 (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 (ab
(i.e $ \dim_F D = n^2 $). 
 It is known that $ D$ contains an unramified extension $ E$ of $ F$, such that $ east / F$ is cyclic Galois. It take after (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 limitation 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$ matrix over $ E$, via the left regular military 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).$$ 
 Once this is done, one can specify the reduced norm on $ D$ by $ $ Nrd_{D / F}(x)=\det\lambda_x.$$ 
 My pastime specificaly is in the group $ 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 specify 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 matrix with $ 1$s along the diagonal and at most one additional non - zero entry), it is not difficult 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 generate 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 awkward.. ): 
 We name a group $ G$ positively finitely generate (ab
 (i.e $\dim_F D=n^2$). 
It is known thau $D$ contains an uueamifixd extehsion $E$ uf $F$, such that $E/F$ is cyclic Jalous. It follows (not trivially) that $D$ clntains qn eowment $u\in V^\fimes$, akb thaf thexe is some $\pi\in N$ such that $D=\bigoplus_{j=0}^{n-1}u^j E$, $u^u=\pi$ and such that the restriction of the mal $d\mapsto u^{-1}xu$ to $E$ ix a gsnerator of the Galois group $\mathbr{G}(E/F)$.
Thuv, $D$ embeds inyo the $F$ algebra $M_n(E)$ of $n\tlmes n$ matrices over $E$, via the ledt rqtular action $x\mapsto \lambda_x$ where $\lambda_x:E^n\to E^n$ is defined by $$\lxmbda_r(y_0,\ldots,y_{n-1})=x\ceor(\sul_{b=0}^{n-1}u^jy_j).$$
Once vhis if done, one can defina the rrduced norm on $D$ uy $$Ned_{D/F}(x)=\det\lambda_x.$$
My intxrest specificaly is in the gsobps $G:=SL_1(D):=\lbrace x\in D\mud Nrd_{D/X}(x)=1\rbsace$. U'm gryjnj tk undegstend whether this group is finitely generaued ir not. 
It is dlear ehwt the embedding defined above maps $G$ itto the group $SL_n(E)$. Once we use the fact that thg group $SL_g(E)$ is generated by transvections (i.e matrices with $1$s alkvg uhc auahonal and at most one additional non-zero entrr), iu ix not hard to mse the local struvtkrr of $E$ to provg that $SL_n(S)$ is finitely genegated. 
Ugfortynately, a subbroup of a finitely generatwd group neeb nit be finitely genzrated in geueral. Of anuone knows of a criteriun tgat might pgove why ffis case is speckal tvat would be a great help.
Og additioial ppssibilkty, os to tse some hfavy bmols from asymptotlc grlu[ theory (egen though I must admit this tucis out a bit vukbegsome..):
We ccll a nroup $G$ positivqly finitely ggnerate (ay
(i.e $\dim_F D=n^2$). It is known that an extension $E$ $F$, such that follows trivially) that $D$ an element $u\in and that there is some $\pi\in such that $D=\bigoplus_{j=0}^{n-1}u^j E$, $u^n=\pi$ and such that the restriction of the map u^{-1}xu$ to $E$ is a generator of the Galois group $\mathbf{G}(E/F)$. Thus, $D$ into $F$ $M_n(E)$ $n\times n$ matrices over $E$, via the left regular action $x\mapsto \lambda_x$ where $\lambda_x:E^n\to E^n$ is 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 the reduced norm on by $$Nrd_{D/F}(x)=\det\lambda_x.$$ My interest specificaly in groups $G:=SL_1(D):=\lbrace D\mid I'm to understand whether group is finitely generated or not. It is clear that the embedding defined above maps $G$ into group $SL_n(E)$. use the that group is generated by matrices with $1$s along the diagonal one additional non-zero entry), it is not hard use the structure of $E$ to prove that is finitely generated. Unfortunately, a subgroup of a generated group need not be finitely generated in general. If anyone knows of a criterion prove why this case special that would a help. additional is to some heavy tools from asymptotic group theory (even though I must this turns out a bit cumbersome..): We call a group finitely (ab
 (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 (ab
 (i.e $\dim_F D=n^2$). 
It  is knownthat$D$  co nt ains  anunramified ext e nsio n $E$ of $F$, such tha t $E/ F$ is c y cl ic Ga lois. I t  f o l low s(n ottr i vi ally)  th at $D$contains a n e le ment $u\in D ^ \t imes$, and  th at there issom e $\pi \i n F $  such  th at $D =\bigo p lus_{j =0}^{n-1} u^ j  E$, $ u ^n=\pi$ a nd  suc h that the restri c ti o n of the map $ x\maps to u^ { - 1}x u$to $E$ isagener a tor oft he G a loi s  group $\math bf{G}(E/F)$ . 
Th us, $D $emb e ds int o the  $ F $ a lgebra $M_n (E)$  of $n\ti mes n$ matrice s  over $ E$, vi a t heleft re gu lar  a c tio n  $ x\m a pst o \lambd a_ x$  wher e $\ l a m b da_x :E^ n\to  E^n$  is defined b y $ $\la m bda _x(y_ 0,\ld ots, y_ {n-1} )=x\cd ot(\s um _{j=0}^{n-1}u^j y_j) .$$
Oncethi sisdo ne, o n e candef ine  the re duced n o rmon $ D $by $$Nrd_{D/F}(x)= \d e t \l ambda_x. $$
Myi nt er e st speci fi cal y is i n the  gro u ps  $G:=SL_ 1(D):= \ lb ra ce x\in  D \mid N rd _{D /F} (x)=1 \ rbra ce$. I 'm tryin g tou nderstand whet h er this group is f in i tely  ge nerated ornot. 
Itis c l ea r t h at th e emb ed d in g  defined above maps  $ G$ int o the  group $SL_n( E)$. Oncew e use thefact th a t the group $S L_n(E )$ is gene r ated bytrans vections  (i.e mat r i ces with  $1 $salo ngt h ediagonal anda t  mos tone add iti onal no n-z ero  en try ),  it is no t hard t ous eth e l ocals tructure  o f $ E$  to  prov e  that$SL_n (E)$  i sf ini tely ge n er a t ed.
U nf ortu nat el y, asubg r oup  of a f initely g ene r ated  g ro up need  not be finit el y generate dingenera l .  If anyo ne knows of a criterion that mi ght  prov e wh y this ca seis spe cia l  thatwouldbe agr eat h elp.O n  a ddi ti onal possi b i lit y, is  t o us e someheavy tools from a s ymp totic group t heo ry ( e v en  th o ug h  Imu s t a d m it this turns o ut a bit c um b er some..):
W e  ca ll  a grou p $G$ p ositi v ely fin itely gen erate (ab 
 (i.e_$\dim_F D=n^2$)._
It is known that_$D$ contains_an_unramified extension_$E$_of $F$, such_that $E/F$ is_cyclic Galois. It follows_(not trivially) that_$D$_contains an element $u\in D^\times$, and that there is some $\pi\in F$ such that_$D=\bigoplus_{j=0}^{n-1}u^j_E$, $u^n=\pi$_and_such_that the restriction of the_map $x\mapsto u^{-1}xu$ to $E$_is a_generator of the Galois group $\mathbf{G}(E/F)$.
Thus, $D$ embeds_into_the $F$ algebra_$M_n(E)$ of $n\times n$ matrices over $E$, via the_left regular action $x\mapsto \lambda_x$ where_$\lambda_x:E^n\to E^n$ is_defined_by_$$\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 (ab