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$(Z / 12)^{\times} = {[1], [5], [7], [11]}$ and $φ (12) = 4$

Example

$(Z / 12)^{\times} = {[1]_{12}, [5]_{12}, [7]_{12}, [11]_{12}}, φ (12) = 4.$

Every displayed unit is its own inverse.

Facts & Assumptions

Given: The quotient $Z / 12$ and its unit group.

[L1]

The classes modulo $12$ have the unique standard representatives $0, \dots, 11$ (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 $Z$ ).

[L2]

The class $[a]_{12}$ is a unit exactly when $\gcd (a, 12) = 1$ , and $φ (12)$ counts these units (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ , The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L3]

Euler's finite prime-divisor product gives $φ (n) = \prod_{p^{k} ∥ n} (p^{k} - p^{k - 1})$ (Euler's product formula $φ (n) = n \prod_{p ∣ n} (1 - 1 / p) = \prod_{p^{k} ∥ n} (p^{k} - p^{k - 1})$ for $n \geq 1$ , stated through a finite injective list of its prime divisors).

[F1]

Equality of residue classes is congruence of representatives (The congruence class $[a]_{n}$ and the quotient set $Z / n$ ), and congruence means divisibility of their difference (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

[F2]

Products of residue classes are computed by multiplying representatives: $[a]_{12} [b]_{12} = [a b]_{12}$ (Addition and multiplication on $Z / n$ by $[a]_{n} + [b]_{n} = [a + b]_{n}$ and $[a]_{n} [b]_{n} = [a b]_{n}$ ).

Verification

technique · direct

1.1

Among $0, \dots, 11$ , exactly $1, 5, 7, 11$ have gcd $1$ with $12$ : every other representative is divisible by $2$ or $3$ . Thus [L1] and [L2] give the displayed unit group and $φ (12) = 4$ .

L1L2

1.2

The congruences $5^{2} = 25 \equiv 1$ , $7^{2} = 49 \equiv 1$ , and $11^{2} = 121 \equiv 1 (m o d 12)$ show that the three nonidentity units, as well as $[1]_{12}$ , are self-inverse.

L2F1F2

2.1

Since $12 = 2^{2} \cdot 3$ , [L3] independently gives $φ (12) = (2^{2} - 2) (3 - 1) = 2 \cdot 2 = 4$ , agreeing with the list.

step 1.1L3∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Mathematics LibreTexts, Euler's phi Function (standard reference, not scraped)
UW Oshkosh, Number Theory Notes 3 (standard reference, not scraped)

(Z/12)×={[1],[5],[7],[11]}(Z/12)×={[1],[5],[7],[11]} and φ(12)=4φ(12)=4