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I have read the fact that the principal congruence subgroups $\Gamma(N)$ of $\mathrm{GL}_n({\mathbb{Z}})$ are torsion free for $N \geq 3$ several times, but only saw proofs for very specific situations. Could someone give me a hint where to find a general proof?

I am interested in this, because i need something similar in a situation where this fact can not be applied directly and I want to see if the proof still works. That is, for an imaginary quadratic number field $\mathbb{Q}(\sqrt{-k})$ and the corresponding algebraic group $\mathrm{SU}(n,1)$, one can attach an arithmetic subgroup $\Gamma$ to a choice of parahoric Subgroups $P_v$ for all nonarchemedian places $v$. However, this $\Gamma$ can not be imbedded into some $\mathrm{GL}_m(\mathbb{Z})$, but rather into $\mathrm{GL}_m(\mathbb{Z}[\frac{1}{k}])$. Is there a reasonable way to define $\Gamma(N)$, for example, when $N$ and $k$ are coprime, as the kernel of the reduction modulo $N$ map (where $\frac{1}{k}$ is just considere as $k^{-1}$ modulo $N$), such that the groups $\Gamma(N)$ are torsion free (for $N\geq 3$) and normal in $\Gamma$? Help would be much appreciated!

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First observe that $\Gamma(p^l) \subseteq \Gamma(p)$ for any prime number $p$ and any $l \geq 0$. For prime numbers $p \geq 3$ the group $\Gamma(p)$ is torsion-free. See this: https://math.stackexchange.com/q/3979175.

Now, $N = \prod_{p| N} p^{\nu_p}$ ($p$ prime number) and by Chinese remainder theorem it is $\mathbb{Z}/N\mathbb{Z} \cong \prod_{p | N} \mathbb{Z}/p^{\nu_p}\mathbb{Z}$ and therefore $\Gamma(N) = \bigcap_{p | N}\Gamma(p^{\nu_p})$. If there is some prime $p \geq 3$ with $p | N$, then $\Gamma(N) \subseteq \Gamma(p^{\nu_p}) \subseteq \Gamma(p)$. Since $\Gamma(p)$ is torsion-free, $\Gamma(N) \subseteq \Gamma(p)$ is torsion-free.

So, the only thing remaining to prove is that $\Gamma(4)$ is torsion-free. This is also at the end of https://math.stackexchange.com/q/3979175.

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  • $\begingroup$ Thank you very much! $\endgroup$ Commented Feb 6 at 12:10

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