Prove that every element $S \in SO(n)$ is a product of even numbers of reflections Prove that every element $S \in SO(n)$ is a product of even numbers of reflection.
I proved that for SO(2),SO(3) and 2 reflections. How is it with SO(n)?
 A: *

*Show that there's a reflection $R_n$ that carries $e_n$ to $f_n = S(e_n)$. Note that if $e_n = f_n$, then this "reflection" may have to be the identity map. 

*Now look at the plane of reflection, which contains the vector $ v_n = \frac{e_n + f_n}{2}$. Note that $S(v_n) = v_n$. The hyperplane $H_n$ orthogonal to $v_n$ is mapped to itself under multiplication by $SR_n$ (because multiplication by $S$ preserves inner products, and hence so does $S R_n$). 

*So restrict attention to $H_n$ (which is $n-1$-dimensional); the restriction of $x \mapsto SR_n x$ on this plane is also an orthogonal map, so you can do the same operation: pick a basis $e'_1, e'_2, \ldots, e'_{n-1}$ of $H_n$, find the vector to which $SR_n$ sends $e'_{n-1}$, and build a reflection $R_{n-1}$, construct a hyperplane of dimension $n-2$ called $H_{n-1}$, and repeat. 
Eventually, you have
$$
S R_n R_{n-1} \cdots R_1
$$
which is the identity (because it sends $e_n$ to $f_n$ and then back to $e_n$, and sends $e'_{n-1}$ to $f'_{n-1}$ and back again, and so on). Since these vectors that are mapped to themselves form an orthogonal basis for $\mathbb R^n$, the map must be the identity.  So 
$$
S = R_1^{-1} R_2^{-1} \cdots R_n^{-1}.
$$
So this expresses $S$ as a product of "reflections" (one or more of which may be the identity). Eliminating all the identity maps, you get $S$ written as a product of $k$ reflections, each of which has determinant $-1$. Since 
$$
det(S)
$$
is the product of the determinants of the rotations, you have $(-1)^k = 1$, so $k$ must be even. 
