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This question is inspired by Upper bound on order of finite subgroups of GL_n(Z_p)?. It's showed that the supremum of orders of finite subgroups of ${\rm GL}_n(\mathbb{Z}_p)$ is finite and can be explicitly bounded, and there are arbitrarily large finite subgroups of ${\rm GL}_n(\mathbb{F}_q[[T]])$ for all prime powers $q$ and all $n\geq 2$. However, it's not hard to show that there is always a bound of the order of torsion elements in ${\rm GL}_n(\mathbb{F}_q[[T]])$, cf. Lemma 5.5. in A topological Tits alternative. My question is the following:

Let $n\geq 2$, $R=\mathbb{Z}_p$ or $R=\mathbb{F}_p[[T]]$, and let $B(n,p)$ denote the maximum order of torsion elements in ${\rm GL}_n(R)$. What is $B(n,p)$? Can we get some estimates on it?

Any references is highly appreciated.

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    $\begingroup$ For $R=\mathbb F_p[[T]]$ the order $p^n-1$ is certainly achievable, since $\mathbb F_{p^n}^\times$ is a subgroup. $\endgroup$ Commented Apr 21, 2023 at 0:51

1 Answer 1

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Lemma Let $F/\mathbb Q_p$ be a $p$-adic field, with residue field $k_F$ of size $q$. Then the group of roots of unity in $F^\times$ is cyclic of order $p^e(q-1)$, where $e\ge0$ is the minimal integer such that $F\supset\mathbb Q_p(\zeta_{p^e})$.

Proof: Since $x^N=1$ we have $|x|=1$, so $x\in\mathcal O_F^\times\cong k_F^\times\times(1+\mathfrak p_F)$. Here $k_F^\times$ is cyclic of order $q-1$, and $1+\mathfrak p_F$ is a pro $p$-group, so the Lemma follows.


Corollary Given an integer $N$, the maximal order of a torsion element in $F^\times$ for a field extension $F/\mathbb Q_p$ of degree $N$ is $p^N-1$.

Proof: For a field $F/\mathbb Q_p$ of degree $N$, let $e\ge0$ be the integer as in the Lemma above. Then the maximal possible order of an element in $F^\times$ is $$p^e(q-1)\le p^e(p^{N/p^{e-1}(p-1)}-1)\le p^N-1,$$ since $\mathbb Q_p(\zeta_{p^e})/\mathbb Q_p$ has degree $p^{e-1}(p-1)$, and $q\le p^{[F:\mathbb Q_p(\zeta_{p^e})]}$. Moreover, $p^N-1$ is achieveable by letting $F/\mathbb Q_p$ be the unique degree $N$ unramified extension.


Now let $g\in\mathrm{GL}_n(\mathbb Q_p)$ be of finite order. Then $g$ has characteristic polynomial $P=P_1\cdots P_r$, where $P_i$ is irreducible, say of degree $n_i$. Then $g$ can be viewed as an element of the group of units of the etale algebra $F_1\times\cdots\times F_r$, where $F_i=F[X]/P_i$ is a degree $n_i$ extension of $F$. Thus, the maximal possible order is $(p^{n_1}-1)\cdots(p^{n_r}-1)\le p^n-1$. We have seen that $p^n-1$ is achieveable (over $\mathbb Z_p$ as well), so it is the answer.

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  • $\begingroup$ When $p=2$, the group ${\rm SL}_2(\mathbb{Z})$ (and hence ${\rm SL}_2(\mathbb{Z}_2)$ ) has an element of order $4$ (resp. $6$). In particular, your claim is not true for $p=2$. $\endgroup$
    – stupid boy
    Commented Nov 18, 2023 at 19:27

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