How to show the commutator of $SO(n)$ is itself?

Question

I am not very familiar with Lie groups, but I want to show that the commutator subgroup of $SO(3)$ is itself. I have looked up many different sources, and it seems to me that almost all of them require some notion of Lie algebra, so I am wondering if it is possible to show this without much knowledge of Lie groups.

Answer 1

You can show directly that the commutator subgroup of $SO(n)$ is $SO(n)$ only for $n\ge 3$.

Now assume $n\ge 3$. In general, we know that every $A \in SO(n)$ can be written as $A = P BP^T$, where $P \in SO(n)$ and $B$ is block diagonal, with block of the form

$$E(\theta) = \begin{bmatrix} \cos \theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$$

Geometrically, it just mean that all $SO(n)$ elements corresponds to rotations of two planes $L_1, L_2, \cdots$, where each $L_i$'s are mutually orthogonal.

Now it suffices to show that $B$ lie in the commutator subgroup. By induction, it suffices to show that

$$C = \begin{bmatrix} E(\theta) & 0 \\ 0 & I_{n-2} \end{bmatrix}$$

lies in the commutator subgroup. It is easy to show that $E(\theta)$ can be written as $ABAB$, where $A, B$ are reflections. Then $A = A^{-1}$, $B= B^{-1}$ and so $E(\theta) = [A, B]$. Then we have $C = [\tilde A, \tilde B]$, where

$$\tilde A = \begin{bmatrix} A & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & I_{n-3} \end{bmatrix} \in SO(n)$$

(Note that $n\ge 3$ is used so that you can write down that $-1$. This is necessary as $\det A = -1$)