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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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First isomorphism theorem for groups: G/ker⁡f≅im⁡f

Statement

First isomorphism theorem for groups: G/ker⁡f≅im⁡f.

For every homomorphism f:G→H, the rule gker⁡f↦f(g) is an isomorphism from G/ker⁡f onto im⁡f.

Facts & Assumptions

Given: A group homomorphism f:G→H.

[L1]

A homomorphism killing a normal subgroup factors uniquely through the quotient (A homomorphism that kills a normal subgroup factors uniquely through the quotient group).

[L2]
[L3]

A homomorphism is injective exactly when its kernel is trivial (A group homomorphism is injective if and only if its kernel is trivial).

[L5]

An isomorphism is a bijective group homomorphism (Group isomorphisms, automorphisms and the set Aut⁡(G)).

Proof

technique · direct
1.1

By [L2] and [L1], fˉ:G/ker⁡f→im⁡f, fˉ(gker⁡f)=f(g), is a well-defined homomorphism; [L4] also gives representative independence directly.

L1L2L3L4L5givenconstruct
2.1

Its image is all of im⁡f, and fˉ(gker⁡f)=eH implies f(g)=eH, hence gker⁡f=ker⁡f; therefore its kernel is trivial.

step 1.1L1L2L3L4L5givenalgebra
3.1

The trivial-kernel conclusion of step 2.1 makes fˉ an isomorphism.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

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Sources