$ \sum_{a \mid n} \Bigl( f(a) \sum_{b \mid \frac{n}{a}} g(b) \Bigr) = \sum_{a \mid n} \Bigl( g(a) \sum_{b \mid \frac{n}{a}} f(b) \Bigr) $ Is the following equation true?
$$
\sum_{a \mid n} \Bigl( f(a) \sum_{b \mid \frac{n}{a}} g(b) \Bigr) =
\sum_{a \mid n} \Bigl( g(a) \sum_{b \mid \frac{n}{a}} f(b) \Bigr)
$$
If this equation is true, how can I prove it?
 A: Hint: both sides are equal to $\displaystyle\sum_{\substack{a,b,c\\abc=n}}f(a)g(b)$.
A: We can conveniently prove this identity between arithmetical functions $f$ and $g$ using the Dirichlet convolution
\begin{align*}
(f\ast g)(n)=\sum_{d|n}f(d)g\left(\frac{n}{d}\right)
\end{align*}
The Dirichlet convolution is commutative $f\ast g=g\ast f$ and associative  $f\ast(g\ast h)=(f\ast g)\ast h$. We also use the constant function  $\zeta$ which is defined as
\begin{align*}
\zeta(n)=1\qquad\qquad n\geq 1
\end{align*}

We obtain
\begin{align*}
\color{blue}{\sum_{a|n}f(a)\sum_{b|\frac{n}{a}}g(b)}&=\sum_{a|n}f(a)\sum_{b|\frac{n}{a}}g(b)\zeta\left(\frac{\frac{n}{a}}{b}\right)\\
&=\sum_{a|n}f(a)\left(g\ast\zeta\right)\left(\frac{n}{a}\right)\\
&=\left(f\ast\left(g\ast\zeta\right)\right)(n)\\
&=\left(\left(g \ast\zeta\right)\ast f\right)(n)\\
&=\left(g \ast\left(\zeta\ast f\right)\right)(n)\\
&=\left(g \ast\left(f\ast \zeta\right)\right)(n)\\
&=\sum_{a|n}g(a)\left(f\ast \zeta\right)\left(\frac{n}{a}\right)\\
&=\sum_{a|n}g(a)\sum_{b|\frac{n}{a}}f(b)\zeta\left(\frac{\frac{n}{a}}{b}\right)\\
&\,\,\color{blue}{=\sum_{a|n}g(a)\sum_{b|\frac{n}{a}}f(b)}\\
\end{align*}
and the claim follows.

