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The generators of a cyclic group of order m are the ga with gcd⁡(a,m)=1, so there are φ(m) of them

Statement

Let G=⟨g⟩ be cyclic of order m. The generators of G are exactly the elements ga with gcd⁡(a,m)=1, for a taken modulo m. Consequently G has φ(m) generators.

Facts & Assumptions

Given: A cyclic group G=⟨g⟩ of finite order m.

[L1]

ord⁡(ga)=m/gcd⁡(a,m) (In a cyclic group of order m, ga has order m/gcd⁡(a,m)).

[L2]

For x of finite order r, the powers x0,…,xr−1 are pairwise distinct and ⟨x⟩={xs:s<r}, so ⟨x⟩ is finite with ∣⟨x⟩∣=ord⁡(x) (If ord⁡(g)=n then gk=e iff k is an integer multiple of n, the powers g0,…,gn−1 are distinct, and ⟨g⟩ has exactly n elements; if g has infinite order then gj=gk only for j=k).

[L3]

The unit classes modulo m are exactly the classes represented by integers coprime to m (For n≥1, [a]n is a unit if and only if gcd⁡(a,n)=1).

Proof

technique · direct
1.1L1L2algebra

By [L2], ∣⟨ga⟩∣=ord⁡(ga) and ∣G∣=∣⟨g⟩∣=ord⁡(g)=m. Since ⟨ga⟩⊆G and G is finite, ga generates G exactly when ∣⟨ga⟩∣=m, that is exactly when ord⁡(ga)=m. By [L1] this says m/gcd⁡(a,m)=m, equivalently gcd⁡(a,m)=1.

1.2L2

By [L2] the powers g0,…,gm−1 are pairwise distinct, so distinct exponent classes modulo m give distinct powers of g.

2.1step 1.1step 1.2L3L4∎

By [L3] and [L4], exactly φ(m) exponent classes satisfy the condition in step 1.1, proving the count.

Depends on

Used by

Dependency tree · two levels

38 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