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Let $G$ be a diagonalizable algebraic group and $X$ be the character group of $G$. Let $Y$ be a subgroup of $X$. We define $Y^{\perp}$ to be all the $x\in G$ such that $\chi(x)=1$ for all $\chi\in Y$. If $X/Y$ has no $p$-torsion, show that $(Y^{\perp})^{\perp}=Y$.

I'm wondering how to proceed with this problem. I think that the hard part is the inclusion of $(Y^{\perp})^{\perp}$ into $Y$. Would it be simpler to do it for tori first as I know that there a diagonalizable group is the direct product of a torus and a finite group?

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A few missing bits? Is $p$ the characteristic of the ground field? You are supposed to show that $(Y^\perp)^\perp$ is... what? $Y$? If $H$ is a subgroup of $G$ does $H^\perp$ mean the subgroup of $X$ that is trivial on $H$? Natural guesses - just wanting to make sure :-) –  Jyrki Lahtonen Feb 26 '13 at 18:31
    
And I suppose $X$ is the character group of $G$ ? –  user18119 Feb 26 '13 at 21:33
    
Yes; I do want to show that it is equal to $Y$. and $X$ is supposed be the character group of $G$. And if $H$ is a subgroup of $G$, then $H^{\perp}$ is meant to be the subgroup of $X$ that is trivial on $H$. I've changed it now thanks. –  user64099 Feb 26 '13 at 22:44

1 Answer 1

I will suppose $G$ is connected, hence isomorphic to a torus $\mathbb G_m^n$. By the classification of finitely generated abelian groups, there exists a basis $e_1, e_2, \dots, e_n$ of $X$ and $a_1, a_2, ..., a_d\in \mathbb N_{>0}$ prime to $p$ such that $a_1e_1, \dots, a_de_d$ is a basis of $Y$. Now write the ring of $G$ as $k[T_1^{\pm 1}, \dots, T_n^{\pm 1}]$ with $e_i(T)=T_i$ (in terms of map on the points, $(t_1,\dots, t_n)\mapsto t_i$, so $a_ie_i: (t_1,\dots, t_n)\mapsto t_i^{a_i}$). Then $$Y^{\perp}= \mu_{a_1} \times \dots \times \mu_{a_d} \times \mathbb G_m^{n-d}$$ and $(Y^{\perp})^{\perp}=(\oplus_{i\le d} a_ie_i\mathbb Z) \oplus (\oplus_{j>d} e_j\mathbb Z)=Y$.

If $G$ is not connected, let $\pi_0(G)$ be its group of connected components and let $G^0$ be the neutral component of $G$. Then $G\simeq G^0\times \pi_0(G)$ because $G^0$, being a torus, is divisible. So the character group of $G$ is just the product $X(G^0)\times X(\pi_0(G))$. This should be enough to handle the general case.

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