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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

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    You are trying to show that $T_x R$ and $T_y$ commute, but they don't! Try to show that $(T_x R)^{-1}T_y(T_xR)$ is a translation. – Angina Seng May 02 '17 at 18:04
  • @LordSharktheUnknown Just did that and it does work - I took $(T_xR)^{-1}$ to be $R^{-1}T_x^{-1}$. But I'm puzzled why normality and commutativity are not the same thing. Thanks –  May 02 '17 at 18:12
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    As you have discovered, $(T_xR)^{-1}T_y(T_xR)$ is a translation but it's not the same translation as $T_x$. What you were effectively doing, was trying to prove it was the same. – Angina Seng May 02 '17 at 18:13
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    @LordSharktheUnknown Maybe you want to make your comment into an answer so I can accept it? –  May 02 '17 at 18:15

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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 $R^{-1}(y)$.

Angina Seng
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