About a right $K$-invariant function on the motion group

Question

The motion group of $\mathbb R^2$, noted by $G = M(2)$ is the semi-direct product of $\mathbb R^2$ with the special orthogonal group $K = SO(2)$. A typical element $g\in G$ is written as $g = (z, e^{i\alpha}), z \in \mathbb C, \alpha\in\mathbb R$. The group law in G is given by $$(z, e^{i\alpha})(w, e^{i\beta})=(z+e^{i\alpha} w, e^{i(\alpha+\beta)}).$$ I would like to know why a right $K$-invariant function on $G$ can be thought of as a function on $\mathbb C$ and viceversa ?

Thank you in advance

Answer 1

Hint This is mostly a matter of unwinding definitions. By definition $f : G \to X$ is a right $K$-invariant function iff $$f(g \cdot k) = f(g)$$ for all $g \in G$, $k \in K$.

Now, the right action of a typical element $k = (0, e^{i \beta}) \in K$ on a typical element $g = (z, e^{i \alpha})$ is $$g \cdot k (z, e^{i \alpha}) \cdot (0, e^{i \beta}) = (z, e^{i (\alpha + \beta)}) .$$

Additional Hint So, $f$ is right $K$-invariant iff $$f(z, e^{i (\alpha + \beta)}) = f(z, e^{i \alpha})$$ for all $z, \alpha, \beta$.