PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The canonical projection $π : G \to G / N$ , $π (g) = g N$ , is a surjective group homomorphism

Let $N ⊴ G$ . The canonical projection

$π : G ⟶ G / N, π (g) : = g N,$

is a surjective group homomorphism.

Given: A group $G$ , a normal subgroup $N ⊴ G$ , and the quotient group $G / N$ .

[L1]

[F1]

A group homomorphism $f : G \to G^{'}$ satisfies $f (g h) = f (g) f (h)$ for all $g, h \in G$ (Monoid homomorphism and group homomorphism).

[F2]

A function $f : A \to B$ is surjective if every $b \in B$ equals $f (a)$ for some $a \in A$ (Injection, surjection, bijection).

[F3]

Every left coset of $N$ has the form $g N$ for a representative $g \in G$ (Left and right cosets $g H$ and $H g$ of a subgroup).

technique · direct

1.1L1F1

For $g, h \in G$ , one has $π (g h) = g h N = (g N) (h N) = π (g) π (h)$ , so $π$ is a group homomorphism.

1.2F2F3

Every element of $G / N$ is a coset $g N = π (g)$ for some $g \in G$ , so $π$ is surjective.

2.1step 1.1step 1.2∎

Hence the canonical projection is a surjective group homomorphism.

Depends on