Exercise 4.4.15 ( $(\mathbb{Z}/5\mathbb{Z})^ imes,\ (\mathbb{Z}/9\mathbb{Z})^ imes$ and $(\mathbb{Z}/18\mathbb{Z})^ imes$ are cyclic)

Question

Prove that each of the following (multiplicative) group is cyclic: $(\mathbb{Z}/5\mathbb{Z})^	imes,\ (\mathbb{Z}/9\mathbb{Z})^	imes$ and $(\mathbb{Z}/18\mathbb{Z})^	imes$.

Richard Ganaye · Accepted Answer

\begin{proof} Let $[a]_n$ denote the class of $a$ modulo $n$.
\begin{itemize}
\item $\overline{2} = [2]_5$ is a generator of $(\mathbb{Z}/5\mathbb{Z})^	imes$:
$$\{2^k \mod 5 \mid 0 \leq k <4\} = \{1,2,4,3\}.$$
Since $|(\mathbb{Z}/5\mathbb{Z})^	imes| = \varphi(5) = 4 = |\{2^k \mod {5}\mid 0 \leq k <4 \} |$, $\overline{2}$ is a generator of $(\mathbb{Z}/5\mathbb{Z})^	imes$, so
$$(\mathbb{Z}/5\mathbb{Z})^	imes = \langle [2]_5 angle$$
is cyclic.

\item $2$ is a generator of $(\mathbb{Z}/9\mathbb{Z})^	imes$:
$$\{2^k \mod {9} \mid 0 \leq k <6\} = \{1, 2, 4, -1, -2, -4\}.$$
Since $|(\mathbb{Z}/9\mathbb{Z})^	imes| = \varphi(9) = 6 = |\{2^k \mod {9} \mid 0 \leq k \} |$, $\overline{2}$ is a generator of $(\mathbb{Z}/9\mathbb{Z})^	imes$, so
$$(\mathbb{Z}/9\mathbb{Z})^	imes = \langle [2]_9 angle$$
is cyclic.

\item $11$ is a generator of $(\mathbb{Z}/18\mathbb{Z})^	imes$ (the other generator is $5$):
$$\{11^k  \mod {18} \mid 0 \leq k <6\} = \{1, 11, 13, -1, -11, -13\}.$$
Since $|(\mathbb{Z}/18\mathbb{Z})^	imes| = \varphi(18) = 6 = |\{11^k \mod {18}\mid 0 \leq k \} |$, $\overline{11}$ is a generator of $(\mathbb{Z}/18\mathbb{Z})^	imes$, so
$$(\mathbb{Z}/18\mathbb{Z})^	imes = \langle [11]_{18} angle$$
is cyclic.
\end{itemize}
\end{proof}
With sagemath:
\begin{verbatim}
sage: [Integers(n).multiplicative_generator() for n in [5,9,18]]
[2, 2, 11]
sage: n = 5; a = Mod(2,n); [a^k for k in range(euler_phi(n))]
[1, 2, 4, 3]
sage: n = 9; a = Mod(2,n); [a^k for k in range(euler_phi(n))]
[1, 2, 4, 8, 7, 5]
sage: n = 18; a = Mod(11,n); [a^k for k in range(euler_phi(n))]
[1, 11, 13, 17, 7, 5]
\end{verbatim}

Exercise 4.4.15 ( $(\mathbb{Z}/5\mathbb{Z})^\times,\ (\mathbb{Z}/9\mathbb{Z})^\times$ and $(\mathbb{Z}/18\mathbb{Z})^\times$ are cyclic)

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Exercise 4.4.15 ( $(\mathbb{Z}/5\mathbb{Z})^\times,\ (\mathbb{Z}/9\mathbb{Z})^\times$ and $(\mathbb{Z}/18\mathbb{Z})^\times$ are cyclic)

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