1. ## Subgroup Test

I want to use the "one step subgroup test".

The identity is

$e=I= \begin{bmatrix}1 & 0 \\0 & 1\end{bmatrix}$

And

$det(I)=1-0=1=(\sqrt{2})^0$

and $0 \in \mathbb{Z}$

Therefore H is not empty since $I \in H$.

Now I have to show that for any two elements $x_1,x_2 \in H$, $x_1x_2^{-1}$ is in H.

Let $\begin{bmatrix}a & b \\c & d\end{bmatrix}, \begin{bmatrix}a' & b' \\c' & d'\end{bmatrix} \in H$

$x_2^{-1}= \frac{1}{(\sqrt{2})^k} \begin{bmatrix}d' & -b' \\-c' & a'\end{bmatrix}$

So we have:

$x_1x_2^{-1}= \begin{bmatrix}a & b \\c & d\end{bmatrix} \frac{1}{(\sqrt{2})^k} \begin{bmatrix}d' & -b' \\-c' & a'\end{bmatrix}$

$=\frac{1}{(\sqrt{2})^k} \begin{bmatrix}-d'a+-c'b & -b'a'+b'a' \\cd'+-c'd & -b'c+a'd\end{bmatrix}$

And $det (x_1x_2^{-1})=\frac{1}{(\sqrt{2})^k}(-d'a+-c'b)(b'c+a'd)-(-b'a'+b'a')(cd'+-c'd)$

If my working is correct so far, could anyone please show me how to manipulate this determinant to show that it satisfies the given condition, and $x_1x_2^{-1} \in H$.

2. You don't need any of that.

$det(AB) = det(A)det(B)$ and $det(A^{-1}) = \frac{1}{det(A)}$ are all you need.

3. Thank you!

So we have: $det(x_1x_2^{-1})= \frac{(ad-bc)}{(a'd'-c'b')}=\frac{(\sqrt{2})^{k_1}}{(\sqrt{2})^{k_2}}$

where $k_1,k_2 \in \mathbb{Z}$

So how could I now simplify this to show that the whole thing can be written as $(\sqrt{2})^k$??

4. $det(x_1x_2^{-1}) = \frac{(\sqrt{2})^{k_1}}{(\sqrt{2})^{k_2}} = (\sqrt{2})^{k_1-k_2}$ and $k_1 - k_2 \in \mathbb{Z}$.