Let $\pi: G \rightarrow S$ be a finite flat group scheme over a locally noetherian connected base scheme $S$.
Its degree is defined as the rank of the locally free $\mathcal O\_S$-module $\pi\_\* \mathcal O\_G$.
Let $H \subset G$ be a closed subgroup scheme of $G$ which is also finite flat over $S$.


I want to show that the degree of $H$ divides the degree of $G$. How does one do this?
I guess this must be easy but I'm somehow stuck.



Degree of finite group schemes
This can be seen from the existence of the quotient $G/H$ as a finite flat $S$-scheme (and an $H$-torsor). One shows that the natural map $G \rightarrow G/H$ is finite flat of order $[H : S]$; the conclusion then follows from the product formula $[G : S] = [G : G/H] [G/H : S]$. Let me give you a reference where all this is spelled out in detail (and which I am basically copying):


Tate, John. Finite flat group schemes. Modular forms and Fermat's last theorem (Boston, MA, 1995), 121–154, Springer, New York, 1997.


What you need is section (3.5) (see also (3.4) and (3.1)).


