Multiplicative functions form abelian group under Dirichlet product

Statement

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

$f (1) = 1$ ,
$f (m n) = f (m) f (n)$ whenever $m, n$ are relatively prime.

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

$(f * g) (n) = \sum_{d | n} f (d) g (n / d)$ .

Then, $M$ forms an abelian group under the Dirichlet product, with multiplicative identity given by the function $I$ that takes the value $1$ at $1$ and $0$ 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 $f, g$ .

To prove: $f * g$ is multiplicative.

Proof: Suppose $m$ and $n$ are relatively prime natural numbers. Consider:

$(f * g) (m n) = \sum_{d | m n} f (d) g (m n / d)$ .

Given any divisor $d$ of $m n$ , $d$ can be expressed uniquely as $d_{1} d_{2}$ , where $d_{1} | m$ and $d_{2} | n$ . Conversely, given $d_{1} | m$ and $d_{2} | n$ , $d_{1} d_{2} | m n$ . Thus, we have:

$(f * g) (m n) = \sum_{d_{1} | m} \sum_{d_{2} | n} f (d_{1} d_{2}) g ((m n) / (d_{1} d_{2}))$ .

Next, since $d_{1}$ and $d_{2}$ are relatively prime and $m / d_{1}$ and $n / d_{2}$ are also relatively prime, and $f$ and $g$ are multiplicative:

$(f * g) (m n) = \sum_{d_{1} | m} \sum_{d_{2} | n} f (d_{1}) f (d_{2}) g (m / d_{1}) g (n / d_{2})$ .

Next, we rearrange:

$(f * g) (m n) = (\sum_{d_{1} | m} f (d_{1}) g (m / d_{1})) (\sum_{d_{2} | n} f (d_{2}) g (n / d_{2}))$ .

Thus:

$(f * g) (m n) = (f * g) (m) (f * g) (n)$ .

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 $I$ 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 $f$ , the inverse of $f$ can be defined inductively as follows:

$g (1) = 1, g (n) = - \sum_{d | n, 1 \leq d < n} g (d) f (n / d)$ .

Since $f (1) = 1$ , this clearly satisfies:

$g (1) f (1) = 1, \sum_{d | n} g (d) f (n / d) = 0 \forall n > 1$ .

Thus, $g * f = I$ . Since the multiplication is commutative, $g$ is a two-sided inverse for $f$ .

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