Multiplicative Group of a Ring
Definition
For any ring with identity \(R\), \(R^\ast\) denotes the set of units in \(R\).
Theorem
\(R^\ast\) is a group under multiplication.
Proof
This proof is fairly simple, and really comes down to proving the a product of units is a unit.
First note that \(1 \in R\) is a unit, since \(1 \cdot 1 = 1 \implies 1 = 1^{-1}\).
Now, for \(a, b \in R^{\ast}\) we have that:
\[\begin{align*}
(ab)(b^{-1}a^{-1}) &= a(b b^{-1}) a^{-1} \\
&= a(1)a^{-1} \\
&= aa^{-1} \\
&= 1 \\
\end{align*}\]
hence \((ab)^1\) exists and is equal to \((b^{-1}a^{-1}\) (note that showing it is a left inverse is basically identical to the above).
Now, \(a^{-1} \in R^{\ast}\) since \(a^{-1}a = 1 \implies (a^{-1})^{-1} = a\).
Finally, associativity is inherited from the properties of \(R\).