📂Number Theory

Arithmetic Functions of Abelian groups

Theorem ¹

Given a set of arithmetic functions $A = \left\{ f : \mathbb{N} \to \mathbb{C} \mid f(1) \ne 0 \right\}$ other than $f(1) \ne 0$ and binary operation $\ast$, $(A,*)$ is an Abelian group.

Description

Strictly speaking, not all sets of arithmetic functions can be Abelian groups, due to the last condition for an algebraic structure to be a group, the existence of an inverse element. Fortunately, this condition is not too difficult and is sufficiently satisfied by $f(1) \ne 0$.

Proof

A monoid $\left< G, \ast\ \right>$ with an element $a$ and identity element $e$ is defined as a group if there exists an $a '$ that satisfies $a \ast\ a’ = a’ \ast\ a = e$. Thus, a group is a binary operation structure that satisfies the following properties:

• (i): Associative law applies to the operation.
• (ii): An identity element exists for every element.
• (iii): An inverse element exists for every element.

Additionally, if the next condition is satisfied, it is called an Abelian group.

• (iv): Commutative law applies to the operation.

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Part (i), (iv). Associative and Commutative Law

Properties of Convolution

• Associative Law: $\left( f \ast g \right) \ast k = f \ast (g \ast k)$
• Commutative Law: $f \ast\ g = g \ast\ f$

All arithmetic functions satisfy associative and commutative laws.

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Part (0). Closure with respect to $\ast$

If $f,g \in A$ and $h = f \ast\ g$, then since $f(1) \ne 0$ and $g(1) \ne 0$, $$h(1) = \left( f \ast g \right)(1) = \sum_{d \mid 1} f(d) g \left( {{ 1 } \over { d }} \right) = f(1)g(1) \ne 0$$ Hence, $f \ast g = h \in A$ is true.

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Part (ii). Identity Element

Identity: An arithmetic function $I$ defined as follows is called the identity function. $$I(n) := \left[ {{ 1 } \over { n }} \right]$$

Since $I(1) = 1 \ne 0$, then $I \in A$ is true. For all arithmetic functions, the Identity $I$ exists as the identity element of $( A,*)$ by satisfying the following. $$I \ast\ f = f \ast\ I = f$$

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Part (iii). Inverse Element

Inverse of Convolution: If an arithmetic function $f$ satisfies $f(1) \ne 0$, then its inverse $f^{-1}$ uniquely exists.

Given the premise of $f(1) \ne 0$, the inverse $f^{-1}$ uniquely exists.

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