LemmaStatement: AI-adaptedProof: AI-generatedprecheck 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 $n \geq 1$

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 $n \geq 1$ 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 $g^{σ (k)} = g^{k} g$ (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 $m \in N$ , that is, there is a bijection $β : G \to m$ (Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A ⪯ B$ ).

[L2]

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

[L3]

On $N$ the order is membership, so the elements of the natural number $σ (m)$ are exactly the natural numbers $k < σ (m)$ , and the elements of $m$ are exactly the natural numbers $k < m$ (On $N$ the order is membership: $m < n ⟺ m \in n$ , The natural numbers $N$ (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 $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[L6]

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

[L7]

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

Proof

technique · direct

1.1L1choose

Fix a bijection $β : G \to m$ with $m \in N$ , available because $G$ is finite.

2.1step 1.1L3given

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

3.1step 2.1L2choose

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

4.1step 1.1step 3.1L4

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

4.2step 3.1L7

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

5.1step 4.1step 4.2L5L6given

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

6.1step 4.2step 5.1L7∎

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

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 $N \to G$ and contradict finiteness. It uses claim 1 of The pigeonhole principle on $N$ directly on the map $k \mapsto β (g^{k})$ from $σ (m)$ 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 $ord (g)$ (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), and that step needs the well-ordering principle, not this lemma.
The hypothesis of finiteness cannot be dropped: in $(Z, +)$ the element $1$ satisfies $n \cdot 1 = n \neq 0$ for every $n \geq 1$ .

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

In a finite group, every element gg satisfies gn=egn=e for some natural n≥1n≥1

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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