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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 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 (not trivially) 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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  • $\begingroup$ kneidell: I suppose you created the tag (subgroup-growth) in this question. Could you please also create at least some short tag-excerpt/tag-wiki to indicate what is the intended meaning of the tag. If some clarification is needed, feel free to drop me line in chat.) $\endgroup$ Commented Nov 9, 2014 at 17:17

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Unless I've misunderstood, you are looking at the subgroup of norm $1$ elements in $D^{\times}$. This is an infinite profinite topological group, and so is not finitely generated.

It is topologically finitely generated. $\mathrm{SL}_1(D)$ is not just a profinite group, but is in fact a compact $p$-adic Lie group. Any such group is topologically f.g., as one can see e.g. by considering the exponential map from its Lie algebra. I think this should be in Serre's lectures on Lie groups and Lie algebras. As a technical point, note that any open subgroup of a compact group has finite index, so you can always check topological finite generation by going to an open subgroup (which you will have to do from the Lie algebra point of view, because the image of the exponential map won't necessarily be the whole 𝑝-adic analytic group --- think about the exponential map from $\mathbb{Z}_p$ to $\mathbb{Z}_p^\times$).

You might want to think about the case of $\mathrm{SL}_2(\mathbb Z_p)$ first, which is a simpler analogue.

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  • $\begingroup$ Yes, of course. I'm sorry- I meant topologically finitely generated. $\endgroup$
    – kneidell
    Commented Sep 27, 2013 at 21:19
  • $\begingroup$ about $SL_2(\mathbb Z_p)$: it has $SL_2(\mathbb Z)$ as a dense subgroup, which is finitely generated. But I don't really know how to generalize this to SL1 of a quaternion algebra, for example :-/.. $\endgroup$
    – kneidell
    Commented Sep 27, 2013 at 21:30
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    $\begingroup$ @kneidell: Dear kneidell, $SL_1(D)$ is not just a profinite group, but is in fact a compact $p$-adic Lie group. Any such group is top. f.g., as one can see e.g. by considering the exponential map from its Lie algebra. I think this should be in Serre's lectures on Lie groups and Lie algebras. There is a theorem of Lazard that characterizes $p$-adic analytic groups as those which contain an open s.g. which is a powerful f.g. pro-$p$-group. (See e.g. Theorem A on p. 1 of this Thesis, although it is the converse part of this ... $\endgroup$
    – Matt E
    Commented Sep 28, 2013 at 12:44
  • $\begingroup$ ... result that is tricky; proving that there is a top. f.g. open s.g. should be just a matter of using the exponential map.) As a technical point, note that any open s.g. of a compact group has finite index, so you can always check top. f.g. by going to an open s.g. (which you will have to do from the Lie algebra point of view, because the image of the exponential map won't necessarily be the whole $p$-adic analytic group --- think about the exponential map from $\mathbb Z_p$ to $\mathbb Z_p^{\times}$). Regards, $\endgroup$
    – Matt E
    Commented Sep 28, 2013 at 12:46
  • $\begingroup$ @kneidell: P.S. Sorry for being pedantic about top. f.g. vs. f.g.! (In stating Lazard's result above I'm using the same abuse of language, which I guess is standard --- although from some reason I've never really used it much myself.) One last remark: when I first started working seriously with $p$-adic Lie groups and such, I found Serre's lecture notes very helpful. The version I had from the library was a big old volume published by Benjamin, I think; I don't know exactly what form they appear in now. The good thing is that they treat the $p$-adic theory somewhat in parallel ... $\endgroup$
    – Matt E
    Commented Sep 28, 2013 at 12:51

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