Why $f \mid V(m) \mid U(m) = f$ for modular form $f$, where $U, V$ are $U, V$-operators?

Question

I think this has an easy answer, but I am having brain fart. If I have a modular form $f = \sum_{n \gg -\infty} a(n)q^{n}$, and $U, V$ are the $U$ and $V$-operators: $$f \mid U(m) := \sum_{n} a(mn)q^{n},$$ $$f \mid V(m) := \sum_{n} a(n)q^{mn},$$ then why is it true that $f \mid V(m) \mid U(m) = f$?

Wouldn't $f \mid V(m) \mid U(m) = \sum_{n} a(mn)q^{mn} \neq \sum_{n} a(n)q^{n}$?

Answer 1

Write $$f \mid V(m) = a(0) + a(1)q^m + a(2)q^{2m} +... =\sum b(n)q^n $$ where $b(n) = a(n/m)$ if $m \mid n$ and $b(n) = 0$ otherwise.

Then $$\left(\sum b(n)q^n \right)\mid U(m) = \sum b(mn)q^n = b(0) + b(m)q + b(2m)q^2 + ... = \sum a(n)q^n.$$