proof that special orthogonal group SO(2) is abelian group 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:


*

*$A^{-1}$ exists $\forall A \in \operatorname{SO}(2),$

*if $A,B \in \operatorname{SO}(2)$, then $AB\in \operatorname{SO}(2),$

*$\forall A,B \in \operatorname{SO}(2),\ AB=BA.$


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 :)
 A: 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.
A: Convince yourself that the rotation around the origin by an angle $\alpha$ in the counterclockwise sense is given by multiplying a vector in the plane by 
$$R_{\alpha}=\begin{pmatrix} \cos \alpha & -\sin \alpha \\\sin\alpha & \cos \alpha\end{pmatrix}. $$
Rotation by $-\alpha$ is the inverse of rotation by $\alpha$, so $R_{\alpha}^{-1} = R_{\alpha}$. Rotation by $\alpha$ followed by rotation by $\beta$, $R_{\beta}R_{\alpha}$, is the same as rotation by $\beta$ followed by rotation by $\alpha$, $R_{\alpha}R_{\beta}$, because both compositions yield  rotation by $\alpha + \beta$, $R_{\alpha + \beta}$. Since $R_{0}= I$, this suffices to show that these matrices constitute an abelian group.
A: I think the easiest way is realising $\mathbb{C} \cong \mathbb{R}^2$ and identifying $z=a+bi$ as
$$
Z=\begin{pmatrix}a & -b \\ b & a \end{pmatrix}.
$$
Also note that $\det{Z}=a^2+b^2=|z|^2$. Next, one can identify $SO(2)$ as the unit circle $S^1=\{z=\exp{ix}:0 \leq x <2\pi\}$, which is clearly abelian.
