Let $f,g:\Bbb N\to \Bbb C$ be multiplicative arithmetic functions, i.e. $$\gcd(m,n)=1\implies f(mn)=f(m)f(n)$$ and same for $g$. We can also assume $f(1)\neq 0$ and $g(1)\neq 0$ if necessary. How can I see that the convolution $f*g$ defined by $$f*g(n)=\sum_{d\mid n}f(d)g(n/d)$$ is again multiplicative?
My computation got ugly very quick and I feel I am missing something.
Try it from the right to the left: $$f \ast g(n) f \ast g(m) = \sum_{d_1 \mid n} f(d_1) g(n/d_1) \sum_{d_2 \mid m} f(d_2) g(m/d_2) = \sum_{d_1 \mid n} \sum_{d_2 \mid m} f(d_1) f(d_2) g(n/d_1) g(m/d_2).$$ Now using multiplicativity of $f,g$ and you have to think about why the double sum and $\sum_{d \mid nm}$ coincide.