if I denotes the principal congurence group of level 2 i.e. $I=\{ M \in SL(2,Z) ; \:M \:\:\text{congruent to I} \mod(2)\}$.

or I= \begin{bmatrix}2\mathbb{Z}+1&2\mathbb{Z}\\2\mathbb{Z}&2\mathbb{Z}+1\end{bmatrix}

H be its subgroup \begin{bmatrix}2\mathbb{Z}+1&4\mathbb{Z}\\2\mathbb{Z}&2\mathbb{Z}+1\end{bmatrix}

how to prove above subgroup is of index 2 in I.

it seems that $4\mathbb{Z}$ and $4\mathbb{Z}+2$ will be two members of the cosets where entries written by me are a12 position of the matrices?

am i right?


$X\sim Y$ iff $XH=YH$ iff $XY^{-1}\in H$. If $X=\begin{pmatrix}p&q\\r&s\end{pmatrix},Y=\begin{pmatrix}u&v\\w&x\end{pmatrix}\in I$, then $X\sim Y$ iff $-pv+qu=0 \mod 4$.

Let $H_1$ be the equivalence class of $X=I_2$: the condition on $Y$ is $v=0\mod 4$. Let $H_2$ be the equivalence class of $X=\begin{pmatrix}1&2\\0&1\end{pmatrix}$: the condition on $Y$ is $-v+2u=0\mod 4$. Since $2u=2\mod 4$, the condition reduces to $v=2\mod 4$. Since $v$ is even, $H_1,H_2$ is a partition of $I$ and we are done.


take an element of I not in H.


writing I=H ∪ Ha


Ha=\begin{bmatrix}2\mathbb{Z}+1&2+2\mathbb{Z}\\2\mathbb{Z}&2\mathbb{Z}+1\end{bmatrix} as for any x in I


we have only 2 possibilities for b

if b is even then x is element of H and if it is odd then x is element of Ha proving I as union of two cosets

proving further they are disjoint is easy as

so clearly intersection is empty.


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