Is $({F^{\times})^{ diag}}\backslash(GL_2 \times E^{\times})_{det=\mathbb{N}}$ a unitary group? Let $F$ be an p-adic field, and $E$ be a quadratic extension of $F$, then is $({F^{\times})^{ diag}}\backslash(GL_2 \times E^{\times})_{det=\mathbb{N}}$ isomorphic to some unitary group $U_{E/F}(2)$? Here, $\mathbb{N}(e)=e *\sigma(e)$, where $\sigma$ is the nontrivial element in $Gal(E/F)$. And $(GL_2 \times E^{\times})_{det=\mathbb{N}} :=\{(g,e)\in GL_2 \times E^{\times}|det(g)=\mathbb{N}(e)\}$.
 A: I cannot answer this question for now. However, I have made some relevant calculations that are too long for a comment.
We set
$$G_0=(GL_2 \times E^{\times})_{{\rm det}={\rm Nm}}\,,\quad G_1=G_0/(F^\times)^{\rm diag},\quad G_2=U_{E/F}(1,-\lambda),$$
where $\lambda\in F^\times$ and  $U_{E/F}(1,-\lambda)$ denotes the unitary group of the diagonal  hermitian form
$$h(z_1,z_2)={\rm Nm}(z_1)-\lambda{\rm Nm}(z_2).$$
We wish to know whether $G_1$ and $G_2$ are isomorphic.
If $\lambda\notin{\rm im}({\rm Nm})$, then the hermitian form $h$ does not represent 0, the topological group $G_2$ is compact, and hence, it is not isomorphic to the noncompact group $G_1$
(this has been already noticed by Paul Broussous).
Therefore, let us assume that $\lambda\in{\rm im}({\rm Nm})$; then we may assume that $\lambda=1$ and $G_2=U_{E/F}(1,-1)$.
Consider the homomorphism
$$\alpha\colon G_0\to F^\times,\quad (g,e)\mapsto {\rm Nm}(e).$$
Clearly, ${\rm im}(\alpha)={\rm im}({\rm Nm})\subset F^\times$.
The homomorphism $\alpha$ induces a homomorphism
$$\beta\colon G_1\to F^\times/F^{\times\,2},$$
and we have
$${\rm im}(\beta)= {\rm im}({\rm Nm})/F^{\times\,2}.$$
We set
$$U_1=\{e\in E\ |\ {\rm Nm}(e)=1\}.$$
We have a canonical homomorphism
$$\gamma\colon{\rm SL}(2,F)\times U_1\to G_1\quad (g,e)\mapsto [g,e],$$
where $[g,e]$ denotes the class of the pair $(g,e)\in G_0$.
Then
$$\ker(\gamma)=\{\pm 1\},\quad {\rm im}(\gamma)=\ker(\beta).$$
Thus we obtain a short exact sequence
$$1\to ({\rm SL}(2,F)\times U_1)/\{\pm 1\}\to G_1\to {\rm im}({\rm Nm})/F^{\times\,2}\to 1.$$
On the other hand, we have a canonical homomorphism
$$\delta\colon {{\rm SU}_{E/F}}(1,-1)\times U_1\to G_2 \quad (s,e)\mapsto se,$$
whose kernel is $\{\pm 1\}$ and whose cokernel is $U_1/U_1^2$, where we write $U_1^2$ for the subgroup of squares in $U_1$.
Thus we obtain a  short exact sequence
$$ 1\to({{\rm SU}_{E/F}}(1,-1)\times U_1)/\{\pm 1\}\to G_2\to U_1/U_1^2\to 1.$$
We have ${{\rm SU}_{E/F}}(1,-1)\simeq{\rm SL}(2,F)$  (see, for instance, the book on classical groups by Jean Dieudonné or the book Geometric Algebra by Emil Artin).
Thus we obtain a short exact sequence
$$ 1\to({\rm SL}(2,F)\times U_1)/\{\pm 1\}\to G_2\to U_1/U_1^2\to 1.$$
Comparing the short exact sequences for $G_1$ and for $G_2$, we see that if the 2-groups ${\rm im}({\rm Nm})/F^{\times\,2}$ and $U_1/U_1^2$ are non-isomorphic (that is, have different orders),
then $G_1$ and $G_2$ are not isomorphic "in a nice way".
I do not know whether the 2-groups ${\rm im}({\rm Nm})/F^{\times\,2}$ and $U_1/U_1^2$ are isomorphic or not. I suggest for OP to ask this on MathOverflow.
