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Let $G$ be a finite group, and let $A$ and $B$ be two abelian subgroups of $G$. Let $K$ be a number field such that all characters of $A$ and of $B$ take values in $K$. Let $\mathcal{O}_K$ be the ring of algebraic integers in $K$. Write $\{e_1,\ldots, e_n\}$ for the set of primitive idempotents of $KA$ and $\{f_1,\ldots, f_m\}$ for the set of primitive idempotents of $KB$.

Assume now that the $\mathcal{O}_K$-algebra generated by $\{e_1,\ldots, e_n,f_1,\ldots f_m\}$ is a finite $\mathcal{O}_K$-module.

Question: Must $A$ and $B$ commute under this assumption?

Edit: here is a motivating example: When both $A$ and $B$ are of order 2, the group they generate is isomorphic to $D_{2n}$ for some $n$. By considering some irreducible representations, the finiteness of the algebra becomes equivalent to the integrality of $\cos(\pi/n)$, and I believe this is not an algebraic integer for $n>2$.

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  • $\begingroup$ Do you have some nice example where it is not finite? $\endgroup$
    – Benjamin Steinberg
    Commented Aug 13, 2021 at 20:09
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    $\begingroup$ How about $S_3$, $A=<(1,2)>$, $B=<(1,3)>$, $K=\mathbb Q$. $\endgroup$
    – markvs
    Commented Aug 13, 2021 at 22:02
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    $\begingroup$ This $S_3$ case is not integral- when you apply the 2-dimensional irreducible character of $S_3$ to $1/2(1+(12))(1/2(1+(23))$ you get 1/4, so the $\mathbb{Z}$-algebra they generate cannot be a finite $\mathbb{Z}$-module $\endgroup$
    – Ehud Meir
    Commented Aug 14, 2021 at 12:08
  • $\begingroup$ What happens in the Heisenberg case? That should be pretty typical. $\endgroup$
    – Kapil
    Commented Aug 14, 2021 at 16:12
  • $\begingroup$ @Kapil If you take the two non-commuting abelian subgroups then the algebra they generate is not a finite $\mathcal{O}_K$-module. One can see this by taking a matrix representation on which the center acts non-trivially $\endgroup$
    – Ehud Meir
    Commented Aug 14, 2021 at 16:13

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