1. ## Subgroup Test

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

The identity is

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

And

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

and $\displaystyle 0 \in \mathbb{Z}$

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

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

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

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

So we have:

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

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

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

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 $\displaystyle x_1x_2^{-1} \in H$.

2. You don't need any of that.

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

3. Thank you!

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

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

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

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