I am reading Serre's a course in arithmetic and I am very confused about the invariance of the order of a modular function (as a meromorphic function) under $SL_2(\mathbb Z)$ action:
The order at $p$ of a meromorphic function $f$ is $v_p(f):=n$, where $n$ is the integer such that $f/(z-p)^n$ is analytic and nonzero at $p$. The following is a screenshot of the book:
Serre says the invariance follows from the identity in the definition of modular forms, but I don't see why. Specifically,
let $g=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$ be a matrix in $SL_2(\mathbb Z)$, let $v_p(f)=n$, the goal is to show that
$$\frac{f(z)}{(z-g.p)^n}=\frac{f(\frac{az+b}{cz+d})}{(cz+d)^{2k}(z-\frac{ap+b}{cp+d})^n}$$
is nonvanishing and holomorphic at $p$. But it is very unclear to me why this should be true. Maybe I am looking at it in the wrong way. Thanks for help!
Update:
From Parthiv's answer below it seems that the interpretation should really be for $g \in SL_2(\mathbb Z)$
(1) $v_p(f)=v_p(f\circ g)$
instead of
(2) $v_p(f)=v_{g(p)}(f)$
But now I wish to see a counterexample for (2) (I don't have a handy example for modular functions).
A modular function of weight $k$ (odd weighted modular functions are identically zero but a priori we don't know that) is meromorphic on the upper half-plane and satisfies $f(gz) = (cz+d)^k f(z)$ for all $g \in SL_2(\mathbb{Z})$. Since $cz+d$ is holomorphic and not equal to zero on the upper half-plane, we have that $(cz+d)^k f(z)$ and $f(z)$ have the same order. By $n = v_{g(p)}(f)$, Serre means $n$ such that $\frac{f(gz)}{(z-p)^n} = (cz+d)^k \frac{f(z)}{(z-p)^n}$ is holomorphic and nonzero in $p$.