TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31

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For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold

Statement

For every $n \in N$ , with addition and multiplication as in Addition and multiplication on $Z / n$ by $[a]_{n} + [b]_{n} = [a + b]_{n}$ and $[a]_{n} [b]_{n} = [a b]_{n}$ :

$(Z / n, +, [0]_{n})$ is an abelian group (Group and abelian group), with $- [a]_{n} = [- a]_{n}$ ;
$(Z / n, \cdot, [1]_{n})$ is a commutative monoid (Semigroup and monoid);
multiplication distributes over addition on both sides.

The assertions include $n = 0$ and $n = 1$ . At $n = 1$ , the two distinguished identities coincide because $[0]_{1} = [1]_{1}$ .

Facts & Assumptions

Given: A natural number $n$ and classes $[a]_{n}, [b]_{n}, [c]_{n}$ in $Z / n$ .

[F1]

$[u]_{n} + [v]_{n} = [u + v]_{n}$ and $[u]_{n} [v]_{n} = [u v]_{n}$ , and these values are independent of representatives (Addition and multiplication on $Z / n$ by $[a]_{n} + [b]_{n} = [a + b]_{n}$ and $[a]_{n} [b]_{n} = [a b]_{n}$ ).

[L1]

$Z$ is a commutative ring: addition and multiplication are associative and commutative, $0$ and $1$ are identities, every integer has an additive inverse, and multiplication distributes over addition (The integers form a commutative ring).

[F2]

An abelian group is an associative commutative binary operation with an identity and inverses; a commutative monoid is an associative commutative binary operation with an identity (Group and abelian group, Semigroup and monoid).

Proof

technique · direct

1.1F1L1

Associativity and commutativity of addition follow from $([a]_{n} + [b]_{n}) + [c]_{n} = [(a + b) + c]_{n} = [a + (b + c)]_{n} = [a]_{n} + ([b]_{n} + [c]_{n})$ and $[a + b]_{n} = [b + a]_{n}$ .

1.2F1L1

The class $[0]_{n}$ is an additive identity, and $[- a]_{n}$ is an additive inverse of $[a]_{n}$ , since $[a + 0]_{n} = [a]_{n}$ and $[a + (- a)]_{n} = [0]_{n}$ .

1.3F1L1

Associativity and commutativity of multiplication follow from $[(a b) c]_{n} = [a (b c)]_{n}$ and $[a b]_{n} = [b a]_{n}$ , while $[1]_{n}$ is a multiplicative identity because $[a \cdot 1]_{n} = [a]_{n}$ .

1.4F1L1

Left distributivity is $[a]_{n} ([b]_{n} + [c]_{n}) = [a (b + c)]_{n} = [a b + a c]_{n} = [a]_{n} [b]_{n} + [a]_{n} [c]_{n}$ ; right distributivity follows identically, or from commutativity.

2.1step 1.1step 1.2step 1.3step 1.4F2∎

Steps 1.1 and 1.2 verify the abelian-group clauses, step 1.3 verifies the commutative-monoid clauses, and step 1.4 gives both distributive laws.

Depends on

Used by

…and 43 more results.

Dependency tree · two levels

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Sources

K. Conrad, Modular Arithmetic (standard reference, not scraped)

For every natural nn, (Z/n,+)(Z/n,+) is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold