I was reading a paper recently concerning a non-commutative version of the matrix determinant. On the third page, it stated a fact without providing a proof or a reference:
If $D$ is a division ring, let $D^\times$ be its multiplicative group, then the abelianisation of $D^\times$, $\frac{D^\times}{[D^\times,D^\times]}$, is isomorphic to $Z(D)^\times$ - the centre of $D^\times$.
This result doesn't seem at all trivial, although of course it holds for any commutative field. I had considered that perhaps the natural map $Z(D)^\times\to \frac{D^\times}{[D^\times,D^\times]},c\mapsto c [D^\times,D^\times]$ could be proved to be an isomorphism. But I can see no reason why this should be be the case.
Is this result really true? If so, does anyone have any idea why? Is the natural map an isomorphism, or is there some other non-canonical isomorphism? Otherwise, can anyone think of a counterexample?
This seems to be false.
An counter-example would be the division algebra $Q_\mathbb{R}$ of real quaternions. The center of $Q_\mathbb{R}$ is the field of real numbers $\mathbb{R}$, and the derived group $[Q_\mathbb{R}^\times, Q_\mathbb{R}^\times]$ is the group of elements $w=a+bi+cj+dk$ with $n(w):=a^2+b^2+c^2+d^2 = 1$ (a proof of this can be found here). Since $n$ is multiplicative, we get that $$ Q_\mathbb{R}^\times/[Q_\mathbb{R}^\times, Q_\mathbb{R}^\times] \cong \mathbb{R}^\times_{>0}. $$
This isomorphism is proved on page 12 of this Intelligencer article of Helmer Aslaksen (with thanks to @rschwieb for pointing out this reference).
However, $Z(Q_\mathbb{R})^\times = \mathbb{R}^\times$. But $\mathbb{R}^\times$ is not isomorphic to $\mathbb{R}^\times_{>0}$ (the former contains two elements that are their own inverses, and the latter only one).
Therefore $Q_\mathbb{R}^\times/[Q_\mathbb{R}^\times, Q_\mathbb{R}^\times]$ is not isomorphic to $Z(Q_\mathbb{R})^\times$.