Let $f,g$ be arithmetic functions. According to Wikipedia, $(f*g)^{-1} = f^{-1} * g^{-1}$ if $f(1) \neq 0$ and $g(1) \neq 0$.
However, it is not clear to me why this is true, and the statement does not provide a reference text for a proof. Would anyone be able to provide a link to one and/or prove it?
Actually we can look at this question from the group theory perspective. If $a,b$ are elements of an abelian group $G$, then $(a*b)^{-1}$ is $a^{-1} * b^{-1}$. So we just need to prove that all arithmetic functions $f$ such that $f(1) \neq 0$ forms an abelian group.
By definition, the Dirichlet product is commutative and associative. The neutral (i.e. identity) element is
$$I(n) = \begin{cases} 1 & n = 1 \\ 0 & n \neq 1 \end{cases}.$$
The inverse of the function $f$ can be found recursively by $f^{-1}(n)$ = $-1/f(1)$ * $\displaystyle\sum_{d|n , d<n}^\mathbb{}f(d)*f^{-1}(n/d)$. Hence we have an abelian group, and our identity is correct because for every group: $(a*b)^{-1}$ * $(a*b) = e = a^{-1} * b^{-1}$ * $(a*b)$ so we have $(a*b)^{-1}$ = $a^{-1} * b^{-1}$