Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02
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For n≥2, reduction Z→Z/n has kernel nZ and realises Z/n by the first isomorphism theorem

Example

For n≥2, reduction Z→Z/n has kernel nZ and realises Z/n by the first isomorphism theorem.

Facts & Assumptions

Given: An integer n≥2 and ρn:Z→Z/n, ρn(a)=[a]n.

[L1]

The first isomorphism theorem identifies a group modulo a homomorphism kernel with its image (First isomorphism theorem for groups: G/ker⁡f≅im⁡f).

[L2]
[L3]

A group homomorphism preserves the operation (Monoid homomorphism and group homomorphism).

[L4]

The kernel is the inverse image of the identity (The kernel and image of a group homomorphism).

[L5]

The integers form a commutative ring, hence an additive group (The integers form a commutative ring).

Verification

technique · direct
1.1

Since [a+b]n=[a]n+[b]n, ρn is a homomorphism of additive groups.

L1L2L3L4L5givenalgebra
2.1

Its kernel is {a:[a]n=[0]n}=nZ, and every residue class is ρn(a).

step 1.1L1L2L3L4L5givenalgebra
3.1

Therefore the kernel and image calculation yields Z/nZ≅Z/n.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

34 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