# The Set of Units of a Ring forms a Group under *

Recall from the Units (Multiplicatively Invertible Elements) in Rings page that if $(R, +, *)$ is a ring with multiplicative identity $1$ then an element $a \in R$ is said to be a unit (or a multiplicatively invertible element) in $R$ if there exists an element $b \in R$ such that $a * b = 1$ and $b * a = 1$.

We will now prove an important result which says that given any ring $(R, +, *)$ that the set of units forms a group with respect to the operation $*$.

Theorem 1: Let $(R, +, *)$ be a ring and let $R^{\times}$ be the set of units in $R$. Then $(R^{\times}, *)$ is a group. |

**Proof:**The associativity of $*$ on elements in $(R^{\times}, *)$ is inherited from the associativity of $*$ on elements in $(R, +, *)$.

- Furthermore, the identity element $1 \in R$ is also contained in $R^{\times}$ since $1 = 1 * 1$.

- Let $a \in R^{\times}$. Then there exists an element $b \in R^{\times}$ such that $a * b = 1$ and $b * a = 1$. So $a^{-1} = b$ is the multiplicative inverse for $a$.

- Lastly we show that $(R^{\times}, *)$ is closed under $*$. Let $a_1, a_2 \in R^{\times}$. Then there exists $b_1, b_2 \in R^{\times}$ such that $a_1 * b_1 = 1$, $b_1 * a_1 = 1$, $a_2 * b_2 = 1$ and $b_2 * a_2 = 1$. So:

(1)

\begin{align} \quad (a_1 * a_2) * (b_2 * b_1) = a_1 * 1 * b_1 = 1 * 1 = 1 \end{align}

(2)

\begin{align} \quad (b_2 * b_1) * (a_1 * a_2) = b_2 * 1 * a_2 = 1 * 1 = 1 \end{align}

- So $a_1 * b_1 \in R^{\times}$.

- Hence $(R^{\times}, *)$ is a group. $\blacksquare$

Definition: Let $(R, +, *)$ be a ring. Then the Group of Units for this ring is denoted by $R^{\times}$. |