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For n≥2n\ge2, reduction Z→Z/n\mathbb Z\to\mathbb Z/n has kernel nZn\mathbb Z and realises Z/n\mathbb Z/n by the first isomorphism theorem

Example

For n≥2n\ge2, reduction Z→Z/n\mathbb Z\to\mathbb Z/n has kernel nZn\mathbb Z and realises Z/n\mathbb Z/n by the first isomorphism theorem.

Facts & Assumptions

Given: An integer n≥2n\ge2 and ρn:Z→Z/n\rho_n:\mathbb Z\to\mathbb Z/n, ρn(a)=[a]n\rho_n(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⁡fG/\ker f\cong\operatorname{im}f).

[L2]

(Z,+)/nZ(\mathbb Z,+)/n\mathbb Z is the congruence-class group (Z/n,+)(\mathbb Z/n,+) (For every n∈Nn\in\mathbb N, the congruence-class group (Z/n,+)(\mathbb Z/n,+) is the quotient group (Z,+)/nZ(\mathbb Z,+)/n\mathbb Z).

[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[a+b]_n=[a]_n+[b]_n, ρn\rho_n is a homomorphism of additive groups.

L1L2L3L4L5givenalgebra
2.1

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

step 1.1L1L2L3L4L5givenalgebra
3.1

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

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 62 results over 17 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