I know that the units of a ring form a multiplicative group because it inherits the associativity from the ring; every element has its inverse by definition; it is closed under multiplication: indeed, if you take $x,y \in U(R)$ there exists $x',y' \in U(R)$ such that
$$
(x \cdot y) \cdot (y' \cdot x') = x \cdot (y \cdot y') \cdot x' = x \cdot x = 1 \implies x \cdot y, y' \cdot x' \in U(R);
$$
And it has the identity element since it itself is a unit ($1 \cdot 1 = 1$). The issue I'm having is to prove that it is abelian, and I suspect that it isn't always true.
Does anyone have a counter example or a proof for this statement?
Best Answer
No. For example multiplication in Hamilton's quaternions $\mathbb H$ is not commutative, so its group of units (which is everything except $0$) is not Abelian.
You can embed whatever group you want into the units of a ring using a construction called a group ring, and in particular you could put in any nonAbelian group that you wanted.