This is a question about the class of arithmetic groups. I am using the definition in Serre's survey: $\Gamma$ is arithmetic if it can be embedded into $G_\mathbb{Q}$ for some algebraic subgroup $G \subset GL_n$ defined over $\mathbb{Q}$ so that $\Gamma$ is commensurable with $G_\mathbb{Z}$. This question came up when looking at groups of homotopy classes of homotopy automorphisms, which are known to be arithmetic groups for simply-connected finite CW-complexes by work of Sullivan.

Suppose that I have an arithmetic group $\Gamma$ and finite subgroup $H \subset \Gamma$. Is it true that the normalizer $N(H) \subset \Gamma$ is again an arithmetic group? Of course the interesting case is when $N$ is not normal.

  • 3
    $\begingroup$ I'm not sure there's a single universally accepted definition of "arithmetic group", could you provide a definition? Do you consider $\mathrm{SL}_n(\mathbf{Z})\ltimes\mathbf{Z}^n$ as arithmetic group? Do you consider a Zariski-dense finitely generated free subgroup of infinite index of $\mathrm{SL}_2(\mathbf{Z})$ as an arithmetic group? $\endgroup$
    – YCor
    Feb 5, 2021 at 15:32
  • $\begingroup$ Yes, I consider both to be arithmetic (see e.g. Example (7) of Serre's survey). $\endgroup$
    – skupers
    Feb 5, 2021 at 15:44
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    $\begingroup$ The normalizer of $N(H)$ is a $\mathbf{Q}$-defined subgroup (for every finite subgroup $H$ of $\mathrm{GL}_n(\mathbf{Q})$), and so is its intersection with $G$. $\endgroup$
    – YCor
    Feb 5, 2021 at 15:49
  • $\begingroup$ So that means the answer to my question is "yes" with Serre's definition, correct? $\endgroup$
    – skupers
    Feb 5, 2021 at 15:53
  • $\begingroup$ @JHM H is finite, so in particular arithmetic. The OP asked whether N(H) is arithmetic, not whether it is an arithmetic subgroup of G. $\endgroup$ Feb 5, 2021 at 19:12

1 Answer 1


The definition of "arithmetic groups" depends essentially on the definition of group of $\mathbb{Z}$ points of a linear algebraic group scheme $\textbf{G}$. The group of $\mathbb{Z}$ points needs be made precise. Up to finite index, we can say the group of $\mathbb{Z}$ points is the stabilizer of a $\mathbb{Z}$-module $\Lambda\approx \mathbb{Z}^n$, e.g. $\Gamma=PGL(\mathbb{Z}^n)$. Compare Platonov-Rapinchuk's "Algebraic Groups and Number Theory", pp.172--173, especially Proposition 4.2. All the standard arithmetic groups have this form ($PGL, Sp, O$, etc.).

We find the answer to your question is "yes" in a certain sense, if we take the lattice-stabilizer definition of arithmetic groups. In this case, we can identify $Z_\Gamma H$ as the stabilizer of a $\mathbb{Z}[H]$-lattice:

We assume that $\Gamma$ is defined as the stabilizer of a lattice $\Lambda$ in some representation $V$ of $G$ defined over $\mathbb{Q}$. It is enough to show the centralizer $Z_\Gamma H$ is "arithmetic", since the indices $[Z_\Gamma H : N_\Gamma H]$ and $[Z_G H : N_G H]$ are finite whenever $H$ is finite.

If $H$ is finite subgroup of $\Gamma$, then the lattice, that is, $\mathbb{Z}$-module $\Lambda$ defining $\Gamma$ naturally becomes a $\mathbb{Z}[H]$-module. Likewise the $\mathbb{Q}$-vector space $V$ defining the representation $G$ naturally becomes a $\mathbb{Q}[H]$-module. Then $Z_\Gamma H$ is "arithmetic" in that $Z_\Gamma H$ consists of all $\mathbb{Q}[H]$-linear automorphisms which stabilize the $\mathbb{Z}[H]$-module $\Lambda$. Thus the $\mathbb{Z}[H]$-module $\Lambda$ becomes the lattice of "integer points" inside the $\mathbb{Q}[H]$-module $V$ defining the arithmeticity of $Z_\Gamma H$. Even more we see $Z_\Gamma H$ is basically the "$\mathbb{Z}$-points" of $Z_G H$, i.e. those $\mathbb{Q}[H]$-automorphisms which stabilize the lattice $\Lambda$.


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