I'm reading some notes(*) about arithmetic lattices in $\operatorname{SU}(n,1)$. I'm trying to understand the data that classifies (up to commensurability) the arithmetic lattices of the "first type" in this group. Fix a totally real number field $F$ with a totally imaginary quadratic extension $E$. Then these lattices are defined as the integer points of a unitary group preserving a Hermitian form $H$ of signature $(n,1)$ defined over $E$, i.e. $\operatorname{SU}(H;\mathcal O_E)$, which is positive- or negative-definite at all other infinite places.

One invariant you might try to pick up from such a construction is the determinant of $H$ (thought of as an $(n+1) \times (n+1)$ Hermitian matrix), which lies in $F^\times$. But this determinant is not actually invariant under a change of basis. It turns out that the relevant invariant lies in $F^\times/N_{E/F}(E^\times)$. Now here's my confusion: in section 6.6 ("Parity and describing commensurability classes") of the cited notes, the author seems to be claiming that the group $F^\times/N_{E/F}(E^\times)$ is a group with only two elements. He refers to the "trivial and nontrivial class[es]". I can see that all elements of this group have order 2, but I don't see why the group itself has order 2.

Is it true that $F^\times/N_{E/F}(E^\times)$ is a group of order 2?

(*) D.B. McReynolds. *Arithmetic lattices in $\operatorname{SU}(n,1)$*.

(I should point out that these notes are clearly unfinished and may not have been intended for publishing. I don't want to call to question the author's work.)

sometrue assertion there, and could conceivably actually suffice for the issues there. But, you're right... as in @KConrad's answer... the assertion is not literally correct, etc. $\endgroup$