group of units
Theorem.
The set $E$ of units of a ring $R$ forms a group with respect to ring multiplication.
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 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.
When $R=\mathbb{Z}$, then $E=\{1,1\}$.

2.
When $R=\mathbb{Z}[i]$, the ring of Gaussian integers^{}, then $E=\{1,i,1,i\}$.

3.
When $R=\mathbb{Z}[\sqrt{3}]$, then (http://planetmath.org/UnitsOfQuadraticFields) $E=\{\pm {(2+\sqrt{3})}^{n}\mathrm{\vdots}n\in \mathbb{Z}\}$.

4.
When $R=K[X]$ where $K$ is a field, then $E=K\setminus \{0\}$.

5.
When $R=\{0+\mathbb{Z},\mathrm{\hspace{0.17em}1}+\mathbb{Z},\mathrm{\dots},m1+\mathbb{Z}\}$ 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 $\mathrm{gcd}(l,m)=1$.
Title  group of units 
Canonical name  GroupOfUnits 
Date of creation  20130322 14:41:32 
Last modified on  20130322 14:41:32 
Owner  pahio (2872) 
Last modified by  pahio (2872) 
Numerical id  24 
Author  pahio (2872) 
Entry type  Theorem 
Classification  msc 16U60 
Classification  msc 13A05 
Synonym  unit group 
Related topic  CommutativeRing 
Related topic  DivisibilityInRings 
Related topic  NonZeroDivisorsOfFiniteRing 
Related topic  PrimeResidueClass 
Defines  group of units of ring 
Defines  multiplicative group of field 