Exercise 0.3.11 ($(\mathbb{Z}/n\mathbb{Z})^ imes$ is closed under multiplication)

Question

Prove that if $\overline{a}, \overline{b} \in (\mathbb{Z}/n\mathbb{Z})^	imes$, then $\overline{a}\cdot \overline{b} \in (\mathbb{Z}/n\mathbb{Z})^	imes$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} 
By definition, if $\overline{a}, \overline{b} \in (\mathbb{Z}/n\mathbb{Z})^	imes$, then there are classes $ \overline{c}, \overline{d} \in \mathbb{Z}/n\mathbb{Z})^	imes$ such that 
$$\overline{a} \cdot  \overline{c} = \overline{1},\qquad \overline{b} \cdot \overline{d} = \overline{1}.$$
Then 
$$(\overline{a}\, \overline{b}) \cdot (\overline{c}\, \overline{d}) = (\overline{a} \cdot  \overline{c} ) \cdot (\overline{b} \cdot \overline{d}) = \overline{1}.$$
So $\overline{a}\cdot \overline{b}$ has an inverse in $\mathbb{Z}/n\mathbb{Z}$. This shows that
$$\overline{a}\cdot \overline{b} \in  (\mathbb{Z}/n\mathbb{Z})^	imes.$$

\end{proof}

Exercise 0.3.11 ($(\mathbb{Z}/n\mathbb{Z})^\times$ is closed under multiplication)

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Exercise 0.3.11 ($(\mathbb{Z}/n\mathbb{Z})^\times$ is closed under multiplication)

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