LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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In a finite group, every element $g$ satisfies $g^{n} = e$ for some natural $n \ge 1$

Statement

Let $G$ be a group (Group and abelian group) whose underlying set is finite (Finite, countably infinite, countable, uncountable), and let $g \in G$ . Then there is a natural number $n \ge 1$ with $g^{n} = e$ , the power being the natural power of Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ .

Facts & Assumptions

Given: A group $G$ with identity $e$ whose underlying set is finite, and an element $g \in G$ ; natural powers $g^{k}$ with $g^{0} = e$ and $g^{\sigma(k)} = g^{k} g$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L1]

$G$ finite means $G \approx m$ for some $m \in \mathbb{N}$ , that is, there is a bijection $\beta : G \to m$ (Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A \preceq B$ ).

[L2]

Claim 1 of the pigeonhole principle: for every $m \in \mathbb{N}$ there is no injection $\sigma(m) \to m$ (The pigeonhole principle on $\mathbb{N}$ ).

[L3]

On $\mathbb{N}$ the order is membership, so the elements of the natural number $\sigma(m)$ are exactly the natural numbers $k < \sigma(m)$ , and the elements of $m$ are exactly the natural numbers $k < m$ (On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$ , The natural numbers $\mathbb{N}$ (von Neumann)).

[L4]

A map $f$ is injective when $f(x) = f(y)$ forces $x = y$ ; a bijection is injective (Injection, surjection, bijection).

[L5]

$g^{i+k} = g^{i} g^{k}$ for natural $i, k$ , in any monoid (Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$ , and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

[L6]

Cancellation in a group: $gx = gy$ implies $x = y$ (Cancellation in a group: $gx = gy$ or $xg = yg$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $gx = h$ and $xg = h$ each have exactly one solution).

[L7]

On $\mathbb{N}$ : exactly one of $i < j$ , $i = j$ , $j < i$ holds (Trichotomy of the order on $\mathbb{N}$ ); $i \le j$ means $i + k = j$ for some $k$ (Order on the natural numbers); every $k \ne 0$ is a successor $\sigma(t) = 1 + t$ (Every nonzero natural number is a successor, Addition is commutative), so $k \ne 0$ implies $1 \le k$ .

Proof

technique · direct

1.1

Fix a bijection $\beta : G \to m$ with $m \in \mathbb{N}$ , available because $G$ is finite.

L1choose

2.1

Define $F : \sigma(m) \to m$ by $F(k) := \beta(g^{k})$ . This is a function: every element $k$ of $\sigma(m)$ is a natural number, so the natural power $g^{k}$ is defined and lies in $G$ , and $\beta$ sends it into $m$ .

step 1.1L3given

3.1

$F$ is not injective, since there is no injection $\sigma(m) \to m$ . Hence there are $i, j \in \sigma(m)$ with $i \ne j$ and $F(i) = F(j)$ .

step 2.1L2choose

4.1

From $\beta(g^{i}) = \beta(g^{j})$ and injectivity of $\beta$ we get $g^{i} = g^{j}$ .

step 1.1step 3.1L4

4.2

By trichotomy and $i \ne j$ , one of $i < j$ and $j < i$ holds; interchanging the names if necessary, assume $i < j$ . Then $i + k = j$ for some $k \in \mathbb{N}$ , and $k \ne 0$ , since $k = 0$ would give $i = j$ .

step 3.1L7

5.1

Hence $g^{i} g^{k} = g^{i+k} = g^{j} = g^{i} = g^{i} e$ , and cancelling $g^{i}$ on the left gives $g^{k} = e$ .

step 4.1step 4.2L5L6given

6.1

Finally $k \ne 0$ gives $1 \le k$ , so $n := k$ is a natural number with $n \ge 1$ and $g^{n} = e$ .

step 4.2step 5.1L7∎

Remarks

The route avoids "a subset of a finite set is finite". That statement is not available at this point in the reading order, so the argument does not build an injection $\mathbb{N} \to G$ and contradict finiteness. It uses claim 1 of The pigeonhole principle on $\mathbb{N}$ directly on the map $k \mapsto \beta(g^{k})$ from $\sigma(m)$ to $m$ : the $m+1$ exponents $0, 1, \dots, m$ cannot receive $m+1$ distinct values in a set of $m$ elements.
The conclusion is one-sided on purpose. It asserts that some positive power is the identity, not which one. Picking the least such power is what defines $\operatorname{ord}(g)$ (The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity), and that step needs the well-ordering principle, not this lemma.
The hypothesis of finiteness cannot be dropped: in $(\mathbb{Z},+)$ the element $1$ satisfies $n \cdot 1 = n \ne 0$ for every $n \ge 1$ .

Depends on

Used by

The order |G| of a finite group and the order ord(g) of an element, with ord(g) = ∞ when no positive power of g is the identity Definition

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 67 results over 23 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Order (group theory) (Wikipedia) (standard reference, not scraped)
Pigeonhole principle (Wikipedia) (standard reference, not scraped)

In a finite group, every element ggg satisfies gn=eg^{n} = egn=e for some natural n≥1n \ge 1n≥1