TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02

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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 $f : G \to H$ , the rule $g \ker f \mapsto f (g)$ is an isomorphism from $G / \ker f$ onto $im f$ .

Facts & Assumptions

Given: A group homomorphism $f : G \to H$ .

[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]

$\ker f$ is normal and $im f$ is a subgroup (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup).

[L3]

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

[L4]

Equal images are exactly equal kernel cosets (Two elements have the same image under a homomorphism if and only if they lie in the same coset of its kernel).

[L5]

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

Proof

technique · direct

1.1L1L2L3L4L5givenconstruct

By [L2] and [L1], $\overset{ˉ}{f} : G / \ker f \to im f$ , $\overset{ˉ}{f} (g \ker f) = f (g)$ , is a well-defined homomorphism; [L4] also gives representative independence directly.

2.1step 1.1L1L2L3L4L5givenalgebra

Its image is all of $im f$ , and $\overset{ˉ}{f} (g \ker f) = e_{H}$ implies $f (g) = e_{H}$ , hence $g \ker f = \ker f$ ; therefore its kernel is trivial.

3.1step 2.1∎

The trivial-kernel conclusion of step 2.1 makes $\overset{ˉ}{f}$ an isomorphism.

Depends on

Dependency tree · 2 levels

Sources

Judson, Abstract Algebra: Theory and Applications, Isomorphism Theorems