LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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In a cyclic group of order $m$ , $g^{a}$ has order $m / \gcd (a, m)$

Statement

Let $G = ⟨ g ⟩$ be cyclic of finite order $m \geq 1$ . For every integer $a$ ,

$ord (g^{a}) = \frac{m}{\gcd (a, m)} .$

Facts & Assumptions

Given: A generator $g$ of a cyclic group of order $m$ and an integer $a$ .

[L1]

The order of an element is the least positive exponent giving the identity (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).

[L2]

Since $ord (g) = m$ , one has $g^{u} = 1$ exactly when $m ∣ u$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

[L3]

$(g^{a})^{t} = g^{a t}$ for every integer $t$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[L4]

Put $d = \gcd (a, m)$ ; then $d$ divides both $a$ and $m$ (Common divisor, and the greatest common divisor $\gcd (a, b)$ , with the convention $\gcd (0, 0) : = 0$ ).

[L5]

After dividing two integers by their nonzero greatest common divisor, the resulting quotients are coprime (If $d = \gcd (a, b)$ is nonzero then $a / d$ and $b / d$ are coprime).

[L6]

If $\gcd (r, s) = 1$ and $r ∣ s t$ , then $r ∣ t$ (If $\gcd (a, b) = 1$ and $a ∣ b c$ then $a ∣ c$ ; and if $a ∣ c$ , $b ∣ c$ and $\gcd (a, b) = 1$ then $a b ∣ c$ ).

Proof

technique · direct

1.1L4L5algebra

Since $m \geq 1$ , the common divisor $d$ is nonzero. Write $a = d a^{'}$ and $m = d m^{'}$ ; [L5] gives $\gcd (a^{'}, m^{'}) = 1$ .

2.1L2L3step 1.1algebra

By [L2] and [L3], $(g^{a})^{t} = 1$ is equivalent to $m ∣ a t$ , hence to $m^{'} ∣ a^{'} t$ .

3.1step 2.1L6

By [L6] and $\gcd (a^{'}, m^{'}) = 1$ , the condition in step 2.1 is equivalent to $m^{'} ∣ t$ .

4.1step 3.1L1∎

Thus the least positive $t$ with $(g^{a})^{t} = 1$ is $m^{'} = m / d$ , which is the asserted order by [L1].

Depends on

Used by

Dependency tree · two levels

43 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

Peter Hackman, Elementary Number Theory, §C.I (standard reference, not scraped)
William Stein, Elementary Number Theory, §2.5 (standard reference, not scraped)

In a cyclic group of order mm, gaga has order m/gcd⁡(a,m)m/gcd(a,m)