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The Jordan property for finite subgroups of ${\rm GL}_n(\mathbb{C})$ says that there exists a constant $c(n)$ so that for any finite subgroup $G$ of ${\rm GL}_n(\mathbb{C})$ there is a normal abelian subgroup $A$ of $G$ of index at most $c(n)$, and we can embed $A$ in a maximal torus of ${\rm GL}_n(\mathbb{C})$.

I was wondering if there exists some sort of "quotient version". I am expecting some result of the form: There exists a constant $c(n)$ satisfying the following: Let $G$ be a subgroup of ${\rm GL}_N(\mathbb{C})$ and $T_1$ be a subtorus of rank $N-n$ of a maximal torus $\mathbb{G}_m^N$ of ${\rm GL}_N(\mathbb{C})$. Assume that there exists an exact sequence $1\rightarrow T_1\rightarrow G\rightarrow G/T_1\rightarrow 1$, where $G/T_1$ is a finite group. Then, there exists a normal abelian subgroup $A\leqslant G/T_1$ of index at most $c(n)$ so that its pre-image on $G$ lies in the maximal torus (or some conjugate).

If $T_1$ is the trivial torus, this gives back the Jordan property. I am not an expert on representation theory. Some results in algebraic geometry I am working on point out to a statement like this, so I wanted to know if there is anything in the literature.

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    $\begingroup$ I don't understand what you mean by "suppose that there exists an exact sequence $1\to T_1\to G$". Do you assume that $T_1\subset G$? Or is $T_1\to G$ allowed to be an arbitrary group homomorphism, unrelated to the context? that would sound weird. $\endgroup$
    – YCor
    Jul 8, 2020 at 6:37
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    $\begingroup$ It's a short exact sequence, so $T_1$ is a subgroup of $G$. In trying to write down a counterexample I see what the OP might be getting at. If a subgroup $G$ of $X=\mathrm{GL}_N(\mathbb{C})$ contains a big subtorus $T_1$ of $X$, then $G/T_1$ is either 'small' (dependent on the codimension of the subtorus in the maximal torus) or it contains a torus. I've not seen anything like this. Can I not take the product of the normalizer of $T_1$ with the Weyl group of the remaining part of the torus? Actually, I don't even need that small symmetric group. Does $G$ need to act irreducibly? Surely. $\endgroup$ Jul 8, 2020 at 8:33
  • $\begingroup$ OK, a clearer statement is "let $G$ be a subgroup of $\mathrm{GL}_N(\mathbb{C})$ with a normal subgroup of finite index $T_1$ that is a subtorus (...)". $\endgroup$
    – YCor
    Jul 8, 2020 at 10:59
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    $\begingroup$ I think it's known that $G=T_1F$ for some finite subgroup $F$ of $G$ (with possibly nontrivial intersection). One can apply Jordan to $F$, and also the fact that finite subgroups of $\mathrm{GL}_{N-n}$ have bounded order, to infer that there is an abelian subgroup $F'$ of $F$ centralizing $T_1$, of bounded index. So $F'T_1$ is abelian and has bounded index. Nevertheless here 'bounded' also depends on $N$, not only on $n$, and OP didn't quantify on $N$. $\endgroup$
    – YCor
    Jul 8, 2020 at 11:06
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    $\begingroup$ But if you have a bound on $N$ and $n$ then you can just apply standard Jordan. So I think this is either clear or false. $\endgroup$ Jul 8, 2020 at 21:46

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