We want to show that the multiplication operation in a finite division ring is abelian.

We denote the centralizer in $D$ of an element $x$ as $C_D(x)$ .

Lemma. The centralizer is a subring.

$0$ and $1$ are obviously elements of $C_D(x)$ and if $y$ and $z$ are, then $x(-y) = -(xy) = -(yx) = (-y)x$ , $x(y+z)=xy+xz=yx+zx=(y+z)x$ and $x(yz)=(xy)z=(yx)z=y(xz)=y(zx)=(yz)x$ , so $-y,y+z$ , and $yz$ are also elements of $C_D(x)$ . Moreover, for $y\neq 0$ , $xy=yx$ implies $y^{-1}x=xy^{-1}$ , so $y^{-1}$ is also an element of $C_D(x)$ .

Now we consider the center of $D$ which we'll call $Z(D)$ . This is also a subring and is in fact the intersection of all centralizers.

$$ Z(D)=\bigcap_{x\in D}C_D(x) $$

$Z(D)$ is an abelian subring of $D$ and is thus a field. We can consider $D$ and every $C_D(x)$ as vector spaces over $Z(D)$ of dimension $n$ and $n_x$ respectively. Since $D$ can be viewed as a module over $C_D(x)$ we find that $n_x$ divides $n$ . If we put $q:=|Z(D)|$ , we see that $q\geq 2$ since $\{0,1\}\subset Z(D)$ , and that $|C_D(x)|=q^{n_x}$ and $|D|=q^n$ .

It suffices to show that $n=1$ to prove that multiplication is abelian, since then $|Z(D)|=|D|$ and so $Z(D)=D$ .

We now consider $D^*:=D-\{0\}$ and apply the conjugacy class formula.

$$ |D^*| = |Z(D^*)| + \sum_{x} [D^*:C_{D^*}(x)] $$

which gives

$$ q^n-1 = q-1 + \sum_{x} \frac{q^n-1}{q^{n_x}-1}$$ .

By Zsigmondy's theorem, there exists a prime $p$ that divides $q^n-1$ but doesn't divide any of the $q^m-1$ for $0<m<n$ , except in 2 exceptional cases which will be dealt with separately. Such a prime $p$ will divide $q^n-1$ and each of the $\frac{q^n-1}{q^{n_x}-1}$ . So it will also divide $q-1$ which can only happen if $n=1$ .

We now deal with the 2 exceptional cases. In the first case $n$ equals $2$ , which would mean $D$ is a vector space of dimension 2 over $Z(D)$ , with elements of the form $a+b\alpha$ where $a,b\in Z(D)$ . Such elements clearly commute so $D=Z(D)$ which contradicts our assumption that $n=2$ . In the second case, $n=6$ and $q=2$ . The class equation reduces to $64-1=2-1+\sum_{x} \frac{2^6-1}{2^{n_x}-1}$ where $n_x$ divides 6. This gives $62=63x+21y+9z$ with $x,y$ and $z$ integers, which is impossible since the right hand side is divisible by 3 and the left hand side isn't.