LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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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If $K\mathrel{\trianglelefteq}G$ , $N\mathrel{\trianglelefteq}G$ and $K\subseteq N$ , then $N/K\mathrel{\trianglelefteq}G/K$

Statement

If $K\mathrel{\trianglelefteq}G$ , $N\mathrel{\trianglelefteq}G$ and $K\subseteq N$ , then $N/K\mathrel{\trianglelefteq}G/K$ .

Facts & Assumptions

Given: Normal subgroups $K,N\mathrel{\trianglelefteq}G$ 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 $(gK)(hK)=ghK$ (The quotient group $G/N$ and coset product $(gN)(hN)=ghN$ ).

[L3]

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

[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 $N/K=\{nK:n\in N\}$ is a subgroup of $G/K$ by the quotient product rule.

L1L2L3L4givenalgebra

2.1

For $gK\in G/K$ and $nK\in N/K$ , $(gK)(nK)(gK)^{-1}=(gng^{-1})K$ belongs to $N/K$ because $N\mathrel{\trianglelefteq}G$ .

step 1.1L1L2L3L4givenalgebra

3.1

Thus the conjugation calculation gives $N/K\mathrel{\trianglelefteq}G/K$ .

step 2.1∎

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 19 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

If K⊴GK\mathrel{\trianglelefteq}GK⊴G, N⊴GN\mathrel{\trianglelefteq}GN⊴G and K⊆NK\subseteq NK⊆N, then N/K⊴G/KN/K\mathrel{\trianglelefteq}G/KN/K⊴G/K