4
$\begingroup$

I am reading M. Brandenburg's paper and came across the following result which is a generalization of Lagrange's theory in group theory:

Let $\mathcal C$ be a $\otimes$-category and $A\to B$ a morphism of algebras (in $\mathcal C$). Then if $M$ is an $A$-module isomorphic to $T\otimes B$ as a $B$-module and $B_{|A}\cong S\otimes A$ as an $A$-module, then $M_{|A}\cong (T\otimes S)\otimes A$.

The proof is a a simple one-liner. My question is how do we obtain Lagrange's theorem from it? At first, I thought we take $\mathcal C$ to be the cartesian closed category $\mathbf{Set}$ and let $A\to B$ be the inclusion of a subgroup in a finite group. But we are assuming $B_{|A}\cong S\times A$ and so there would be nothing to show since that would imply the order of $A$ divided the order of $B$. I know this should be easy, any help?

$\endgroup$
  • $\begingroup$ Are you sure the algebras are commutative? I don't think they necessarily are (otherwise you would only get abelian groups). $\endgroup$ – Najib Idrissi Oct 19 '15 at 11:45
  • 3
    $\begingroup$ My impression is that'll take @MartinBrandenburg himself to answer what he meant by that remark. :) $\endgroup$ – Omar Antolín-Camarena Oct 19 '15 at 21:35
  • $\begingroup$ @NajibIdrissi Yes, you are correct. $\endgroup$ – Rachmaninoff Oct 20 '15 at 0:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.