Show the set of translations is a normal subgroup of the Euclidean group, $E(n)$

Question

I know a similar question was answered here for $E(2)$

Group of translations is a normal subgroup of group $E(2)$ of isometries of $\mathbb{R}^2$?

Yet I would appreciate help showing this (and correcting my error) from a different perspective.

This question is from Hall's "Lie groups." (Ex. 17 on page 40.)Every element of the Euclidean group $E(n)$ is proved to be a unique orthogonal linear transformation followed by a translation,i.e., of the form $T_xR$ with $x\in \mathbb{R}^n$ and $R\in O(n)$, the orthogonal group.

Letting $T'_y$ be a translation, $T_xR$ be any element of $E(n)$ and $z\in \mathbb{R}^n$, to show normality, I would like to show

$$T'_y (T_xR)z=(T_xR)T'_yz$$

For the LHS I get $Rz+x+y$. But for the RHS I get $R(z+y)+x$.

So I must be making some mistake. Thanks

Answer 1

Let $T_x$ denote translation through $x$. Let $R$ be any rotation/reflection and $A=T_x R$. Consider $$A^{-1}T_yA=R^{-1}T_x^{-1}T_y T_xR=R^{-1}T_yR.$$ Then $$R^{-1}T_yR(z)=R^{-1}(R(z)+y)=R^{-1}R(z)+R^{-1}(y)=z+R^{-1}(y)$$ so $$A^{-1}T_yA=T_{R^{-1}(y)},$$ translation through $R^{-1}(y)$.