PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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For every $n \in N$ , the congruence-class group $(Z / n, +)$ is the quotient group $(Z, +) / n Z$

Statement

For every $n \in N$ , view $n$ as its canonical nonnegative integer and put $n Z : = {n k : k \in Z}$ . Then the left cosets of $n Z$ in $(Z, +)$ are exactly the congruence classes modulo $n$ , and coset addition is the published addition of congruence classes. Thus

$(Z, +) / n Z = (Z / n, +)$

as the same group on the same underlying set. This includes $n = 0$ and $n = 1$ .

Facts & Assumptions

Given: A natural number $n$ , viewed in $Z$ under the canonical embedding, and the set $n Z = {n k : k \in Z}$ .

[L1]

The integers form a commutative ring with identity (The integers form a commutative ring), and the canonical embedding of $N$ preserves addition and multiplication (The naturals embed in the integers).

[F1]

A subset of a group is a subgroup when it contains the identity and is closed under the operation and inverses (Subgroup).

[L2]

Every subgroup of an abelian group is normal (Every subgroup of an abelian group is normal).

[F2]

The congruence $x \equiv a (m o d n)$ means that $x - a = n q$ for some $q \in Z$ (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

[F3]

The congruence class is $[a]_{n} = {x \in Z : x \equiv a (m o d n)}$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ ).

[F4]

Addition modulo $n$ is $[a]_{n} + [b]_{n} = [a + b]_{n}$ (Addition and multiplication on $Z / n$ by $[a]_{n} + [b]_{n} = [a + b]_{n}$ and $[a]_{n} [b]_{n} = [a b]_{n}$ ).

[L3]

The cosets of a normal subgroup form a group under $(a + N) + (b + N) = (a + b) + N$ (For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$ ).

[L4]

For every $n \in N$ , $(Z / n, +, [0]_{n})$ is an abelian group, including at $n = 0$ and $n = 1$ (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

Proof

technique · direct

1.1L1F1algebra

The set $n Z$ contains $0 = n 0$ ; if $n k, n ℓ \in n Z$ , then $n k + n ℓ = n (k + ℓ)$ and $- (n k) = n (- k)$ also lie in $n Z$ . Hence $n Z \leq (Z, +)$ .

1.2F2F3algebra

For $a, x \in Z$ , one has $x \in a + n Z$ if and only if $x = a + n q$ for some $q \in Z$ , if and only if $x \equiv a (m o d n)$ , if and only if $x \in [a]_{n}$ . Therefore $a + n Z = [a]_{n}$ .

2.1step 1.1L2

Since $(Z, +)$ is abelian, the subgroup $n Z$ is normal.

2.2step 1.2F4L3

Under the equality in step 1.2, [L3] and [F4] give $(a + n Z) + (b + n Z) = (a + b) + n Z = [a + b]_{n} = [a]_{n} + [b]_{n}$ .

3.1step 2.1step 1.2step 2.2L4∎

Steps 2.1, 1.2, and 2.2 show that the quotient group and the group of congruence classes have the same underlying set and operation; [L4] confirms the published group convention, including $n = 0$ and $n = 1$ .

Depends on

Dependency tree · 2 levels

Sources

Ernst, An Inquiry-Based Approach to Abstract Algebra, Quotients of Groups