= Solution
Choose an angle $2\pi\alpha$ with $\alpha\notin\mathbb Q$. Rotation through that angle has infinite order, so the <subgroup> it generates is isomorphic to $\mathbb Z$.
The <group> also contains a copy of $\mathbb Z^2$. Choose real numbers $\alpha,\beta$ such that $1,\alpha,\beta$ are linearly independent over $\mathbb Q$, for example $\alpha=\sqrt2$ and $\beta=\sqrt3$. Define
$$
\phi:\mathbb Z^2\longrightarrow SO(2),
\qquad
\phi(m,n)=R_{2\pi(m\alpha+n\beta)}.
$$
This is a homomorphism. If $\phi(m,n)=I$, then $m\alpha+n\beta\in\mathbb Z$, and the stated rational independence forces $m=n=0$. Thus $\phi$ is injective and its image is isomorphic to $\mathbb Z^2$.
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