[Math] proof that special orthogonal group SO(2) is abelian group

abelian-groupsabstract-algebragroup-theory

I have to show that $\operatorname{SO}(2)$ defined as:

$$\operatorname{SO}(2)=\{ \left(\begin{array}{cc}
\cos\phi& -\sin\phi\\
\sin\phi&\cos\phi
\end{array}\right)\in M_2(\mathbb R)\,|\, \phi \in [0,2\pi]\}$$

is an abelian group by proving these points:

  1. $A^{-1}$ exists $\forall A \in \operatorname{SO}(2),$
  2. if $A,B \in \operatorname{SO}(2)$, then $AB\in \operatorname{SO}(2),$
  3. $\forall A,B \in \operatorname{SO}(2),\ AB=BA.$

The first point is easy:

$$\forall A \in \operatorname{SO}(2): \det(A)=(\sin\phi)^2+(\cos\phi)^2=1 \implies \det(A)\neq 0 \to \exists A^{-1}.$$

The third one is also true, you just have to multiply $AB$ and $BA$ and you will get:
$$\forall A,B \in \operatorname{SO}(2): AB=BA.$$

I'm having trouble with the second point, I tried just multiplying but I don't think it's enough to show that $(AB)_{11}=(AB)_{22}$ and $(AB)_{12}=-(AB)_{21}$. One has to show that this applies to the same $\phi$ for both equations. But I don't know how.

Thanks for your tips and help in advance 🙂

Best Answer

You haven't really proved 1., you've shown that $A$ is invertible matrix, you didn't show that the inverse is contained in $\operatorname{SO}(2)$.

For all three points, the same hint applies, write

$$A = \begin{pmatrix} \cos \alpha & -\sin\alpha\\ \sin \alpha & \cos\alpha \end{pmatrix},\ B = \begin{pmatrix} \cos \beta & -\sin\beta\\ \sin \beta & \cos\beta \end{pmatrix}$$

and after you multiply them, use trigonometric addition formulas.

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