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.