$f$ is multiplicative $\implies f^{-1}$ is multiplicative.

Question

Let $f$ be a multiplicative function i.e. $f(mn)=f(m)f(n)$ for all $m,n$ satisfying $\gcd(m,n)=1$ and $f\not\equiv 0$. Define $f^{-1}$ to be the function $g$ such that $f*g=I$ where $I(n)=1$ if $n=1$ and $0$ otherwise. Here $*$ represents dirichlet convolution defined by $(f*g)(n)=\sum\limits_{d|n} f(d)g(\frac{n}{d})$. Now I need to show that $g$ is multiplicative. How to show that? Can someone help me?

Answer 1

From here we know that

$$g(n)=-\sum_{d|n,d<n}f\left(\frac{n}{d}\right)g(d)$$

for all $n>1$ (with $g(1)=1$). Then we can prove $g(n)$ is multiplicative inductively on the prime factors of $n$. Suppose that $p$ is a prime such that $p\not|n$. Then for any $k$ we have

$$g(np^k)=-\sum_{d|np^k,d<np^k}f\left(\frac{np^k}{d}\right)g(d)$$

$$=-\sum_{r=0}^k\sum_{d|n,d<n}f\left(\frac{np^k}{dp^r}\right)g(dp^r)-\sum_{r=0}^{k-1}f\left(\frac{np^k}{np^r}\right)g(np^r)$$

$$=-\left[\sum_{r=0}^kf(p^{k-r})g(p^r)\right]\left[\sum_{d|n,d<n}f\left(\frac{n}{d}\right)g(d)\right]-g(n)\sum_{r=0}^{k-1}f\left(p^{k-r}\right)g(p^r)$$

Note that in this last step we are using our induction assumption to say that $g(dp^r)=g(d)g(p^r)$ and $g(np^r)=g(n)g(p^r)$. To make this formal, one would need to do a double induction proofover the unique prime factors and the multiplicities of the prime factors of the input argument, but the gist of the logic is here and can be easily modified. We then have

$$=-\left[\sum_{r=0}^kf(p^{k-r})g(p^{r})\right]\left[g(n)+\sum_{d|n,d<n}f\left(\frac{n}{d}\right)g(d)\right]+g(n)g(p^k)$$

Notice we have subtracted and added the term $g(n)g(p^k)$ to get this arrangement. Then shifting indices gives

$$=-\left[\sum_{r=0}^kf(p^{r})g(p^{k-r})\right]\left[g(n)+\sum_{d|n,d<n}f\left(\frac{n}{d}\right)g(d)\right]+g(n)g(p^k)$$

However, the left most term in square brackets is simply $I(p^k)=0$. Thus, this simplifies to $g(n)g(p^k)$ as desired.