Generator of $SL(2,\mathbb{R})$ Find all 2x2 matrices $ A \in SL(2,\mathbb {R}) $such that  $det(A) =1$ and $A\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1  \end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1  \end{pmatrix} $. Do these matrices form a group?
My thoughts:
I have considered the matrices \begin{pmatrix} 1 & 0\\ x & 1  \end{pmatrix} and \begin{pmatrix} 1 & x\\ 0 & 1  \end{pmatrix}  I have realised that the matrices will not satisfy the condition if $x \neq 0$. These two have the identity element, and are invertible. Another matrix of interest was \begin{pmatrix} 0 & -1\\ 1 & 2  \end{pmatrix}. My problem is, how does one compactly answer this question, and I believe that the group is finite. 

A nonempty subset $S$ of a group $G$ forms a subgroup if for all $A,B\in S$, $AB^{-1}\in S$.  In this case, $S$ is nonempty because the identity is in $S$.  You then assume that $A$ and $B$ are two matrices with the given property and consider (the $\frac{1}{\sqrt{2}}$ doesn't matter in this problem since it's a scaling factor).
$$
AB^{-1}\begin{pmatrix}1\\-1\end{pmatrix}.
$$
Since 
$$
B\begin{pmatrix}1\\-1\end{pmatrix}=\begin{pmatrix}1\\-1\end{pmatrix},
$$
$$
\begin{pmatrix}1\\-1\end{pmatrix}=B^{-1}\begin{pmatrix}1\\-1\end{pmatrix}.
$$
Moreover, since
$$
A\begin{pmatrix}1\\-1\end{pmatrix}=\begin{pmatrix}1\\-1\end{pmatrix},
$$
the product has the correct property and the set $S$ forms a group.  I'm leaving the details at this point for you to fill in.
If you want to describe the group, write a general group element as
$$
\begin{pmatrix}a&b\\c&d\end{pmatrix}
$$
and observe that the following three equalities must hold:
\begin{align*}
a-b&=1\\
c-d&=-1\\
ad-bc&=1
\end{align*}
You can write $a=1+b$ and $c=-1+d$.  By substituting these into the determinant equation, you can solve for $c$ in terms of $a$.  This gives a one-parameter family of solutions for different $a$'s.
