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In $(Z / p)^{\times}$ , inversion pairs every class except $[1]_{p}$ and $[- 1]_{p}$ , which are the only self-inverse classes

Statement

Let $p$ be prime. In the finite unit group $(Z / p)^{\times}$ , inversion partitions all classes other than $[1]_{p}$ and $[- 1]_{p}$ into disjoint pairs ${u, u^{- 1}}$ with distinct members. The only self-inverse classes are $[1]_{p}$ and $[- 1]_{p}$ .

When $p = 2$ these two displayed classes coincide, and the unit group has that single element.

Facts & Assumptions

Given: A prime $p$ and the unit group $(Z / p)^{\times}$ .

[L1]

The quotient $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

[L2]

Units form a finite group under multiplication and every class has one standard representative; moreover $[a]_{p} = [b]_{p}$ exactly when $a \equiv b (m o d p)$ , which means $p ∣ (a - b)$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ , For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ , The congruence class $[a]_{n}$ and the quotient set $Z / n$ , Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

[L3]

An invertible element has a unique inverse, and every one-sided inverse equals that inverse (In a monoid, a left inverse and a right inverse of the same element are equal; hence an invertible element has exactly one inverse, and it is two-sided).

Proof

technique · direct

1.1

If a unit $u$ is self-inverse, then $u^{2} = [1]_{p}$ , so $(u - [1]_{p}) (u + [1]_{p}) = [0]_{p}$ . In a field a product is zero only if a factor is zero: if the first factor is nonzero, multiply by its inverse. Hence $u = [1]_{p}$ or $u = - [1]_{p} = [- 1]_{p}$ .

1.2

Conversely, $[1]_{p}^{2} = [1]_{p}$ and $[- 1]_{p}^{2} = [1]_{p}$ , so both displayed classes are self-inverse.

1.3

If $p = 2$ , then $2 ∣ (1 - (- 1))$ , so $[1]_{2} = [- 1]_{2}$ ; the unique nonzero standard class is $[1]_{2}$ , and it is the only unit.

L1L2

2.1

On the remaining finite set, inversion has no fixed point by steps 1.1 and 1.2. Since $u$ is an inverse of $u^{- 1}$ , uniqueness in [L3] gives $(u^{- 1})^{- 1} = u$ . Therefore the inversion orbits are disjoint pairs ${u, u^{- 1}}$ with distinct members.

step 1.1step 1.2L2L3

3.1

Steps 1.1 through 2.1 prove the pairing and its boundary case.

step 1.1step 1.2step 2.1step 1.3∎

Depends on

Used by

Wilson's theorem: for every prime p, (p-1)!≡-1 (mod p) Theorem

Dependency tree · two levels

26 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

Mathematics LibreTexts, Wilson's Theorem (standard reference, not scraped)

In (Z/p)×(Z/p)×, inversion pairs every class except [1]p[1]p​ and [−1]p[−1]p​, which are the only self-inverse classes