Answer to Question #272610 in Combinatorics | Number Theory for Prathibha Rose

Question

Prove that the set of arithmetic functions form an abelian group under the

mapping Dirichlet multiplication with f1≠0

Answer 1

Let f and g be arithmetic functions. We define the Dirichlet product of f and g by

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

Let S be the set of arithmetic functions f such that $f(1)≠0$

Since $f * g (1) = f (1) g (1)$ , then S is closed under Dirichlet multiplication. Furthermore, simple algebraic manipulation shows * to be both commutative and associative.

Let 

$e(n)=\begin{cases} 1 &\text{if } n=1 \\ 0 &\text{otherwise } \end{cases}$

It is clear that $e * f = f$ for all arithmetic functions f, so that e is the identity element. Finally, we must show that, given f ∈ S, there exists an f⁻¹∈ S such that $f * f ^{−1} = e$ . Given f, we will construct f⁻¹ inductively. First, we need $f *f ^{−1} (1) = e (1)$ , which occurs if and only if

$f (1) f^{ −1} (1) = 1$ . Since $f (1) \neq 0$ , f^-1 is uniquely determined. Now, assume that n > 1 and f⁻¹ has been determined for all k < n. Then we have

$f*f^{-1}(n)=\displaystyle \sum_{d|n}f(d)f^{-1}(n/d)\implies -f(1)*f^{-1}(n)=\displaystyle \sum_{d|n,d>1}f(d)f^{-1}(n/d)$

This uniquely determines f^-1(n). Thus, we may uniquely determine an f^-1  for all f ∈ S.