p-1) \cdot (2p-2) \cdots (p+1)$ ways. Two $p$-cycles are equivalent if one is obtained from the other by cyclic rotation. So we divide out by $p$, the number of cyclic rotations, to obtain:
$$2p(2p-1) \cdots (p+1)/p$$
such ways to choose the first $p$-cycle. Similarly, the second $p$-cycle is chosen in $p!/p = (p-1)!$ ways. The selections are independent. So by the rule of product, we have $n_{p} = (2p)!/p^{2}$. So $|N(H)| = p^{2}$.
