Cyclicity Condition for Units of Ring of Integers Modulo n
Theorem
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $\struct {\Z / n \Z, +, \times}$ be the ring of integers modulo $n$.
Let $U = \struct {\paren {\Z / n \Z}^\times, \times}$ denote the group of units of $\struct {\Z / n \Z, +, \times}$.
Then $U$ is cyclic if and only if either:
- $n = p^\alpha$
or:
- $n = 2 p^\alpha$
where $p \ge 3$ is prime and $\alpha \ge 0$.
Proof
Sufficient condition
Let $U$ be cyclic.
Let $n \ge 0$ be an integer.
Let $n = p_1^{e_1} \cdots p_r^{e_r}$, be the decomposition of $n$ into distinct prime powers given by the Fundamental Theorem of Arithmetic.
Then by Chinese Remainder Theorem: Corollary we have an isomorphism:
- $\Z / n \Z \simeq \Z / p_1^{e_1} \Z \times \cdots \times \Z / p_r^{e_r} \Z$
By Units of Ring Direct Product are Ring Direct Product of Units we have:
- $\paren {\Z / n \Z}^\times \simeq \paren {\Z / p_1^{e_1} \Z}^\times \times \cdots \times \paren {\Z / p_r^{e_r} \Z}^\times$
Suppose that $r \ge 2$, and choose $i, j \in \set {1, \ldots, r}$ such that $i \ne j$.
Now $\paren {\Z / n \Z}^\times$ has a subgroup
- $\set {a \in \paren {\Z / n \Z}^\times:
a \equiv 1\bmod p_k^{e_k}\text{ for all } k \ne i, k \in \set {1, \ldots, r}}$
which is isomorphic to $\paren {\Z / p_i^{e_i} \Z}^\times$.
By Subgroup of Cyclic Group is Cyclic, if $\paren {\Z / p_i^{e_i} \Z}^\times$ or $\paren {\Z / p_j^{e_j} \Z}^\times$ is not cyclic, then $\paren {\Z / n \Z}^\times$ cannot be cyclic.
Therefore suppose that $\paren {\Z / p_i^{e_i} \Z}^\times$ and $\paren {\Z / p_j^{e_j} \Z}^\times$ are cyclic.
By Order of Group of Units of Integers Modulo n these groups have orders:
- $\map \phi {p_i^{e_i} }$
and:
- $\map \phi { p_j^{e_j} }$
respectively, where $\phi$ is the Euler $\phi$ function.
By Euler Phi Function of Integer we have:
- $\map \phi {p_i^{e_i} } = p_i^{e_i - 1} \paren {p_i - 1}$
and
- $\map \phi {p_j^{e_j} } = p_j^{e_j - 1} \paren {p_j - 1}$
If $p_i, p_j$ are odd, $2$ divides $p_i - 1$ and $p_j - 1$.
Therefore $2$ divides $\map \phi {p_i^{e_i} }$ and $\map \phi {p_j^{e_j} }$.
In particular, $\map \phi {p_i^{e_i} }$ and $\map \phi {p_j^{e_j} }$ are not coprime.
Now by Group Direct Product of Cyclic Groups, $\paren {\Z / n \Z}^\times$ is not cyclic.
Let $p_i$ or $p_j$ be even.
Without loss of generality, we can assume $p_i = 2$.
Then:
- $\map \phi {p_i^{e_i} } = \map \phi {2^{e_i} } = p_i^{e_i - 1} \paren {p_i - 1}$
So if $e_i \ge 2$, then $2$ divides $\map \phi {p_i^{e_i} }$ and $\map \phi {p_j^{e_j} }$.
In particular $\map \phi {p_i^{e_i} }$ and $\map \phi {p_j^{e_j} }$ are not coprime.
Again by Group Direct Product of Cyclic Groups, $\paren {\Z / n \Z}^\times$ is not cyclic.
Thus if $\paren {\Z / n \Z}^\times$ is cyclic, then $n = 2^e \times p^\alpha$ with $e = 0$ or $e = 1$, $\alpha \ge 0$ and $p \ge 3$ prime.
$\Box$
Necessary Condition
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