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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Third isomorphism theorem for groups: $(G / K) / (N / K) ≅ G / N$

Statement

Third isomorphism theorem for groups: $(G / K) / (N / K) ≅ G / N$ .

If $K \subseteq N$ are normal subgroups of $G$ , then

$(G / K) / (N / K) ≅ G / N .$

Facts & Assumptions

Given: $K \subseteq N$ with $K, N ⊴ G$ .

[L1]

$N / K ⊴ G / K$ (If $K ⊴ G$ , $N ⊴ G$ and $K \subseteq N$ , then $N / K ⊴ G / K$ ).

[L2]

The first isomorphism theorem identifies a quotient by a kernel with the image (First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

[L3]

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

[L4]

Quotient groups use coset multiplication (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

Proof

technique · direct

1.1

Define $ϕ : G / K \to G / N$ by $ϕ (g K) = g N$ ; it is well defined because $K \subseteq N$ , and [L4] shows it is a homomorphism.

L1L2L3L4givenconstruct

2.1

The map is onto and $ϕ (g K) = N$ exactly when $g \in N$ , so $\ker ϕ = N / K$ .

step 1.1L1L2L3L4givenalgebra

3.1

The kernel and image calculation yields $(G / K) / (N / K) ≅ G / N$ .

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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

Judson, Abstract Algebra: Theory and Applications, Isomorphism Theorems (standard reference, not scraped)

Third isomorphism theorem for groups: (G/K)/(N/K)≅G/N(G/K)/(N/K)≅G/N