Sum (over divisors of a number) manipulation Prove or disprove
$$
\sum_{d \mid m}\sum_{k \mid d}f\left(\frac{m}{d}\right)g(k) = \sum_{k \mid m}\sum_{d \mid (m/k)}f\left(\frac{m}{d}\right)g(k)
$$
where k and d are index variables, m is a positive integer and $d \mid m$  means d divides m.
If the above equality doesn't hold, what can be written for the left hand side of the equation after manipulating the sum ? and prove it.
 A: Let us denote $\sum_{d\mid n} f(d)g(n/d)$ as $(f*g)(n)$. $*$ is the multiplicative/Dirichlet convolution between arithmetic functions, and it is both associative and commutative. Let us denote as $\mathbb{1}$ the function such that $\mathbb{1}(n)=1$ for any $n\in\mathbb{N}^+$. The LHS equals
$$ \sum_{d\mid m} f\left(\frac{m}{d}\right)(g*\mathbb{1})(d) = (f*g*\mathbb{1})(m)$$
and by reindexing the RHS equals:
$$\begin{eqnarray*} \sum_{k\mid m}\sum_{d\mid(m/k)}f\left(\frac{m}{d}\right)g(k) &\stackrel{d\mapsto \frac{m}{kd}}{=}& \sum_{k\mid m}\sum_{d\mid(m/k)}f\left(kd\right)g(k) \\&\stackrel{k\mapsto\frac{m}{k}}{=} &\sum_{k\mid m}\sum_{d\mid k}f\left(\frac{md}{k}\right)g\left(\frac{m}{k}\right)\\&\stackrel{d\mapsto\frac{k}{d}}{=}&\sum_{k\mid m}\sum_{d\mid k}f\left(\frac{m}{d}\right)g\left(\frac{m}{k}\right)\end{eqnarray*}$$
or, by denoting $f(an)$ as $f_a(n)$,
$$ \sum_{k\mid m}g(k) \left(f_{k}*1\right)\left(\frac{m}{k}\right)\neq \sum_{k\mid m}g(k) \left(f*1\right)\left(\frac{m}{k}\right)=\text{LHS}. $$
