TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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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.
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Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group

Statement

Let $⟨ X ∣ R ⟩$ be a presentation, let $H$ be a group, and let $u : X \to H$ be a function. If the evaluation of every $r \in R$ under $u$ is $e_{H}$ , then there is a unique homomorphism

$\overline{u} : ⟨ X ∣ R ⟩ ⟶ H$

with $\overline{u} ([x]) = u (x)$ for every $x \in X$ . Moreover, $\overline{u}$ is surjective if and only if $u (X)$ generates $H$ .

Facts & Assumptions

Given: A presentation $⟨ X ∣ R ⟩$ , a group $H$ , and a function $u : X \to H$ whose evaluation sends every $r \in R$ to $e_{H}$ .

[L1]

If $N ⊴ G$ , $f : G \to H$ is a homomorphism, and $N \subseteq \ker f$ , then there is a unique homomorphism $\overset{ˉ}{f} : G / N \to H$ with $f = \overset{ˉ}{f} \circ π$ (A homomorphism that kills a normal subgroup factors uniquely through the quotient group).

[L2]

For every group $G$ and every function $u : X \to G$ , there is a unique group homomorphism $\hat{u} : F (X) \to G$ extending $u$ (Free group on a set of generators).

[L3]

For a normal subgroup $N ⊴ G$ , the canonical projection $π : G \to G / N$ is surjective (The canonical projection $π : G \to G / N$ , $π (g) = g N$ , is a surjective group homomorphism).

[F1]

The normal closure of $R$ is the smallest normal subgroup containing $R$ (The normal closure of a subset of a group).

[F2]

The subgroup $⟨ S ⟩$ is the smallest subgroup containing $S$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L4]

For every group homomorphism $f : G \to H$ , one has $im f \leq H$ and $\ker f ⊴ G$ (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup).

[F3]

A group homomorphism preserves products, identities, and inverses, and a composite of group homomorphisms is a group homomorphism (Monoid homomorphism and group homomorphism).

[F4]

The presented group is $⟨ X ∣ R ⟩ = F (X) / ⟨ ⁣ ⟨ R ⟩ ⁣ ⟩_{F (X)}$ (Group presentation by generators and relations).

Proof

technique · constructive

1.1

By [L2], construct the unique homomorphism $f : F (X) \to H$ whose value on each free generator $x$ is $u (x)$ .

L2givenconstruct

1.2

The free generators generate $F (X)$ : if $K = ⟨ X ⟩ \leq F (X)$ , the map $X \to K$ extends by [L2] to $a : F (X) \to K$ , and inclusion $j : K ↪ F (X)$ makes $j \circ a$ agree with ${id}_{F (X)}$ on $X$ , so uniqueness gives $j \circ a = {id}_{F (X)}$ and $K = F (X)$ . By [L3], the canonical quotient map $π : F (X) \to ⟨ X ∣ R ⟩$ is surjective; since it sends $X$ to the classes $[x]$ , [F2] and [F3] show that these classes generate the presented group.

L2L3F2F3construct

2.1

The hypothesis puts every $r \in R$ in $\ker f$ ; [L4] makes the kernel normal, so the minimality in [F1] gives $⟨ ⁣ ⟨ R ⟩ ⁣ ⟩_{F (X)} \subseteq \ker f$ .

F1L4step 1.1given

3.1

By [F4], apply [L1] to factor $f$ uniquely through $F (X) / ⟨ ⁣ ⟨ R ⟩ ⁣ ⟩ = ⟨ X ∣ R ⟩$ , obtaining $\overline{u}$ with $\overline{u} ([x]) = u (x)$ .

F4L1step 2.1construct

4.1

If $h : ⟨ X ∣ R ⟩ \to H$ also has $h ([x]) = u (x)$ , then [F3] makes $h \circ π : F (X) \to H$ a homomorphism extending $u$ , so [L2] gives $h \circ π = f = \overline{u} \circ π$ ; uniqueness of the factorisation in [L1] gives $h = \overline{u}$ .

L1L2F3step 3.1

5.1

By [L4], $im \overline{u}$ is a subgroup containing every $u (x)$ , so [F2] gives $⟨ u (X) ⟩ \subseteq im \overline{u}$ . Conversely, put $K = ⟨ u (X) ⟩$ . By [F3], ${\overline{u}}^{- 1} (K)$ is a subgroup of the domain, and it contains every $[x]$ ; step 1.2 and [F2] therefore give ${\overline{u}}^{- 1} (K) = ⟨ X ∣ R ⟩$ . Hence $im \overline{u} \subseteq K$ , so $im \overline{u} = ⟨ u (X) ⟩$ . Thus $\overline{u}$ is surjective exactly when $u (X)$ generates $H$ .

F2L4F3step 1.2step 3.1discharge-construct∎

Depends on

Used by

Dependency tree · two levels

25 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

Ashot Minasyan, MATH6138 Geometric Group Theory, §2.2 (standard reference, not scraped)
Nicholas Touikan, An Introduction to Combinatorial and Geometric Group Theory, §1.6 (standard reference, not scraped)