If $P$ is a $p$-Sylow subgroup of $G$ and $N_G(P)\leq U\leq G$, then $|G:U|\equiv 1$ mod $p$.

Question

Hi: $P$ acts on the set $\Omega$ of all cosets $Ug$, $g\in G$. Then $|\Omega|=|G:U|$ and by a previous theorem, $$|C_\Omega(P)|\equiv |\Omega|$$ mod $p$, where $C_\Omega(P)$ is the set of all those elements of $\Omega$ that are fixed by every element of $P$. $U$ is in $C_\Omega(P)$ since $P\leq U$. Let $Ug\in C_\Omega(P)$. Then $UgP=Ug$, and this implies $gPg^{-1}\leq U$. If $Ug\mapsto gPg^{-1}$ were a function I can prove it is a bijection and then $|C_\Omega(P)|=$ number of Sylow p-subgroups of $U$, that is $|C_\Omega(P)|\equiv 1$ mod $p$ and this would mean $|G:U|\equiv 1$ mod $p$. But I can't prove the function is well defined. Any way to prove the title statement?