I am studying A Classical Introduction to Modern Number Theory by Ireland and Rosen, and this is exercise 9 from chapter 2.
The authors define a function $f$ to be multiplicative if for all $a, b$ with $\gcd(a,b)=1$, $f(ab)=f(a)f(b)$.
The exercise is to show that given a multiplicative function $f$, we can construct another function $$h(n) = \sum_{d\mid n} \mu(n/d) f(d)$$
How do I show that this function is multiplicative? $\mu$ is the Mobius function. Thank you in advance.
More generally, suppose that $g$ is another multiplicative function. You want to see that $$F(n)=\sum_{ab\mid n}f(a)g(b)$$ is multiplicative. The key is to see how the set of divisors $ab\mid mn$ partition when $(m,n)=1$. After that, it should be cake. Good luck.