My question is motivated by this part of 'Basic Notions of Algebra' from Igor R. Shafarevich. He mentions the group of motions of the plane in the following context.
Let denote $SE(2)$ the 2-dimensional special Euclidean group, $T(2)\subseteq SE(2)$ the translation subgroup of it, $G=SE(2)/T(2)$, and $G'\subseteq SE(2)$ the group of rotations around some fixed point (call it 'origin'). We know that $G'\cong SE(2)/T(2)\cong SO(2)$. We also know that $G'$ is a complementary subgroup of $T(2)$, and it is the image of a section $s: G\to SE(2)$ of the bundle $\pi:SE(2)\to G$. Every other section $s'$ can be written in the form
$$s'(g)=f(g)s(g)\label{a}\tag{1}$$
where $f$ is a function $G\to T(2)$. According Shafarevich, a necessary condition for a section $s'$ defined by $f$ to be a subgroup is $$f(g_1g_2)=f(g_1)g_1(f(g_2))\label{b}\tag{2}$$ where $$g_1(f(g_2)):=\gamma f(g_2) \gamma^{-1}\label{c}\tag{3}$$ with any $\gamma\in g_1$, e.g. $\gamma =s(g_1)$ (the result doesn't depend on the choice). We call such $f$ functions 1-cocycles. According to Shafarevich, there is always a 'trivial' choice of $f$, which results in $$s'(g)= f(g)s(g)=as(g)a^{-1}\label{d}\tag{4}$$ with an $a\in T(2)$, that is, $$f(g) = as(g)a^{-1}s(g)^{-1}\label{e}\tag{5}$$ We can check that this $f$ satisfies ($\ref{b}$). These cocycles are called 1-coboundaries.
By some intuition (regarding the motion of the Moon around the Earth) I've found that taking any $a\in T(2)$
$$s'(g) = c_{s(g)}(as(g)a^{-1})=s(g)as(g)a^{-1}s(g)^{-1}\label{f}\tag{6}$$
is also a section with image forming a subgroup of $SE(2)$ complementary to $T(2)$. The 1-cocycle belonging to this $s'$ is $$f(g) = c_{s(g)}(as(g)a^{-1})s(g)^{-1}\label{g}\tag{7}$$ (where $c_{s(g)}$ means conjugation by $s(g)$).
My question: How can we find all cocycles, i.e. all functions $f$ that satisfies ($\ref{b}$). In other words, how can we find all complementary subgroups to $T(2)$ in $SE(2)$?
The question was to describe all subgroups $G<SE(e)$ (the group of orientation-preserving isometries of $E^2$) whose linear part equals $SO(2)$ and whose intersection with ${\mathbb R}^2$ (the subgroup of translations of $SE(2)$) is trivial. The answer is that all such subgroups have the form $K_x$, the group of rotations fixing $x\in E^2$.
In order to prove that these are the only subgroups, observe that if $R_1, R_2$ are two rotations fixing distinct points in $E^2$ then their commutator $[R_1,R_2]$ is a nontrivial translation (this is a simple computation which involves composition of four affine isometries). Therefore, a subgroup $G<SE(e)$ which has trivial intersection with ${\mathbb R}^2$ contains no rotations around distinct points in $E^2$. Hence, such $G$ is contained in one of the groups $K_x$. On the other hand, since the map to the linear part is an isomorphism $K_x\to SO(2)$, we conclude that $G=K_x$. qed