LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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In a group extension the kernel is normal and the quotient recovers the base

Statement

Let

$1 \to N \overset{i}{\to} E \overset{π}{\to} Q \to 1$

be a group extension. Then $i (N) = \ker π$ is a normal subgroup of $E$ , and the quotient $E / i (N)$ is canonically isomorphic to $Q$ .

Facts & Assumptions

Given: The displayed short exact sequence of groups.

[L1]

In a short exact sequence, the image of the first map equals the kernel of the second (Group extensions, sections, complements, and split extensions).

[L2]

The kernel of a group homomorphism is a normal subgroup of its domain (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup).

[L3]

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

Proof

technique · direct

1.1

By [L1], $i (N) = \ker π$ . Since $\ker π$ is normal in $E$ by [L2], the subgroup $i (N)$ is normal in $E$ .

L1L2

2.1

The map $π$ is surjective because the sequence is exact, so its image is $Q$ . Applying [L3] to $π$ and using step 1.1 gives $E / i (N) = E / \ker π ≅ Q$ .

L1L3step 1.1∎

Depends on

Used by

Dependency tree · two levels

13 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

J. S. Milne, Group Theory (standard reference, not scraped)