Applying the Möbius inversion formula to $f(n) = \sum_{p\mid n}g(p)$?

Question

Is there a specific technique that exists to reducing summation over divisors to prime divisors? Specifically I am interested in applying the Möbius inversion formula to an arithemetic function of the form $$g(n) = \sum_{p\mid n}f(p)$$ The Möbius inversion formula states that for $$g(n) = \sum_{d\mid n} f(d)$$ we can get $$f(n) = \sum_{d\mid n} \mu(d) g\left(\frac{n}{d}\right)$$ However this definition is for the divisors of $n$. How would we get a formula including the prime divisors of $n$. Is there a formula or technique to go from $\sum_{d \mid n}$ to $\sum_{p \mid n}$?

Answer 1

To see that there cannot be such a formula, define arithmetic functions $f_1, f_2,$ where $f_1(n) = 1$ for all $n,$ and $f_2(n) = 1$ if $n$ is prime, $f_2(n) = 0$ if $n$ is composite. Then $$ \sum_{p\mid n}f_1(p) = \sum_{p\mid n}f_2(p) $$ for all $n,$ but $f_1 \ne f_2.$ So the function that creates $g$ from $f$ is not injective, and therefore it has no inverse.