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Let $$S^1 := \{z \in \mathbb{C} : |z| = 1\}$$ I know that $$\operatorname{SO}(2) = \left\{\begin{pmatrix} \cos \varphi & -\sin \varphi\\ \sin\varphi & \cos \varphi \end{pmatrix} : \varphi \in \mathbb{R}\right\}$$ So to show that $$(S^1,\cdot) \cong (SO(2),\cdot)$$ it would be natural to consider the mapping induced by the polar representation of a complex number $z \in S^1$ $$z = \cos \varphi + i\sin \varphi \mapsto \begin{pmatrix} \cos \varphi & -\sin \varphi\\ \sin\varphi & \cos \varphi \end{pmatrix}$$ This mapping is clearly surjective. My problem is now, that the maping is not really well defined because of the periodicity of the trigonometric functions. How can I overcome this formally? I thought of using the first isomorphism theorem.

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    You are on the right track: both objects have the property $M(\theta)\cdot M(\phi) = M(\theta+\phi)$. – 2017-02-22
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    If you have two numbers on the complex unit circle that become the same $SO(2)$-matrix under this map, can they possibly be distinct points on the circle? – 2017-02-22
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    @Arthur No this cannot be. – 2017-02-22
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    Then you have shown that the mapping is injective as well, so it is bijective. – 2017-02-23
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    Well, the argument of a complex number is only well-defined up to integer multiples of $2 \pi$, too. If you want to be very formal, you could identify $S^1$ with $\mathbb{R}/2\pi \mathbb{Z}$. – 2017-02-23

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Actually, your map is $$z\mapsto\begin{pmatrix}\Re z&-\Im z\\\Im z&\Re z\end{pmatrix} $$ and hence well-defined (with inverse $$ \begin{pmatrix}a&b\\c&d\end{pmatrix}\mapsto a+ic$$ also well-defined)

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    I also saw this. So I could restrict this mapping to $S^1$ and I would get the desired isomorphism is that right? – 2017-02-22
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You could also consider the following surjective group homomorphisms:

$f: \mathbb{R}^+ \to S^1 \, ; \, t \mapsto e^{ti}$

$g: \mathbb{R}^+ \to SO(2) \, ; \, t \mapsto \left( \begin{matrix} \cos t & -\sin t \\ \sin t & \cos t \end{matrix} \right)$

Each has kernel $K = \{2 \pi n : n \in \mathbb{Z} \}$, so that $S^1 \cong \mathbb{R}^+ / K \cong SO(2)$.