Wedderburn's little theorem
Noncommutativity is hard to sustain in a small space. You might hope to build a tiny finite number system where multiplication is genuinely lopsided yet you can still divide by everything nonzero. Wedderburn's little theorem says you cannot: any such finite system is forced to be commutative. Finiteness, all by itself, irons out the noncommutativity. Every finite division ring is secretly a field.
Precisely: every finite division ring is commutative, hence is a finite field. Combined with the classification of finite fields, this means a finite division ring of order q is the unique field with q elements, where q is a prime power. So there are no finite skew fields at all — the whole phenomenon of noncommutative division requires infinitely many elements.
The classic proof is a gem of the interplay between groups and number theory. One writes the class equation for the multiplicative group of the division ring, expresses the sizes of the conjugacy classes using the centralizers, and observes that the cyclotomic polynomial Phi_n evaluated at q must divide both q^n - 1 and each (q^n - 1)/(q^d - 1) term. A divisibility estimate on the value of Phi_n at q then forces n = 1, meaning the center is everything.
It is worth savoring how sharply finiteness is used. The same statement is false without it: the infinite quaternions are a noncommutative division ring. And the theorem has geometric echoes — by a theorem of Wedderburn and others, every finite projective plane coordinatized by a division ring is Desarguesian and Pappian, because the coordinate division ring is forced to be a commutative field.
A finite division ring of order 8 must be the field with 8 elements, F_8 = F_2[x]/(x^3 + x + 1), which is commutative. There is no noncommutative 8-element division ring — Wedderburn forbids it.
Any order forces commutativity: only the finite field exists.