Let $\Gamma_0(4)$ be a congruence subgroup of $SL(2,\mathbb{Z})$ defined as
$$\Gamma_0(4)=\{M=\begin{pmatrix} a &b\\ c& d \end{pmatrix}\in SL(2,\mathbb{Z})\mid c=0\bmod 4\}.$$
How to prove that this group has no elements of finite order except $-1$ and the action of $\Gamma_0(4)/\pm1$ on the complex upper-half plane $\mathbb{H}$ is free?
If $M$ were of a finite order, say $n$, then the characteristic polynomial $$\chi(T)=T^2-(a+d)T+(ad-bc)$$ of $M$ can only have factors that are cyclotomic polynomials of order $d\mid n$. The cyclotomic polynomial $\Phi_d(T)$ is of degree $\phi(d)$, where $\phi$ is the Euler totient function.
We know that $\phi(n)=1$ only when $n=1,2$, and $\phi(n)=2$ only when $n=3,4,6$. Taking into account the determinant condition $ad-bc=1$ this leaves the following alternatives for $\chi(T)$: $$\chi(T)\in\{T^2+1,T^2\pm T+1, T^2\pm 2T+1\}.$$ The last case implies $M=\mp I$. In the first two cases we have $a+d=0$ or $a+d=\pm1$. Because $bc\equiv0\pmod4$ we must have $ad\equiv1\pmod 4$. This congruence implies that $a+d\equiv2\pmod4$, so we have our contradiction.
From the action of $\Gamma=SL_2(\Bbb{Z})/\{\pm I\}$ on $\Bbb{H}$ we know that the non-trivial stabilizers have orders $2$ or $3$. When lifted to matrices in those stabilizers have elements of order $3,4$ or $6$ - exactly the possibilities we excluded above.
