# Multiplicative functions form abelian group under Dirichlet product

## Contents

## Statement

Suppose is the set of natural numbers and is a commutative unital ring. Let be the set of all multiplicative functions from to , i.e., all the functions satisfying the following:

- ,
- whenever are relatively prime.

Consider the Dirichlet product, a binary operation defined on functions from to :

.

Then, forms an abelian group under the Dirichlet product, with multiplicative identity given by the function that takes the value at and elsewhere.

## Proof

### Closure under Dirichlet product

We first need to show that the Dirichlet product is a well-defined binary operation on multiplicative functions; in other words, that a Dirichlet product of multiplicative functions is multiplicative.

**Given**: Multiplicative functions .

**To prove**: is multiplicative.

**Proof**: Suppose and are relatively prime natural numbers. Consider:

.

Given any divisor of , can be expressed uniquely as , where and . Conversely, given and , . Thus, we have:

.

Next, since and are relatively prime and and are also relatively prime, and and are multiplicative:

.

Next, we rearrange:

.

Thus:

.

### Associativity and commutativity

These follow from the fact that the Dirichlet product is associative and commutative for *all* functions:

### Identity element

This follows from the fact that the function is an identity element for the Dirichlet product for *all* functions: Identity element for Dirichlet product is indicator function for one.

### Inverses

Given a multiplicative function , the inverse of can be defined inductively as follows:

.

Since , this clearly satisfies:

.

Thus, . Since the multiplication is commutative, is a two-sided inverse for .

Next, we need to verify that is multiplicative. *Fill this in later*