Reading through Tom Apostol's "Introduction to Analytic Number Theory, page 162. One finds the following theorem:
Let $s_k(n) = \sum_{d | (n,k)} f(d) g(\frac k d)$ where $f$ and $g$ are multiplicative, then we have $s_{mk}(ab) = s_m(a) s_k(b)$ whenever $(a,k) = (m,k) = 1$.
Note that $s_k$ is a generalization of Ramanujan' sums. I formalized all the proof. But then realized that I was assuming complete multiplicativity (the difference being that in multiplicativity you only assume $f(ab) = f(a)f(b)$ for coprime $a,b$). But then, I am unable to show:
$g(\frac{mk}{d_1d_2}) = g(\frac{m}{d_1}) \cdot g(\frac{k}{d_2})$ for $d_1 | (a,m),d_2 | (b,k)$
this step is necessary for the proof. What is happening here?
$$s_k(n) = \sum_{de = k} f(d)1_{d | n} g(e)$$
Then $s_{mk}(ab) = s_m(a) s_k(b)$ when $(a,k) = (b,m) = 1$ and $f$ is multiplicative and $g$ is completely multiplicative
$$s_m(a)s_k(b) =\sum_{d_1e_1 = m} f(d_1)1_{d_1 | a}g(e_1)\sum_{d_2e_2=k}f(d_2)1_{d_2| b}g(e_2)\\=\sum_{d_1e_1 = m,d_2e_2=k} f(d_1d_2)1_{d_1 | a,d_2| b}g(e_1e_2) \\=\sum_{d_1d_2e = mk, d_1 | m, d_2 | k} f(d_1d_2)1_{d_1 | a,d_2| b}g(e) \\=\sum_{d_1d_2e = mk, d_1 | (a,m), d_2 | (b,k)} f(d_1d_2)g(e) \\=s_{mk}(ab)$$
$e_1,e_2$ are not always coprime : $a = b = 1, k = m=2, d_1=d_2=1,e_1=e_2=2$.
My guess is that Apostol meant $s_k(n) = \sum_{de = k} f(d)1_{d | n,e | n} g(e)$ in which case it is better behaved.