How to show that cusp of congruence subgroup $\Gamma_0(4)$ is $0,\frac{1}{2},\infty$？ 
Let $\Gamma_0(4)$ be a congruence subgroup of $SL(2,\mathbb{Z})$ defined as
  $$\Gamma_0(4)=\Big\{M=\begin{pmatrix}
a &b\\
c& d
\end{pmatrix}\in SL(2,\mathbb{Z}) |  c \equiv 0\bmod 4\Big\}.$$
  How to show that cusp of $\Gamma_0(4)$ is $0,\frac{1}{2},\infty$？I have looked up the answer: Inequivalent cusps of $\Gamma_0(4)$, but I do not understand. 
  We define the group action: if $\forall \gamma \in SL(2,\mathbb{Z})$, 
  $$ \gamma \infty= \begin{cases}  \frac{a}{c}, c\neq 0\\ \infty, c=0 \end{cases}$$
$$ \gamma z= \begin{cases}  \frac{az+b}{cz+d}, cz+d \neq 0\\ \infty, cz+d=0 \end{cases}$$

I try to use the same method of proof $\Gamma_0(2)$ cusp $0,\infty$. 
For $\forall \frac{p}{q}$, if $q$ is even, there exists $r,s\in \mathbb{Z}$, s.t. $rp-sq=1$, we have 
$$\begin{pmatrix}
p &s\\
q& r
\end{pmatrix} \infty = \frac{p}{q},$$
where $\begin{pmatrix}
p &s\\
q& r
\end{pmatrix} \in \Gamma_0(2)$.
if $q$ is odd, there exists $r,s\in \mathbb{Z}$, sinece $\gcd(2p,q)=1$, s.t. $-2rp+sq=1$, we have 
$$\begin{pmatrix}
s &p\\
2r& q
\end{pmatrix} 0 = \frac{p}{q}, $$
where $\begin{pmatrix}
s &p\\
2r& q
\end{pmatrix} \in \Gamma_0(2)$.
Similarly, for $\Gamma_0(4)$, I have known $0, \infty$ and $0,\frac{1}{2}$ are not equivalent. $\textbf{But for $\forall \frac{p}{q}\in \mathbb{Q}$, how to find the matrix $\gamma \in \Gamma_0(4)$}$,  s.t. 
$$ \gamma \frac{1}{2}=\frac{p}{q}.$$
If $q \equiv 0 \mod4$, $\gcd(p,q)=1$, it is easy to seek the 
$$\begin{pmatrix}
p &s\\
q& r
\end{pmatrix} \infty = \frac{p}{q},$$
where $\begin{pmatrix}
p &s\\
q& r
\end{pmatrix} \in \Gamma_0(4)$.
How about the $q \equiv 1,2,3 \mod4$?
 A: I am using definitions from Milne's notes on modular forms. The relevant part is on "classification of linear fractional transformations".
A point $s \in \mathbb{R} \cup \{\infty\}$ is called a cusp of $\Gamma_0(4)$ if it is fixed by a parabolic element $\gamma \in \Gamma_0(4)$ with $s$ fixed by $\gamma$.
I found $\alpha = \begin{pmatrix}-1&1\\-4&3\end{pmatrix}$ is in $\Gamma_0(4)$ because $c = -4 \equiv 0 \mod 4$. It is also parabolic since $\text{Tr}(\alpha) = -1 + 3 = 2$. Moreover, $\alpha\frac{1}{2}$ is $\frac{1}{2}$.
I found this $\alpha$ by first finding a matrix $\beta$ which sends $\frac{1}{2}$ to $\infty$. And then I took $\alpha = \beta^{-1} T \beta$ where $T = \begin{pmatrix}1&1\\0&1\end{pmatrix}$.
The $\beta$ which I found is $\begin{pmatrix}1&0\\-2&1\end{pmatrix}$. An easy calculation shows that the negative of the square of the lower-left entry gives you the lower-left entry of the $\beta^{-1}T\beta$. Thus, you are assured here that $c \equiv 0 \mod 4$.
I'm not sure if you're using an equivalent definition, but that's how I would show that $\frac{1}{2}$ is a cusp.
A: This is addressed in the third erratum of the book at the page 103 note
http://people.reed.edu/~jerry/MF/errata3.pdf
You should check all four errata if you have not done so yet, as there are quite a few mistakes that can make things confusing.
You might also want to look at this 
Modular Forms: Find a set of representatives for the cusps of $\Gamma_0(4)$
