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

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

Statement

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

Facts & Assumptions

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

[L1]

A normal subgroup is invariant under conjugation (Normal subgroup: invariance under conjugation).

[L2]

$G / K$ consists of cosets and has product $(g K) (h K) = g h K$ (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

[L3]

Quotient multiplication is well defined for normal subgroups (For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$ ).

[L4]

A conjugation-stable subgroup is normal (Equivalent characterisations of a normal subgroup by conjugates and left and right cosets).

Proof

technique · direct

1.1

Since $K \subseteq N$ , the subset $N / K = {n K : n \in N}$ is a subgroup of $G / K$ by the quotient product rule.

L1L2L3L4givenalgebra

2.1

For $g K \in G / K$ and $n K \in N / K$ , $(g K) (n K) (g K)^{- 1} = (g n g^{- 1}) K$ belongs to $N / K$ because $N ⊴ G$ .

step 1.1L1L2L3L4givenalgebra

3.1

Thus the conjugation calculation gives $N / K ⊴ G / K$ .

step 2.1∎

Depends on

Used by

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

Dependency tree · two levels

10 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)

If K⊴GK⊴G, N⊴GN⊴G and K⊆NK⊆N, then N/K⊴G/KN/K⊴G/K