For all $g = (\tau, R) \in \operatorname{Dep}(\mathbb{R}^{d})$, we denote by $\phi_{g} : \mathbb{R}^{d} \rightarrow \mathbb{R}^{d}$ the map defined by $\phi_{g}(x) = Rx + \tau$. Let $g \in \operatorname{Dep}(\mathbb{R}^{d})$.
[(a)] Show that $\phi_{g}$ is bijective. We denote by $\phi_{g}^{-1}$ its inverse map.
[(b)] Show that there exists a unique $g^{\prime} \in \operatorname{Dep}(\mathbb{R}^{d})$, which we will express in terms of $g$, such that $\phi_{g^{\prime}} = \phi_{g}^{-1}$. We denote $g^{\prime} = g^{-1}$.
[(c)] Verify that $ge = eg = g$ and then that $gg^{-1} = g^{-1}g = e$.
For all $g = (\tau, R) \in \operatorname{Dep}(\mathbb{R}^{d})$, we denote by $\phi_{g} : \mathbb{R}^{d} \rightarrow \mathbb{R}^{d}$ the map defined by $\phi_{g}(x) = Rx + \tau$. Let $g \in \operatorname{Dep}(\mathbb{R}^{d})$.
\begin{itemize}
\item[(a)] Show that $\phi_{g}$ is bijective. We denote by $\phi_{g}^{-1}$ its inverse map.
\item[(b)] Show that there exists a unique $g^{\prime} \in \operatorname{Dep}(\mathbb{R}^{d})$, which we will express in terms of $g$, such that $\phi_{g^{\prime}} = \phi_{g}^{-1}$. We denote $g^{\prime} = g^{-1}$.
\item[(c)] Verify that $ge = eg = g$ and then that $gg^{-1} = g^{-1}g = e$.
\end{itemize}