 Solution

Write $T_{\theta ,\mathbf{b}}(\mathbf{v})=R_{\theta }\mathbf{v}+\mathbf{b}$. The sine addition formula and cosine addition formula give $R_{\theta }{R}_{\psi }=R_{\theta +\psi }$. Therefore function composition satisfiesThe resulting angle is real and the new translation vector lies in ${\mathbb{R}}^2$, so $G$ is closed under function composition. Associativity holds because, for any functions $S,T,U$ and any $\mathbf{v}$, both $(S\circ T)\circ U$ and $S\circ (T\circ U)$ take $\mathbf{v}$ to $S(T(U(\mathbf{v})))$.

The identity element is $T_{0,0}$, since $R_0=I$. Every element has the two-sided inverse elementIndeed $R_{\theta }{R}_{-\theta }=R_{-\theta }{R}_{\theta }=I$, and substituting into the composition formula makes the translation vector zero in either order. These verify all the group axioms. Thus $G$ is a group under function composition, the special Euclidean group in two dimensions.