7
$\begingroup$

Consider the following subgroups of $\text{SL}(2,\mathbb{Z})$ :

  • $A$ the subgroup of matrices with determinant $1$ :

\begin{bmatrix}4\mathbb{Z}+1&8\mathbb{Z}\\4\mathbb{Z}&4\mathbb{Z}+1\end{bmatrix}

  • $B$ the subgroup of matrices with determinant $1$ :

\begin{bmatrix}2\mathbb{Z}+1&8\mathbb{Z}\\4\mathbb{Z}&2\mathbb{Z}+1\end{bmatrix}

I want some onto homomorphism from $B$ to $A$ whose kernel is \begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}-1&0\\0&-1\end{bmatrix}

How to get this? I have no idea how to find the map.

$\endgroup$
4
  • 1
    $\begingroup$ The upper right entry of $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ doesn’t belong to $8ℤ$, you meant to write the identity matrix, right? $\endgroup$
    – k.stm
    Commented Feb 6, 2016 at 10:33
  • $\begingroup$ ohh sorry yes identity $\endgroup$
    – PAMG
    Commented Feb 7, 2016 at 4:46
  • $\begingroup$ It's not clear to me how you are defining your subgroups. Are you looking elements of the form $\begin{bmatrix}4n+1&8n\\4n&4n+1\end{bmatrix},$ or is that integer allowed to vary by entry? $\endgroup$
    – Couchy
    Commented Apr 13, 2017 at 17:57
  • $\begingroup$ @Couchy311: The integer is allowed to vary by entry. The subgroups are well-defined. This is a good question, it's not immediately clear to me how to prove/disprove it. $\endgroup$ Commented Apr 13, 2017 at 18:00

2 Answers 2

5
+100
$\begingroup$

Let $$\beta=\begin{pmatrix} 2a+1 & 8 b \\ 4c&2d+1 \end{pmatrix} \in B .$$ Then $4 ad+ 2a+2d+1 -32bc = 1$, that is $2ad+a+d = 16 bc$, which yields $a \equiv b \pmod2$. Thus, either the elements on the diagonal of $\beta$ are both congruent to $1$ or both congruent to $3$ modulo $4$. In the first case I set $\varphi (\beta) = \beta$ and I call $\beta$ even, in the second case $\varphi(\beta) = - \beta$ and call $\beta$ odd. Parity behaves as usual and thus $\varphi$ is a homomorphism. Its image is $A$ since $A$ is the subgroup of the even matrices of $B$ and its kernel is $\{\pm I_2\}$. Hence $\varphi$ is the desired homomorphism.

$\endgroup$
2
  • $\begingroup$ You're right, my mistake. Thank you for your explanation, I just deleted my faulty comments $\endgroup$ Commented Apr 14, 2017 at 14:18
  • $\begingroup$ Excellent, I missed that $\endgroup$ Commented Apr 15, 2017 at 10:20
3
$\begingroup$

This is not an answer, but it might help in some way:

Since $C=\left\{\begin{pmatrix} 1&0\\0&1 \end{pmatrix},\begin{pmatrix} -1&0\\0&-1 \end{pmatrix}\right\}$ is a normal subgroup of $B$, there exists a surjective morphism $g:B\rightarrow G$ such that $\frac{B}{C}=\frac{B}{\ker(g)}\cong G$. If the desired map exists, we need to show that $\frac{B}{C}\cong A$ .

Now we want to find an isomorphism $\phi: \frac{B}{C}\rightarrow A:\begin{pmatrix} 2x+1&8a\\ 4b & 2y+1 \end{pmatrix}\cdot C\mapsto ?$ It has to be well-defined, so if $X,Y\in B$ are such that $XY^{-1}\in C$, then they need to get mapped to the same thing. In particular $X$ and $-X$ need to get mapped to the same element (in fact this all you need to get a well-defined function). This might give some hint on how to construct this map (which would also solve the problem). Again, this is exactly the same as the original question, but now we see that under $\phi$, $X$ and $-X$ need to get mapped to the same thing which might help to see some formula's.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .