If $(f \ast g)(n) = \sum\limits_{d|n}f(d)g(\frac{n}{d})$, prove that $\ast$ is commutative.

Question

If $(f \ast g)(n) = \sum\limits_{d|n}f(d)g(\frac{n}{d})$, prove that $\ast$ is commutative. I believe this is convolution.

My attempt at the proof:

$(f \ast g)(n) = \sum\limits_{d|n}f(d)g(\frac{n}{d})$

$= \sum\limits_{d|n} f(\frac{n}{k})g(k)$ ($d|n \Rightarrow \exists k$ s.t. $n = kd$)

$= \sum\limits_{d|n}g(k)f(\frac{n}{k})$

$= (g \ast f)(n)$

Note that I found this question in an abstract algebra book.

Answer 1

Hints: $$(f\ast g)(n)=\sum_{i=1}^{+\infty}\sum_{j=1}^{+\infty}f(i)g(j)\mathbf 1_{ij=n}\qquad\&\qquad\mathbf 1_{ij=n}=\mathbf 1_{ji=n}. $$ Notation: for every property $P$, $\mathbf 1_P=1$ if $P$ is true and $\mathbf 1_P=0$ if $P$ is false.

Answer 2

Write it as $$\sum_{ab=n}f(a)g(b)$$