Noncommutative rings with abelian group of units Every commutative ring has an abelian group of units. So, I want to know if the converse holds or when it holds.
My question is: is commutative every ring with its abelian group of units?, and can you give an example of a noncommutative ring with an abelian group of units (if such ring exists)?
 A: Let $k$ be a field and $k[X,Y]$ be the ring of non-commutative polynomials over $k$. The invertible elements are exactly the elements of $k$.
A: My thinking about this question went as as follows:
1) When I think of non-commutativity, one of the first things that comes to my mind are matrices.
2) I want a ring of matrices whose unit group is nice (thinking: finite of small order).
3) What about the non-commutative ring $R\,$ of $2\times 2$ matrices with entries in $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}.$
4) By definition $R^\times = GL_2(\mathbb{F}_2)$ and a simple counting argument shows that $\#GL_2(\mathbb{F}_2) = 6.$
5) There are 2 groups of order 6 up to isomorphism: $C_6$ (abelian) and $S_3$ (non-abelian).
6) A little messing around showed me that $GL_2(\mathbb{F}_2)$ is not abelian :(
7) If $R^\times$ were smaller, then it would have to be abelian because all groups of order $<6$ are abelian.
8) Let's try to find a non-commutative subring of $R.$
9) What about the subring $S$ of all upper triangular matrices.
10) The unit group consists of all upper triangular matrices which have non-zero diagonal entries. 
11) There are exactly two such matrices and so $\#S^\times = 2$ and we're done! :)
I hope this helps :)
