TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-03

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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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Every group admits a presentation

Statement

Every group $G$ is isomorphic to a group given by generators and relations. More precisely, if $X$ is the underlying set of $G$ , the free-group extension of the identity function $X\to G$ gives a presentation

$G\cong\langle X\mid\ker\pi\rangle.$

Facts & Assumptions

Given: A group $G$ and its underlying set $X$ .

[L1]

The reduced-word construction supplies a free group on $X$ , and its universal property extends every function $X\to G$ uniquely to a group homomorphism (Reduced words form the free group on an alphabet, Free group on a set of generators).

[L2]

The kernel of a group homomorphism is a normal subgroup (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup).

[L3]

The normal closure of a set is the smallest normal subgroup containing it (The normal closure of a subset of a group).

[L4]

The presentation $\langle X\mid R\rangle$ is the quotient of $F(X)$ by the normal closure of $R$ (Group presentation by generators and relations).

[L5]

A homomorphism induces an isomorphism from its quotient by the kernel onto its image (First isomorphism theorem for groups: $G/\ker f\cong\operatorname{im}f$ ).

Proof

technique · direct

1.1

Apply [L1] to the identity function $u:X\to G$ to obtain a homomorphism $\pi:F(X)\to G$ satisfying $\pi(x)=x$ for every $x\in X$ . It is surjective because every element of $G$ is such an $x$ .

L1given

2.1

Put $K:=\ker\pi$ . By [L2], $K\mathrel{\trianglelefteq}F(X)$ ; since $K$ is itself a normal subgroup containing $K$ , the minimality in [L3] gives $\langle\!\langle K\rangle\!\rangle_{F(X)}=K$ .

step 1.1L2L3

3.1

By [L4], $\langle X\mid K\rangle=F(X)/K$ . By [L5] and the surjectivity from step 1.1, $F(X)/K\cong\operatorname{im}\pi=G$ .

step 1.1step 2.1L4L5

4.1

Hence $G\cong\langle X\mid\ker\pi\rangle$ , as required.

step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 48 results over 14 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

M. Brittenham, Group Presentations, Class Notes (standard reference, not scraped)