Let $A$ be a finite dimensional semisimple $F$-algebra and $K\_1(A)$ the Whitehead group of $A$.


By splitting $A$ into its Wedderburn components, the reduced norm map $\operatorname{nr}\_A:K\_1(A)\to Z(A)^\times$ can be defined, where $Z(A)$ is the centre of $A$.


I've read in the literature that if $G$ is a finite group of *odd* order then the norm map $\operatorname{nr}\_{F[G]}$ is *bijective*.


Is there a simple proof of this?



Why is $\operatorname{nr}_{F[G]}:K_1(F[G])\to Z(F[G])^\times$ a bijection?
I am assuming that you meant $F$ to be a number field.
Let $A = M\_n(D)$ where $D$ is a finite dimensional division algebra over $\mathbb{Q}$ and let $Z = Z(A) = Z(D)$. Then $K\_1(A)\cong D^\times/[D^\times,D^\times] = D^\times/SL\_1(D)$. By Eichler's theorem, if $A$ is not ramified at any real place of $Z$, then the reduced norm map $\mathrm{nr} : D^\times \to Z^\times$ is surjective. Now if $G$ is a finite group of odd order, then the degree of the Wedderburn components of $F[G]$ are odd (because they always divide the order of $G$), so they cannot be ramified at real places. This proves the result.


