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

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 primitive element 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:
:$g^d \equiv 1 \pmod {p^2}$
From Congruence by Divisor of Modulus:
:$g^d \equiv 1 \pmod p$
The order of $\eqclass g p$ in $\paren {\Z / p \Z}^\times$ is:
:$\order {\paren {\Z / p \Z}^\times} = p - 1$
We get:
{{begin-eqn}}
{{eqn | n = 1
      | l = p - 1
      | o = \divides
      | r = d
}}
{{end-eqn}}
From Order of Element Divides Order of Finite Group:
{{begin-eqn}}
{{eqn | n = 2
      | l = d
      | o = \divides
      | r = \order {\paren {\Z / p^2 \Z}^\times}
      | rr= = p \paren {p - 1}
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | n = 3
      | l = d
      | o = <
      | r = \order {\paren {\Z / p^2 \Z}^\times}
      | rr= = p \paren {p - 1}
}}
{{end-eqn}}
By $(1)$, $(2)$ and $(3)$:
:$d = p - 1$
so:
{{begin-eqn}}
{{eqn | n = 4
      | l = g^{p - 1}
      | o = \equiv
      | r = 1
      | rr= \pmod {p^2}
}}
{{end-eqn}}
Consider:
:$g' = g + p$
Let $d'$ be the order of $\eqclass {g'} {p^2}$ in $\paren {\Z / p^2 \Z}^\times$.
Then:
:${g'}^{d'} \equiv 1 \pmod {p^2}$
From Congruence by Divisor of Modulus:
:${g'}^{d'} \equiv 1 \pmod p$
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
{{begin-eqn}}
{{eqn | n = 5
      | l = p - 1
      | o = \divides
      | r = d'
}}
{{end-eqn}}
From Order of Element Divides Order of Finite Group:
{{begin-eqn}}
{{eqn | n = 6
      | l = d'
      | o = \divides
      | r = \order {\paren {\Z / p^2 \Z}^\times}
      | rr= = p \paren {p - 1}
}}
{{end-eqn}}
By Binomial Theorem for Integral Index:
{{begin-eqn}}
{{eqn | l = {g'}^{p-1}
      | r = \paren {g + p}^{p - 1}
}}
{{eqn | r = 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}
}}
{{eqn | o = \equiv
      | r = g^{p - 1} + p \paren {p - 1} g^{p - 2}
      | rr= \pmod {p^2}
}}
{{eqn | o = \equiv
      | r = 1 + p \paren {p - 1} g^{p - 2}
      | rr= \pmod {p^2}
      | c = from $(4)$
}}
{{end-eqn}}
so:
:${g'}^{p - 1} \not \equiv 1 \pmod {p^2}$
Therefore:
{{begin-eqn}}
{{eqn | n = 7
      | l = d'
      | o = \ne
      | r = p - 1
}}
{{end-eqn}}
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.
{{qed}}
Category:Ring of Integers Modulo m
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