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Raising to the power $1 + p$ is an automorphism of order $p$ of a cyclic group of order $p^{2}$

Statement

Let $p$ be a prime. In $Z / p^{2}$ the class of $1 + p$ is a unit of multiplicative order $p$ , and $(1 + p)^{k} = 1 + k p$ for every $k \in N$ . Consequently, if $A$ is a cyclic group of order $p^{2}$ , the map $x \mapsto x^{1 + p}$ is an automorphism of $A$ of order $p$ (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

Facts & Assumptions

Given: A prime $p$ , the ring $Z / p^{2}$ , and a cyclic group $A$ of order $p^{2}$ .

[F1]

For every $n \in N$ , $(Z / n, +, [0]_{n})$ is an abelian group, $(Z / n, \cdot, [1]_{n})$ is a commutative monoid, and multiplication distributes over addition on both sides (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[F2]

$ord (g) : = \min {k \in N : k \geq 1 and g^{k} = e}$ when that set is nonempty (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[F3]

$p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L1]

$Aut (C_{n}) ≅ (Z / n)^{\times}$ , and if $C_{n} = ⟨ g ⟩$ the unit class $[a]$ corresponds to the automorphism $g \mapsto g^{a}$ ( $Aut (C_{n}) ≅ (Z / n Z)^{\times}$ ).

[L2]

A cyclic group with a generator of finite order $n$ is isomorphic to $(Z / n, +)$ (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ).

Proof

technique · induction

1.1

At $k = 0$ the claim reads $(1 + p)^{0} = 1 = 1 + 0 \cdot p$ in $Z / p^{2}$ .

F1base

1.2

Assume $(1 + p)^{k} = 1 + k p$ in $Z / p^{2}$ for a given $k \in N$ .

1.3

In $Z / p^{2}$ the class of $p^{2}$ is zero, so $k p = 0$ holds exactly when $p^{2}$ divides $k p$ , that is exactly when $p$ divides $k$ .

F1F3algebra

1.4

An automorphism of a cyclic group of order $p^{2}$ is $x \mapsto x^{a}$ for a unit class $[a]$ of $Z / p^{2}$ , and this correspondence is an isomorphism of groups, so it preserves orders.

L1L2

2.1

Then $(1 + p)^{k + 1} = (1 + k p) (1 + p) = 1 + (k + 1) p + k p^{2} = 1 + (k + 1) p$ , the last equality because $p^{2} = 0$ in $Z / p^{2}$ .

F1step 1.2algebra

3.1

Hence $(1 + p)^{k} = 1 + k p$ for every $k \in N$ ; by step 1.3 this equals $1$ exactly when $p$ divides $k$ , so the least positive such $k$ is $p$ and the class of $1 + p$ is a unit of order $p$ . Under the correspondence of step 1.4 it is the automorphism $x \mapsto x^{1 + p}$ , which therefore has order $p$ .

F2step 1.3step 1.4step 2.1discharge-induction∎

Remarks

The computation holds at $p = 2$ as well: there $1 + p = 3$ and $3^{2} = 9 = 1$ in $Z / 4$ , so the automorphism $x \mapsto x^{3} = x^{- 1}$ of a cyclic group of order four is inversion and has order two.

Depends on

Used by

Dependency tree · two levels

42 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

D. A. Craven, The Theory of p-Groups, Definition 3.3 (standard reference, not scraped)

Raising to the power 1+p1+p is an automorphism of order pp of a cyclic group of order p2p2