Prove that $[O(2),O(2)] = SO(2)$. In words, prove that the commutator subgroup of the orthogonal group of 2x2 matrices is equal to the the Special Orthogonal Group of 2x2 matrices.
I know that the commutator subgroup of a group $G$, denoted by $[G,G]$ is the set of all elements of the form $[x_1,y_1] \cdot [x_2,y_2] \cdot ... \cdot [x_n,y_n]$ for some $x_1,...,x_n,y_1,...,y_n$ in $G$.
Furthermore, I know the $[G,G]$ is a normal subgroup of $G$.
I also know that in $\mathbb{R^2}$, rotations and reflections are represented by 2x2 orthogonal matrices from $O(2)$ and the such reflections and rotations can be composed rather easily.
For example, let $R_\alpha$ and $R_\beta$ be rotations in $\mathbb{R^2}$ by angles $\alpha$ and $\beta$ respectively and $F_l$ be a reflection in some line $l$ through the origin.
Then, $R_\beta \circ R_\alpha \circ (R_\beta)^{-1} = R_\alpha \in SO(2)$
and
$F_l \circ R_\alpha \circ (F_l)^{-1} = R_{-\alpha} \in SO(2)$
But beyond this, I am not sure how to do this problem. I've done problems regarding commutator subgroups before but not with respect to orthogonal groups and I'm a little weak in my linear algebra in this regard.
Can anyone provide some assistance?
Best Regards.
Some hints. If $F_1,F_2$ are reflections then $[F_1,F_2]=F_1F_2F_1^{-1}F_2^{-1}=(F_1F_2)^2$.
If the lines for $F_1$ and $F_2$ make an angle of $\theta$, then $F_1F_2$ is a rotation by what angle?