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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 :)

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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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  • $\begingroup$ Or rather, you have proved your 1., it's just that this is not enough. Subgroup criterion says: $H$ is subgroup of $G$ iff 1. $(\forall h\in H)\ h^{-1}\in H$ and 2. $h,h'\in H\implies hh'\in H$. $\endgroup$
    – Ennar
    May 12 '18 at 12:31
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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.

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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.

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