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Be $R, +, \times$ a Ring with multiplicative identity $1$ and $\#R > 1$. Then, show that $U(R), \times$ is a Group.

For $U(R), \times$ to be a Group, it has to satisfy four conditions: $U(R)$ must have closure under $\times$, $\times$ must be associative in $U(R)$, there must be a neutral element $\in U(R)$ and each element $\in U(R)$ must have an inverse $\in U(R)$.

associativity: $\times$ is associative on $U(R)$ because $\times$ is associative on $R$ (it's a Ring).

neutral element: It is $1$.

inverse: This follows trivially from the definition of $U(R)$.

I don't know how to show closure. My textbook says that it is easily shown that if $u_1, u_2 \in U(R)$ then $u_1 \times u_2 \in U(R)$ as well. It provides $(u_1 \times u_2)^{-1} = u_2^{-1} \times u_1^{-1}$ as a hint.

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The hint tells you that the product of two units is a unit by providing a concrete inverse. That proves closure.

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