Exercise 0.3.14

Question

Conclude from the previous two exercises that $(\mathbb{Z}/n\mathbb{Z})^	imes$ is the set of elements $\overline{a}$ of $\mathbb{Z}/n\mathbb{Z}$ with $(a,n) = 1$ and hence prove Proposition 4. Verify this directly in the case $n = 12$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} 
We have proved in Exercise 10 that for all $a \in \Z$,
\begin{align}
 \overline{a} \in ( \Z/n\Z)^	imes \iff a \wedge n = 1.
 \end{align}
so 
$$(\Z/n\Z)^	imes = \{\overline{a} \in \Z/n\Z \mid a \wedge n = 1\}.$$
This proves Proposition 4 (this is also a direct consequence of Exercises 12 and 13).

If $n = 12$, then the elements $a \in [\![ 1, 12[\![$ relatively prime with $12$ are $\{1, 5, 7, 11\}$.
\begin{itemize}
\item These elements are invertible modulo $12$: 
$$1 \cdot 1 \equiv 5 \cdot 5 \equiv 7 \cdot 7 \equiv 11 \cdot 11 \equiv 1 \pmod {12}.$$

\item The others elements $2, 3,4, 6, 8, 9, 10$ satisfy
$$0 \equiv 2 \cdot 5 \equiv 3 \cdot 4 \equiv 4 \cdot 3 \equiv 6 \cdot 2 \equiv 8 \cdot 3 \equiv 9 \cdot 4 \equiv 10 \cdot 6\pmod {12},$$
so are not invertible modulo $12$ by Exercise 12.
\end{itemize}
\end{proof}

Exercise 0.3.14

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Exercise 0.3.14

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