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

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Second isomorphism theorem for groups: $H / (H \cap N) ≅ H N / N$

Statement

Second isomorphism theorem for groups: $H / (H \cap N) ≅ H N / N$ .

If $H \leq G$ and $N ⊴ G$ , then

$H / (H \cap N) ≅ H N / N .$

Facts & Assumptions

Given: A subgroup $H \leq G$ and a normal subgroup $N ⊴ G$ .

[L1]

$H N \leq G$ and $H \cap N ⊴ H$ (If $H \leq G$ and $N ⊴ G$ , then $H N$ is a subgroup and $H \cap N ⊴ H$ ).

[L2]

A homomorphism modulo its kernel is isomorphic to its image (First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

[L3]

The canonical quotient map has the given normal subgroup as kernel (The canonical projection $π : G \to G / N$ , $π (g) = g N$ , is a surjective group homomorphism).

[L4]

Quotients by normal subgroups are groups of cosets (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

Proof

technique · direct

1.1

Restrict the quotient map $H N \to H N / N$ to $ϕ : H \to H N / N$ , $ϕ (h) = h N$ ; [L1] and [L4] make this a homomorphism.

L1L2L3L4givenconstruct

2.1

Its kernel is ${h \in H : h N = N} = H \cap N$ , while every $h n N = h N$ shows that its image is $H N / N$ .

step 1.1L1L2L3L4givenalgebra

3.1

The kernel and image calculation in step 2.1 gives $H / (H \cap N) ≅ H N / N$ .

step 2.1∎

Depends on

Used by

Dependency tree · two levels

18 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)

Second isomorphism theorem for groups: H/(H∩N)≅HN/NH/(H∩N)≅HN/N