Prove that $T=T_xR$ with $x\in \Bbb R^n$ and $R\in O(n)$

Question

I was studying the book of Lie groups, by Brian C. Hall, there at page 9, he has defined the euclidean group $E(n)$ by the group of all one-to-one, onto, distance preserving maps from $\Bbb R^n \to \Bbb R^n$ s.t $d(x,y)=|x-y|$. Note here $f\in E(n)$ is not in particular linear. Define translation $T_x:\Bbb R^n \to \Bbb R^n$ s.t $T_x(y)=x+y$, the set of translation is also a subgroup of $E(n)$ again $O(n)$ is also a subgroup of $E(n)$.

Now there is a proposition which I am unable to prove that

Every element $T$ of $E(n)$ can be written uniquely as an orthogonal linear transformation followed by a translation, i.e, in the form $T=T_xR$ with $x\in \Bbb R^n$ and $R\in O(n)$

Now I think there are two steps which I am not able to prove here

1. Every $T \in E(n)$ does not fixe a point then it would be of the form $T_x$.
2. Every $1-1$, onto, distance preserving map of $\Bbb R^n$ to itself which fixes the origin must be linear.

If we could prove this then $T_x^{-1}T \in O(n)$ and we are done. Please help in proving these two parts.

Answer 1

Suppose first that $T(0) = 0$. Then $$ | T(x) | = | T(x) - T(0) | = | x - 0 | = | x | $$ for every $x \in \mathbb{R}^n$ which shows that $T$ is preserves the norm $|\cdot|$. For all $x, y \in \mathbb{R}^n$ it follows that \begin{align*} &\, |x|^2 - 2\langle x,y \rangle + |y|^2 \\ =&\, |x - y|^2 = |T(x) - T(y)|^2 \\ =&\, |T(x)|^2 - 2\langle T(x), T(y) \rangle + |T(y)|^2 \\ =&\, |x|^2 - 2\langle T(x), T(y) \rangle + |y|^2 \,, \end{align*} so that $$ \langle T(x), T(y) \rangle = \langle x,y \rangle \,. $$ It follows that \begin{align*} &\, |T(x+y) - T(x) - T(y)|^2 \\ =&\, |T(x+y)|^2 + |T(x)|^2 + |T(y)|^2 \\ &\, - 2\langle T(x+y), T(x) \rangle - 2\langle T(x+y), T(y) \rangle + 2\langle T(x), T(y) \rangle \\ =&\, |x+y|^2 + |x|^2 + |y|^2 - 2\langle x+y, x \rangle - 2\langle x+y, y \rangle + 2\langle x,y \rangle \\ =&\, |(x+y) - x - y|^2 = 0 \end{align*} for all $x, y \in \mathbb{R}^n$, and therefore that $T(x + y) = T(x) + T(y)$. This shows that $T$ is additive. For all $x \in \mathbb{R}^n$ and $\lambda \in \mathbb{R}$ we have that \begin{align*} &\, |T(\lambda x) - \lambda T(x)|^2 = |T(\lambda x)|^2 - 2 \langle T(\lambda x), \lambda T(x) \rangle + |\lambda T(x)|^2 \\ =& |T(\lambda x)|^2 - 2 \lambda \langle T(\lambda x), T(x) \rangle + \lambda^2 |T(x)|^2 = |\lambda x|^2 - 2 \lambda \langle \lambda x, x \rangle + \lambda^2 |x|^2 \\ =& \lambda^2 |x|^2 - 2 \lambda^2 |x|^2 + \lambda^2 |x|^2 = 0 \end{align*} and therefore that $T(\lambda x) = \lambda T(x)$. This shows that $T$ is homogeneous. Alltogether this shows that $T$ is linear, and thus orthogonal (here we use that $\mathbb{R}^n$ is finite-dimensional).

For every $T \in E(n)$ we have for $x = T(0)$ that $(T_x^{-1} T)(0) = 0$. Then $R := T_x^{-1} T$ is orthogonal by the prevous discussion, and $T = T_x R$ is the desired decomposition.

For the uniqueness note that it follows from $T = T_x R$ that $T(0) = T_x(R(0)) = T_x(0) = x$, which shows the uniqueness of $x$. The uniqueness of $R$ then follows from $R = T_x^{-1} T$.