Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ be a ring and $A^\times$ be the collection of unit elements of $A$. If $A$ is a commutative ring, then $A^\times$ is a commutative group. Conversely, if $A^\times $ is a commutative group, does $A$ necessarily be a commutative ring? Is there any counterexample?

share|cite|improve this question
up vote 11 down vote accepted

Let $K$ be a field. Consider the free $K$-algebra on two generators $x,y$. Its unit group is $K^*$, hence commutative, but $x,y$ don't commute.

Here is a more explicit example: Consider the ring of upper-triangular $2 \times 2$-matrices over $\mathbb{F}_2$. It has $8$ elements and it is in fact the smallest noncommutative ring. The unit group has just two elements, namely the identity matrix and $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. A group of order $2$ is commutative.

share|cite|improve this answer

In order for ring $(A, +, \cdot)$ to be a commutative ring, the group $(A, \cdot)$ has to be commutative.

share|cite|improve this answer
$(A,\cdot)$ is not always a group although it is a semigroup. And this solution is fine if "group" is replaced with "semigroup." – rschwieb Jan 28 '14 at 13:17
Hi new user! $$\color{red}{\Large\text{Welcome to MSE!}}$$ Don't worry about it now (since you're new) but you might like to know that we prefer MathJax here :) – Shaun Jan 28 '14 at 13:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.