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 $\eqclass g p$ be a generator of $\paren {\Z / p \Z}^\times$ for some $g \in \Z$.
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:
- $g^d \equiv 1 \pmod p$
The order of $\eqclass g p$ in $\paren {\Z / p \Z}^\times$ is:
- $\order {\paren {\Z / p \Z}^\times} = p - 1$
We get:
| \(\text {(1)}: \quad\) | \(\ds p - 1\) | \(\divides\) | \(\ds d\) |
| \(\text {(2)}: \quad\) | \(\ds d\) | \(\divides\) | \(\ds \order {\paren {\Z / p^2 \Z}^\times}\) | \(\ds = p \paren {p - 1}\) |
We need to prove that:
- $\eqclass g {p^2}$ generates $\paren {\Z / p^2 \Z}^\times \implies \paren {\Z / p^2 \Z}^\times$ is cyclic.
If not, then:
| \(\text {(3)}: \quad\) | \(\ds d\) | \(<\) | \(\ds \order {\paren {\Z / p^2 \Z}^\times}\) | \(\ds = p \paren {p - 1}\) |
By $(1)$, $(2)$ and $(3)$:
- $d = p - 1$
so:
| \(\text {(4)}: \quad\) | \(\ds g^{p - 1}\) | \(\equiv\) | \(\ds 1\) | \(\ds \pmod {p^2}\) |
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
| \(\text {(5)}: \quad\) | \(\ds p - 1\) | \(\divides\) | \(\ds d'\) |
| \(\text {(6)}: \quad\) | \(\ds d'\) | \(\divides\) | \(\ds \order {\paren {\Z / p^2 \Z}^\times}\) | \(\ds = p \paren {p - 1}\) |
By Binomial Theorem for Integral Index:
| \(\ds {g'}^{p-1}\) | \(=\) | \(\ds \paren {g + p}^{p - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds g^{p - 1} + \paren {p - 1} p g^{p - 2} + \underbrace {\sum_{k \mathop \ge 2} \binom {p - 1} k p^k g^{p - 1 - k} }_{\text{divisible by } p^2}\) | ||||||||||||
| \(\ds \) | \(\equiv\) | \(\ds g^{p - 1} + p \paren {p - 1} g^{p - 2}\) | \(\ds \pmod {p^2}\) | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds 1 + p \paren {p - 1} g^{p - 2}\) | \(\ds \pmod {p^2}\) | from $(4)$ |
so:
- ${g'}^{p - 1} \not \equiv 1 \pmod {p^2}$
Therefore:
| \(\text {(7)}: \quad\) | \(\ds d'\) | \(\ne\) | \(\ds p - 1\) |
By $(5)$, $(6)$ and $(7)$:
- $d' = p \paren {p - 1}$
so:
- $\eqclass {g'} {p^2}$ generates $\paren {\Z / p^2 \Z}^\times \implies \paren {\Z / p^2 \Z}^\times$ is cyclic.
$\blacksquare$