Let $F$ be the non-archimedean local field $\mathbb{Q}\_p$ for some prime $p$ and $D$ be a quaternion division algebra over $F$. Let $\mathcal{O}\_D$ and $\mathcal{P}\_D$ denote the ring of integers of $D$ and its unique maximal ideal (respectively). Then, what is the finite group
$$ \frac{D^\*}{F^\*(1+ \mathcal{P}\_D)} = {?}$$ where $D^\*=D-\{0\}$ and $F^\*=F-\{0\}$ are multiplicative groups.


Consider the reduced norm map $N\_\text{rd}:D \rightarrow F$, then $N\_\text{rd}(D^\*)=F^\*$ and if $D^1$ denotes the reduced norm one elements of $D$, then we have an exact sequence
$$1 \rightarrow D^1 \rightarrow D^\* \rightarrow F^\* \rightarrow 1$$ but we have $D^1 \cap F^\*=\{\pm 1\}$. We know from Carl Riehm's article that $$ \frac{D^1}{(1+ \mathcal{P}\_D)} \cong {\_N}(\mathbb{F}\_{p^2})= \text{Finite cyclic group of order } (p+1).$$
Here ${\_N}(\mathbb{F}\_{p^2})$ is the subgroup of $\mathbb{F}\_{p^2}$ consisting of norm 1 elements.


Question: Similarly, can we write $ \frac{D^\*}{F^\*(1+ \mathcal{P}\_D)}$ in terms of finite fields?


Any comments or suggestions will be extremely helpful.



What is the quotient group $D^*/{F^*(1+P_D)}$ for a quaternion division algebra $D$ over a local field $F$?
Yes, we can.$\newcommand{\order}{\mathcal{O}}$ $\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\prim}{\mathcal{P}}$ $\newcommand{\F}{\mathbb{F}}$


First, let me remind you of the following explicit description of $\order\_D$. I won't use it explicitly but it is convenient to check some of my claims below. Let $\pi$ be a uniformiser of $F$ and $L/F$ the unique unramified quadratic extension with Galois group generated by $\sigma$. We have
$$\order\_D = \order\_L + \order\_L \Pi$$
where $\Pi^2 = \pi$ and $\Pi \lambda = \sigma(\lambda) \Pi$ for $\lambda\in L$
(you also trivially get $D$ by extension of scalars).
Let $v$ denote the normalised valuation on $D$, so that $v(\Pi) = 1$ (and $v(\pi) = 2$).


The valuation induces an isomorphism $D^\times / \order\_D^\times \cong \Z$, and therefore $D^\times / F^\times\order\_D^\times \cong \Z/2\Z$.
Since $\order\_D^\times \cap F^\times = \order\_F^\times$ we get an exact sequence
$$ 1 \to \order\_D^\times/\order\_F^\times(1+\prim\_D) \to D^\times/F^\times(1+\prim\_D) \to \Z/2\Z \to 1$$
and this sequence is split by the existence of the element $\Pi$ since $\Pi^2 = \pi \in F^\times$.


Since $\order\_D^\times/(1+\prim\_D) \cong \F\_{q^2}^\times$ and the image of $\order\_F^\times$ is $\F\_q^\times$, we obtain
$$ \order\_D^\times/\order\_F^\times(1+\prim\_D) \cong \F\_{q^2}^\times/\F\_q^\times.$$ Moreover, the action of the nontrivial element of $\Z/2\Z$ is via conjugation by $\Pi$, which is the same as the action of the Frobenius automorphism $x\mapsto x^q$.


We therefore obtain
$$ D^\times/F^\times(1+\prim\_D) \cong \F\_{q^2}^\times/\F\_q^\times \rtimes \mathrm{Gal}(\F\_{q^2}/\F\_q) \cong C\_{q+1} \rtimes C\_2,$$
where in the first semidirect product the action is the natural one, and in the second one the nontrivial element of $C\_2$ acts by inversion on $C\_{q+1}$.


