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I've been struggling with this one all day, and I was wondering if someone can give me a hand with the proof. I'm not even sure if the group in question is finitely generated, so I would appreciate if anyone will tell me otherwise as well.

remark 1 I've posted this question on Math.stackexchange as well, and am posting here since I don't seem to get any replies. I don't know if this question is "MO standard", but I guess there's no harm in trying :-P..

remark 2 For infinite topological groups, I write $G$ is finitely generated to mean that $G$ has a dense subgroup which is finitely generated (i.e. $G$ is topologically finitely generated).

Let $F\supseteq \mathbb{Q}_p$ be a local field, and let $D$ be a finite dimensional division algebra over $F$, of degree $n$ (i.e $\dim_F D=n^2$).

It is known that $D$ contains an unramified extension $E$ of $F$, such that $E/F$ is cyclic Galois. It follows that $D$ contains an element $u\in D^\times$, and that there is some $\pi\in F$ such that $D=\bigoplus_{j=0}^{n-1}u^j E$, $u^n=\pi$ and such that the restriction of the map $x\mapsto u^{-1}xu$ to $E$ is a generator of the Galois group $\mathbf{G}(E/F)$.

Thus, $D$ embeds into the $F$ algebra $M_n(E)$ of $n\times n$ matrices over $E$, via the left regular action $x\mapsto \lambda_x$ where $\lambda_x:E^n\to E^n$ is defined by $$\lambda_x(y_0,\ldots,y_{n-1})=x\cdot(\sum_{j=0}^{n-1}u^jy_j).$$

Once this is done, one can define the reduced norm on $D$ by $$Nrd_{D/F}(x)=\det\lambda_x.$$

My interest specificaly is in the groups $G:=SL_1(D):=\lbrace x\in D\mid Nrd_{D/F}(x)=1\rbrace$. I'm trying to understand whether this group is finitely generated or not.

It is clear that the embedding defined above maps $G$ into the group $SL_n(E)$. Once we use the fact that the group $SL_n(E)$ is generated by transvections (i.e matrices with $1$s along the diagonal and at most one additional non-zero entry), it is not hard to use the local structure of $E$ to prove that $SL_n(E)$ is finitely generated.

Unfortunately, a subgroup of a finitely generated group need not be finitely generated in general. If anyone knows of a criterion that might prove why this case is special that would be a great help.

On additional possibility, is to use some heavy tools from asymptotic group theory (even though I must admit this turns out a bit cumbersome..):

We call a group $G$ positively finitely generate (abbreviated PFG), if there exists a number $k\in\mathbb N$ such that the probability that an arbitrary $k$-tuple $(x_1,\ldots, x_k)\in G^k$ generates $G$, is positive. Clearly, any PFG group is finitely generated (although the converse is not true, e.g the free group in two generators is not PFG).

For any $n\in\mathbb N$ we define $m_n(G)$ to be the number of maximal subgroups of $G$ with index $n$. It is a theorem of Avinoam Mann that the group $G$ is PFG if and only if $m_n(G)$ is bounded by some polynomial in $n$. So it would suffice to prove that the (maximal) subgroups of $G$ have polynomial growth.

As I said, this sort of proof seems a bit of an overkill, but if someone knows of a reference (or a direct proof) to why the group $G$ has polynomial (maximal) subgroup growth that will be swell :-).

Anyway, this question turns out to be a bit lengthy, so I'll stop here. I would very much appreciate If someone can offer any hints as to whether or not $SL_1(D)$ is finitely generated.

Thank you very much.

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1 Answer 1

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Yes, this follows from finiteness properties of arithmetic groups and the strong approximation theorem for simply connected groups.

Choose a global field $K$ with a non-archimedean place $v$ such that $K_v \simeq k$; this can be found via elementary approximation arguments. By our knowledge of the Brauer groups of $K$ and $k$ via class field theory, there is a central division algebra $A$ over $K$ such that $A_v \simeq D$ over $k$.

Let $G$ be the absolutely simple semisimple $K$-group informally denoted as ${\rm{SL}}_1(A)$ (to be precise, $G$ represents the functor $R \rightsquigarrow {\rm{SL}}_1(A \otimes_K R)$ on the category of $K$-algebras), so $G(k) = {\rm{SL}}_1(D)$. Hence, it suffices to show that $G(k)$ contains a dense finitely generated subgroup.

By general "spreading out" arguments, for any connected semisimple $K$-group $H$ there is a finite set of places $S$ of $K$ containing the archimedean places and such that $H$ extends to a semisimple group scheme $\mathcal{H}$ over $O_{K,S}$, so $H$ is quasi-split at all places outside $S$ and hence (by the Borel--Tits structure theory) has positive rank at all places outside $S$. Thus, by enlarging $S$ to contain at least two such places, we can arrange that $$\sum_{w \in S} {\rm{rank}}(H_{k_w}) \ge 2.$$ We apply this with $H = G$ and therefore denote $\mathcal{H}$ as $\mathcal{G}$, and we enlarge $S$ if necessary so that $v \in S$.

Consider the arithmetic $S$-group $\Gamma = \mathcal{G}(O_{K,S})$ inside $G(K)$. In the number field case this is finitely generated, by general finiteness results for arithmetic $S$-subgroups of the group of rational points of absolutely simple semisimple group over number fields. In the function field case the same holds because of the rank-sum condition arranged above (by a theorem of Behr). So it suffices to prove that $\Gamma$ has dense image in $G(k) = G(K_v)$.

Let $S' = S - \{v\}$ (non-empty). Since $G$ is absolutely simple and simply connected, by the strong approximation theorem we know that $G(K)$ is dense in $G(\mathbf{A}_K^{S'}) = G(K_v) \times G(\mathbf{A}_K^S)$, where $\mathbf{A}_K^{S'}$ is the adele ring of $K$ with the $S'$-factors removed. Let $\widehat{O}_{K,S} \subset \mathbf{A}_K^S$ be the open subring $\prod_{w \not\in S} O_{K_w}$, so $\mathcal{G}(\widehat{O}_{K,S})$ is an open subset of $G(\mathbf{A}_K^S)$ that meets $G(K)$ in exactly $\mathcal{G}(O_{K,S}) = \Gamma$. Hence, if $U \subset G(K_v)$ is a non-empty open subset then $U \times \mathcal{G}(\widehat{O}_{K,S})$ is an open subset of $G(\mathbf{A}_K^{S'})$ whose intersection with $G(K)$ is identified (via ${\rm{pr}}_v$) with $U \cap \Gamma$ inside $G(K_v)$. But recall that $G(K)$ is dense in $G(\mathbf{A}_K^{S'})$, so its intersection with $U \times \mathcal{G}(\widehat{O}_{K,S})$ is non-empty. In other words, $U \cap \Gamma$ is non-empty inside $G(K_v)$. Since $U$ was arbitrary, this says that $\Gamma$ is dense in $G(K_v) = G(k)$ as desired.

This argument is quite robust with respect to the permitted generality of the initial connected semisimple $k$-group, the main point being the existence of the "globalization" of some $K$. One is then led to ask: if $G$ is a connected semisimple group over $k$ that is absolutely simple then does $G$ admit a "globalization" relative to some pair $(K,v)$ consisting of a global field $K$ and place $v$ such that $K_v \simeq k$ (and if so then can this be proved without case-checking). But this refinement was not part of the question, so I'll stop here.

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