CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
AI-adapted — a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$g^{|G|}=e$ for every element $g$ of a finite group $G$

Statement

Let $G$ be a finite group with identity $e$ . Then

$g^{|G|}=e$

for every $g\in G$ .

Facts & Assumptions

Given: A finite group $G$ with identity $e$ and an element $g\in G$ .

[L1]

The order $n=\operatorname{ord}(g)$ is finite and $\iota(n)$ divides $\iota(|G|)$ (The order of every element of a finite group divides the order of the group, 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, The naturals embed in the integers).

[L2]

If $\operatorname{ord}(g)=n$ , then for every integer $k$ , $g^k=e$ if and only if $n\mid k$ (If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ , Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , 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).

Proof

technique · direct

1.1

By [L1], $\operatorname{ord}(g)$ divides $|G|$ after both naturals are embedded in $\mathbb Z$ .

2.1

Applying [L2] with $k=\iota(|G|)$ gives $g^{|G|}=e$ .

step 1.1L2∎

Depends on

Used by

The exponent of a finite group Definition
Euler's theorem: if n≥1 and gcd(a,n)=1, then a^φ(n)≡1pmod n Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 76 results over 21 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

Thomas W. Judson, Abstract Algebra: Theory and Applications, Lagrange's Theorem (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §6.2: Lagrange's Theorem (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §4.1: Cyclic Subgroups (standard reference, not scraped)

g∣G∣=eg^{|G|}=eg∣G∣=e for every element ggg of a finite group GGG