The principal congruence subgroup Gamma(2)

The principal congruence subgroup of level $2$ is

$\mathrm{\Gamma }(2):=\{\gamma \in S{L}_2(\mathbb{Z}):\gamma \equiv I(\mathrm{m}\mathrm{o}\mathrm{d}2)\},$

the congruence $\gamma \equiv I(\mathrm{m}\mathrm{o}\mathrm{d}2)$ being entrywise (Congruence modulo an integer: $a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}n)$ when $n\mid (a-b)$, including the moduli $0$ and $1$, The congruence class $[a{]}_n$ and the quotient set $\mathbb{Z}\mathrm{/}n$, The modular group and its action on the upper half-plane).

It is the kernel of the entrywise reduction homomorphism $\rho :S{L}_2(\mathbb{Z})\to S{L}_2({\mathbb{F}}_2)$: reduction of entries is a group homomorphism because addition and multiplication of residues are compatible with the operations on $\mathbb{Z}$, and it lands in $S{L}_2({\mathbb{F}}_2)$ because $\det \gamma =1$ reduces to $1$ (Monoid homomorphism and group homomorphism, The kernel and image of a group homomorphism). Hence $\mathrm{\Gamma }(2)$ is a normal subgroup of $S{L}_2(\mathbb{Z})$ (First isomorphism theorem for groups: $G\mathrm{/}\ker f\cong \mathrm{im}f$) and it contains $\pm I$, since $-1\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}2)$.

The reduction is surjective. Indeed $S{L}_2({\mathbb{F}}_2)=G{L}_2({\mathbb{F}}_2)$ (the only nonzero scalar in ${\mathbb{F}}_2$ is $1$), and it has order $6$: its first column is any of the three nonzero vectors of ${\mathbb{F}}_2^2$, and then the second column is any of the two vectors outside the span of the first, the resulting matrix being automatically invertible. The images $\bar{S},\bar{T}$ of $S,T\in S{L}_2(\mathbb{Z})$ lie in the image of $\rho$ and generate $S{L}_2({\mathbb{F}}_2)$: ${\bar{T}}^2=1$, $(\bar{S}\bar{T}{)}^3=1$ by reduction of $S^2=(ST{)}^3=-I\equiv I$, and $\bar{S}\mathrm{\notin }⟨\bar{T}⟩$, so $⟨\bar{S},\bar{T}⟩$ has order divisible by $2$ and $3$ and at most $6$, hence equals $S{L}_2({\mathbb{F}}_2)$ (The modular group and its action on the upper half-plane). So $\rho$ is onto, and the first isomorphism theorem identifies $S{L}_2(\mathbb{Z})\mathrm{/}\mathrm{\Gamma }(2)$ with $S{L}_2({\mathbb{F}}_2)$; therefore

$[S{L}_2(\mathbb{Z}):\mathrm{\Gamma }(2)]=\mathrm{\mid }S{L}_2({\mathbb{F}}_2)\mathrm{\mid }=6$

(First isomorphism theorem for groups: $G\mathrm{/}\ker f\cong \mathrm{im}f$, If $[G:N]$ is finite then $\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }=[G:N]$; for finite $G$ this equals $\mathrm{\mid }G\mathrm{\mid }\mathrm{/}\mathrm{\mid }N\mathrm{\mid }$, For $N⊴G$, the cosets form a group with identity $N$ and inverse $(gN{)}^{-1}=g^{-1}N$).

Finally let $\bar{\mathrm{\Gamma }}(2)\le PS{L}_2(\mathbb{Z})$ be the image of $\mathrm{\Gamma }(2)$ under the quotient map $S{L}_2(\mathbb{Z})\to PS{L}_2(\mathbb{Z})$. Since $\rho$ kills $-I$, it factors through that quotient and defines a surjective homomorphism $PS{L}_2(\mathbb{Z})\to S{L}_2({\mathbb{F}}_2)$ whose kernel is exactly $\bar{\mathrm{\Gamma }}(2)$. As $\mathrm{\Gamma }(2)$ contains $\ker (S{L}_2(\mathbb{Z})\to PS{L}_2(\mathbb{Z}))=\{\pm I\}$, the correspondence of subgroups in the quotient gives $[PS{L}_2(\mathbb{Z}):\bar{\mathrm{\Gamma }}(2)]=[S{L}_2(\mathbb{Z}):\mathrm{\Gamma }(2)]=6$, and $\bar{\mathrm{\Gamma }}(2)$ is the kernel of that reduction (First isomorphism theorem for groups: $G\mathrm{/}\ker f\cong \mathrm{im}f$).