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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?

Shiquan
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2 Answers2

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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.

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

Shaun
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user124441
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    $(A,\cdot)$ 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
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    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