What is the quotient group $D^*/{F^*(1+P_D)}$ for a quaternion division algebra $D$ over a local field $F$? 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.
 A: To give an alternative answer, let us first recall  the article "Construction of Locally Compact Near-Fields from $p$-Adic
Division Algebras" by Detlef Groger:
Fix a prime element $\pi_F$ of $F$. Then $D$ is generated as a non-commutative $F$-algebra by an unramified extension $E/F$ of degree 2 and an element $\pi$ with $\pi^2=\pi_F$. we consider $\varpi$ for a (fixed) primitive $(p^2  -1)^\text{th}$ root of unity in $E$ and $ \varpi_F=\varpi^\frac{p^2  -1}{p-1} \in F$.
Suppose $U_F$, $U_D$ denote the groups of units in $F$ and $D$ with $U^1_F := 1 + P_F$ and $U^1_D := 1 + P_D$.
Put $C = \langle \varpi, \pi \rangle$ and $C_F=C \cap F^* = \langle \varpi_F, \pi_F \rangle$. Then, $C$ is a complement of $U^1_D$ in the semidirect product $D^*=U^1_D \rtimes C$, whereas the product $F^*=U^1_F \times C_F$ is direct product. Therefore, $F^* U^1_D= U^1_D \times C_F$  and,
$$\frac{D^*}{F^*(1+P_D)} \cong C/C_F \cong \langle\overline{\varpi}\rangle \rtimes \langle\overline{\pi}\rangle \cong Z_{p+1} \rtimes Z_2 .$$
A: 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 $C_{q+1}$.
