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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.

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The proof should be the same as for Lagrange's theorem: $H$ acts freely on $G$ by translations, making $G$ an $H$-torsor over $G/H$. –  Keerthi Madapusi Pera May 18 '12 at 10:28
    
Maybe you want to assume your $\pi$ is finite locally free instead of just finite flat, if your $S$ might not be locally Noetherian. I feel like $\pi_*\mathscr{O}_G$ might not be locally free if $\pi$ is not locally of finite presentation. –  Keenan Kidwell May 18 '12 at 16:24
    
@Keenan: that's a good point, I simply forgot the locally noetherian assumption. –  Veen May 18 '12 at 17:09
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up vote 3 down vote accepted

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)).

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That's the perfect reference, thanks a lot! –  Veen May 19 '12 at 7:27
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