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

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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

We need to prove $\paren {\Z / p^\alpha \Z}^\times$ and $\paren {\Z / 2 p^\alpha \Z}^\times$ are cyclic.

By Chinese Remainder Theorem: Corollary we have an isomorphism:

$\Z / 2 p^\alpha \Z \simeq \Z / 2 \Z \times \Z / p^\alpha \Z$

By Units of Ring Direct Product are Ring Direct Product of Units we have:

$\paren {\Z / 2 p^\alpha \Z}^\times \simeq \paren {\Z / 2 \Z}^\times \times \paren {\Z / p^\alpha \Z}^\times$

But $\paren {\Z / 2 \Z}^\times$ is trivial group, we have:

$\paren {\Z / 2 p^\alpha \Z}^\times \simeq \paren {\Z / p^\alpha \Z}^\times$

So we only need to prove $\paren {\Z / p^\alpha \Z}^\times$ is cyclic.

Case $\alpha = 1$ follows from Ring of Integers Modulo Prime is Field and Group of Units of a Finite Field is Cyclic.

Next we handle $\alpha \ge 2$.

The case $\alpha = 2$ is proved in Group of Units Ring of Integers Modulo $p^2$ is Cyclic.

Let $g$ be a generator of $\paren {\Z / p^2 \Z}^\times$,

then the order of $g$ in $\paren {\Z / p^2 \Z}^\times$ is $p \paren{p - 1}$.

We will prove that $g$ generates $\paren {\Z / p^\alpha \Z}^\times$.

Write $\map {\text{ord}_{p^\alpha}} g$ as the order of $g$ in $\paren {\Z / p^\alpha \Z}^\times$. Note that

$g^{\map {\text{ord}_{p^\alpha}} g} \equiv 1 \pmod{p^\alpha}$

and $\alpha \ge 2$, so

$g^{\map {\text{ord}_{p^\alpha}} g} \equiv 1 \pmod{p^2}$

so

$p \paren{p - 1} \divides \map {\text{ord}_{p^\alpha}} g$.

Also, by Lagrange's Theorem,

\(\text {(⋆)}: \quad\)

\(\ds \map {\text{ord}_{p^\alpha} } g\)

\(\divides \order {\paren {\Z / p^\alpha \Z}^\times}\)

\(\ds \)

\(\ds = p^{\alpha - 1} \paren{p - 1}\)

Thus

$\map {\text{ord}_{p^\alpha}} g = p^{\map e \alpha} \paren{p - 1}$ for some $\map e \alpha$, $1 \leq \map e \alpha \leq \alpha - 1$.

To show $g$ generates $\paren {\Z / p^\alpha \Z}^\times$, it suffices to show that

$\map e \alpha = \alpha - 1$

but $\map e \alpha \leq \alpha - 1$, so it suffices to show that

$\map e \alpha \nleq \alpha - 2$

in other words

\(\text {(1)}: \quad\)

\(\ds g^{p^{\alpha - 2} \paren{p - 1} }\)

\(\not \equiv\)

\(\ds 1 \pmod{p^\alpha}\)

We will prove this by induction on $\alpha$.

Induction Hypothesis

(1) is true for all $\alpha \ge 2$.

Basis for the Induction

By definition, the order of $g$ in $\paren {\Z / p^2 \Z}^\times$ is $p \paren{p - 1}$.

Since $p - 1 < p \paren{p - 1}$,

$g^{p - 1} \not \equiv 1 \pmod{p^2}$

which proves (1) for $\alpha = 2$.

Induction Step

Suppose it is true for $\alpha = r \ge 2$.

From ($\star$),

$g^{p^{r - 2} \paren{p - 1}} \equiv 1 \pmod{p^{r - 1}}$

From (1) for $\alpha = r$:

$g^{p^{r - 2} \paren{p - 1}} \not \equiv 1 \pmod{p^r}$

so

$g^{p^{r - 2} \paren{p - 1}} = 1 + t p^{r - 1}$

for some $t$ not divisible by $p$.

Consider $\alpha = r + 1$. By binomial expansion,

\(\ds g^{p^{r - 1} \paren{p - 1} }\)

\(=\)

\(\ds \paren{1 + t p^{r - 1} }^p\)

\(\ds \)

\(=\)

\(\ds 1 + t p^r + \underbrace{\binom{p}{2} t^2 p^{2 \paren{r - 1} } }_{\text{divisible by } p^{r + 1} } + \sum_{j \ge 3} \binom p j t^j p^{j \paren{r - 1} }\)

since $p \divides \binom p 2$ (for $p>2$), and $1 + 2 \paren {r-1} \geq r+1$ (for $r \ge 2$), so $p^{r + 1} \divides \binom p 2 p^{2 \paren {r-1} }$

\(\ds \)

\(\equiv\)

\(\ds 1 + t p^r + \sum_{j \ge 3} \underbrace{\binom p j t^j p^{j \paren{r - 1} } }_{\text{divisible by } p^{r + 1} }\)

For $j \ge 3$, each term is divisible by $p^{3 \paren{r - 1} }$, and $3 \paren {r - 1} \ge r + 1$ (for $r \ge 2$)

\(\ds \)

\(\equiv\)

\(\ds 1 + \underbrace{t p^r}_{\text{not divisible by } p^{r + 1} }\)

as $t$ is not divisible by $p$

\(\ds \)

\(\not \equiv\)

\(\ds 1 \pmod{p^{r + 1} }\)

This proves the inductive step.

Hence the result by induction.

$\blacksquare$