Proof.  If $u$ and $v$ are two units, then there are the elements $r$ and $s$ of $R$ such that  $ru=ur=1$  and  $sv=vs=1$.  Then we get that $(sr)(uv)=s(r(uv))=s((ru)v)=s(1v)=sv=1$,  similarly  $(uv)(sr)=1$.  Thus also $uv$ is a unit, which means that $E$ is closed under multiplication.  Because  $1\in E$  and along with $u$ also its inverse $r$ belongs to $E$, the set $E$ is a group.

Corollary.  In a commutative ring, a ring product is a unit iff all factors are units.

The group $E$ of the units of the ring $R$ is called the group of units of the ring.  If $R$ is a field, $E$ is said to be the multiplicative group of the field.

Examples

1. 1.

   When  $R=\mathbb{Z}$, then  $E=\{1,\,-1\}$.

2. 2.

   When  $R=\mathbb{Z}[i]$,  the ring of Gaussian integers, then  $E=\{1,\,i,\,-1,\,-i\}$.

3. 3.

   When  $R=\mathbb{Z}[\sqrt{3}]$, then  $E=\{\pm(2\!+\!\sqrt{3})^{n}\,\vdots\,\,\,n\in\mathbb{Z}\}$.

4. 4.

   When  $R=K[X]$  where $K$ is a field, then  $E=K\!\smallsetminus\!\{0\}$.

5. 5.

   When  $R=\{0\!+\!\mathbb{Z},\,1\!+\!\mathbb{Z},\,\ldots,\,m\!-\!1\!+\!\mathbb{Z}\}$  is the residue class ring modulo $m$, then  $E$ consists of the prime classes modulo $m$, i.e. the residue classes $l\!+\!\mathbb{Z}$ satisfying  $\gcd(l,m)=1$.