Group of Units Ring of Integers Modulo p^2 is Cyclic

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Theorem

Let $p$ be a prime.

Let $\struct {\Z / p^2 \Z, +, \times}$ be the ring of integers modulo $p^2$.

Let $U = \struct {\paren {\Z / p^2 \Z}^\times, \times}$ denote the group of units of $\struct {\Z / p^2 \Z, +, \times}$.

Then $U$ is cyclic.


Proof

The case $p = 2$ follows from Isomorphism between Group of Units Ring of Integers Modulo $2^n$ and $C_2 \times C_{2^{n - 2} }$.

Next, we suppose $p > 2$.

From Ring of Integers Modulo Prime is Field and Group of Units of Finite Field is Cyclic, $\paren {\Z / p \Z}^\times$ is cyclic.

Let $g \in \Z$ such that $\eqclass g p$ be a generator of $\paren {\Z / p \Z}^\times$.

Let $d$ be the order of $\eqclass g {p^2}$ in $\paren {\Z / p^2 \Z}^\times$.

Then:

$g^d \equiv 1 \pmod {p^2}$

so:

$g^d \equiv 1 \pmod p$

But the order of $\eqclass g p$ in $\paren {\Z / p \Z}^\times$ is $\order {\paren {\Z / p \Z}^\times} = p - 1$, we get

\(\text {(1)}: \quad\) \(\ds p - 1\) \(\divides\) \(\ds d\)

By Lagrange's Theorem

\(\text {(2)}: \quad\) \(\ds d\) \(\divides\) \(\ds \order {\paren {\Z / p^2 \Z}^\times}\) \(\ds = p \paren {p - 1}\)

If $\eqclass g {p^2}$ generates $\paren {\Z / p^2 \Z}^\times$, then $\paren {\Z / p^2 \Z}^\times$ is cyclic, the proof is done.

If not, then

\(\text {(3)}: \quad\) \(\ds d\) \(<\) \(\ds \order {\paren {\Z / p^2 \Z}^\times}\) \(\ds p \paren {p - 1}\)

By (1)(2)(3), $d = p - 1$, so

\(\text {(4)}: \quad\) \(\ds g^{p - 1}\) \(\equiv\) \(\ds 1 \pmod {p^2}\)

Consider $g' = g + p$.

Let $d'$ be the order of $\eqclass {g'} {p^2}$ in $\paren {\Z / p^2 \Z}^\times$. Then:

${g'}^{d'} \equiv 1 \pmod {p^2}$

so

${g'}^{d'} \equiv 1 \pmod p$

Note that $\eqclass {g'} p = \eqclass g p$. Then the order of $\eqclass {g'} p$ in $\paren {\Z / p \Z}^\times$ is $\order {\paren {\Z / p \Z}^\times} = p - 1$, we get

\(\text {(5)}: \quad\) \(\ds p - 1\) \(\divides\) \(\ds d'\)

By Lagrange's Theorem

\(\text {(6)}: \quad\) \(\ds d'\) \(\divides\) \(\ds \order {\paren {\Z / p^2 \Z}^\times}\) \(\ds = p \paren {p - 1}\)

By binomial expansion:

\(\ds {g'}^{p-1}\) \(=\) \(\ds \paren {g + p}^{p - 1}\)
\(\ds \) \(=\) \(\ds g^{p - 1} + \paren {p - 1} p g^{p - 2} + \underbrace {\sum_{k \mathop \ge 2} \binom {p - 1} k p^k g^{p - 1 - k} }_{\text{divisible by } p^2}\)
\(\ds \) \(\equiv\) \(\ds g^{p - 1} + p \paren {p - 1} g^{p - 2} \pmod {p^2}\)
\(\ds \) \(\equiv\) \(\ds 1 + p \paren {p - 1} g^{p - 2} \pmod {p^2}\) from $(4)$

so:

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

Therefore:

\(\text {(7)}: \quad\) \(\ds d'\) \(\ne\) \(\ds p - 1\)

By (5)(6)(7)

$d' = p \paren{p - 1}$

so $\eqclass {g'} {p^2}$ generates $\paren {\Z / p^2 \Z}^\times$, then $\paren {\Z / p^2 \Z}^\times$ is cyclic.

$\blacksquare$

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