This proof comes from S. Lang's Algebra Revised Third Edition. Honestly 2/3 of this proof makes sense to me, except the application of class formula and the followings. My questions are:
Why do we have $(G:G_x)$ is divisible by $p$? Is it possible to deduce this simply from the fact that $(G:G_x)>1$?
Why $p$ divides the order of $Z$ because $p$ divides the order of $G$?
Appreciate in advance for anything helpful. It seems that I'm missing some critical details of the class formula. Feel free to point out anything wrong with me!
Edit 1: Some denotations.
The set of elements $x \in G$ such that $xs=s$ is obviously a subgroup of $G$, called the isotropy group of $s$ in $G$, and denoted by $G_s$.
Moreover, Lang denoted the orbit of $s$ under $G$ by $Gs$.
$G_x$ is a subgroup of $G$. Since $(G:G_x)>1$, $G_x$ is a proper subgroup of $G$, so that we know that $p$ divides $(G : G_x)$. Equivalently, one could argue that $G_x = G \Leftrightarrow x \in Z$.
Once 1. has been proved, you get that $\sum (G : G_x)$ is divisible by $p$. Since $(G : 1)$ is also divisible by $p$ and $0 \neq (Z : 1) = \sum (G : G_x) - (G : 1)$, this implies that $p$ also divides $(Z : 1)$.