CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-01

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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.
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$g^{∣ G ∣} = e$ 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 $n = ord (g)$ is finite and $ι (n)$ divides $ι (∣ G ∣)$ (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 $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, The naturals embed in the integers).

[L2]

If $ord (g) = n$ , then for every integer $k$ , $g^{k} = e$ if and only if $n ∣ k$ (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 $j = k$ , 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 $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

Proof

technique · direct

1.1

By [L1], $ord (g)$ divides $∣ G ∣$ after both naturals are embedded in $Z$ .

2.1

Applying [L2] with $k = ι (∣ G ∣)$ 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)≡1 (mod n) Theorem

Dependency tree · two levels

39 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

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∣=e for every element gg of a finite group GG