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The order of every element of a finite group divides the order of the group
Statement
If is finite and , then has finite order and
in , where is the canonical embedding.
Facts & Assumptions
Given: A finite group and an element .
The generated set is a subgroup of and equals the set of integer powers of (The subgroup generated by a subset, the cyclic subgroup , and cyclic groups, , and every cyclic group is abelian).
If has finite order, then is finite and ; every element of a finite group has finite order (If then iff is an integer multiple of , the powers are distinct, and has exactly elements; if has infinite order then only for , The order of a finite group and the order of an element, with when no positive power of is the identity).
Lagrange's theorem gives for every subgroup of a finite group and consequently (Lagrange's theorem: for every subgroup of a finite group , Divisibility in : when for some integer , The naturals embed in the integers).
Proof
The element has finite order, and has order .
Apply [L2] to to obtain .
Depends on
- Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$
- 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
- 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 $j = k$
- The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$, and cyclic groups
- $\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$, and every cyclic group is abelian
- Divisibility in $\mathbb{Z}$: $d \mid a$ when $a = dq$ for some integer $q$
- The naturals embed in the integers
Used by
- A finite group of prime order is cyclic and every nonidentity element generates it Corollary
- g^|G|=e for every element g of a finite group G Corollary
- 1→⟨ i⟩→ Q₈→ Q₈/⟨ i⟩→1 does not split, with nonabelian middle group Counterexample
- For an odd prime p and a primitive root g modulo p, at least one of g and g+p is primitive modulo p² Lemma
- For a complex character, χ(1)=dim V, χ is a class function, and |χ(g)|≤χ(1) with equality exactly at scalars Proposition
- For every odd prime p and k≥1, (ℤ/pᵏℤ)^× is cyclic of order pᵏ⁻¹(p-1) Theorem
Dependency tree · two levels
42 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, Cosets and 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)