Which group is isomorphic to $\left\langle\begin{bmatrix}0&1\\1&0\end{bmatrix},\begin{bmatrix}1&-1\\0 & -1 \end{bmatrix} \right\rangle$? Both matrices have determinant equal to -1, so their products are matrices with determinant $\in \{1,-1\}$. Can I conclude that this is isomorphic to $ O_2(\mathbb{R}) $ ?
 A: Let $a$ be the first matrix and $b$ be the second matrix.
Hint: you've probably already worked out that $a^2=1$ and $b^2=1$. So, any element in the group will be a word beginning and ending with an $a$ or $b$ and with alternating $a$s and $b$s in between. Start looking at these words and see if you can find an alternating word which looks like a matrix you already know. What does this tell you about the group?
A: If you set $$a=\begin{pmatrix}0&1\\1&0\end{pmatrix},~~b=\begin{pmatrix}1&-1\\0 & -1 \end{pmatrix}$$ exactly as @Daniel noted; you will find out that $$a^2=b^2=(ab)^3=1$$ The group with this relation has the form $$\langle a,b\mid a^2=1,b^2=1,(ab)^3=1\rangle$$ which is $D_6$ or $S_3$. The following codes written in GAP environment approved our result:
  gap> f:=FreeGroup("a","b"); 
       a:=f.1;;
       b:=f.2;;
          s:=f/[a^2,b^2,(a*b)^3];;
      Size(s);
                                           6
      IsAbelian(s);
                                         false
      StructureDescription(s);
                                           S3

A: Every finitely generated group is at most countable infinite. On the other hand,the group $O_2(\mathbb R)$ is easily seen to be uncountable.
Therefore your group is not isomorphic to $O_2(\mathbb R)$.
