If a,b is in (Z/nZ), then ab is in (Z/nZ)

The Mathematician — Sun, 30 Oct 2022 13:00:07 +0000

If $a,b \in (\mathbb{Z}/n\mathbb{Z})^{\times}$ , then $a,b \in (\mathbb{Z}/n\mathbb{Z})^{\times}$ .

Proof. By definition, if

a,b \in (\mathbb{Z}/n\mathbb{Z})^{\times}

, then

\begin{align*} ac &= 1 \in \mathbb{Z}/n\mathbb{Z} \text{ for some } c \in \mathbb{Z}/n\mathbb{Z} \\ bd &= 1 \in \mathbb{Z}/n\mathbb{Z} \text{ for some } d \in \mathbb{Z}/n\mathbb{Z}. \end{align*}

Obviously, we will get

\begin{align*} 1 = 1 \cdot 1 = ac \cdot bd = ab \cdot cd, cd \in \mathbb{Z}/n\mathbb{Z}, \end{align*}

which implies that

ab \in (\mathbb{Z}/n\mathbb{Z})^{\times}

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