> [!NOTE] **Theorem** (Unit Group is Group) > > Let $(R,+,\times)$ be a [[Rings|ring]]. Then its [[Unit Group of Ring|unit group]] is a [[Groups|group]] under multiplication. > **Proof.** (*Closure*) Suppose $u_{1}, u_{2} \in R$. Since they are units there are $v_{1}, v_{2} \in R$ such that $u_{1}v_{1}=v_{1}u_{1} = 1, \quad u_{2}v_{2} = v_{2}u_{2}=1.$WTS $u_{1}u_{2}$ is a unit. Note that $v_{2}v_{1} \in R$ since $R$ is a ring. We have $\begin{align}(u_{1}u_{2}) (v_{2}v_{1}) &= u_{1}(u_{2} v_{2}) v_{1} & \text{associativity of product} \\ &= u_{1} \cdot 1 \cdot v_{1} & \text{since $u_{2}v_{2}=1$} \\ &= 1 &\text{since $u_{1}v_{1}=1$} \end{align}$Similarly $(v_{2}v_{1})(u_{1}u_{2}) =1$ which shows that $u_{1}u_{2}$ is a unit in $R$, so $u_{1}u_{2} \in R^{*}.$ (Associativity) Multiplication on $R^{*}$ is associative since $R^{* }\subset R$ and multiplication on $R$ is associative as it is a ring. (Identity element) $1.1=1$, is a unit and so $1 \in R^{*}$ (Inverses) Suppose $u \in R^{*}$. Then $uv = vu =1$ for some $v \in R$. Note that this makes $v$ a unit as well. Thus $u$ has a multiplicative inverse in $R^{*}.$