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I'm trying to see that:

The group of translations $T=\{t(x)=x+a : a \in \mathbb{R}^2 \}$ is a normal subgroup of group $E(2)$ of isometries of $\mathbb{R}^2$.

I know the definitions of a normal subgroup, but I don't understand how the translations and isometries interact in this case.

mavavilj
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1 Answers1

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We use the notations of your previous question: The subgroup of translations in $$ E(n)={\rm Isom} ( \mathbb{R}^n) =\left\{\begin{pmatrix} A & v \\ 0 & 1 \end{pmatrix} \mid A\in O_n(\mathbb{R}), v\in \mathbb{R}^n\right\}, $$ given by $$ T(n)=\left\{\begin{pmatrix} I_n & v \\ 0 & 1 \end{pmatrix} \mid v\in \mathbb{R}^n\right\}. $$ is indeed normal. To see this, we compute $$ \begin{pmatrix} A & v \\ 0 & 1 \end{pmatrix} \begin{pmatrix} I_n & w \\ 0 & 1 \end{pmatrix} \begin{pmatrix} A & v \\ 0 & 1 \end{pmatrix} ^{-1}= \begin{pmatrix} I_n & Aw \\ 0 & 1 \end{pmatrix}, $$ which is again in $T(n)$. Hence $T(n)$ is a normal subgroup.

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Dietrich Burde
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  • Another way to phrase essentially the same argument: The function from $E(n)$ to $O_n(\mathbb R)$ that sends $\begin{pmatrix}A&v\0&1\end{pmatrix}$ to $A$ is a homomorphism, and the translations form its kernel. – Andreas Blass May 05 '16 at 20:40
  • Could you clarify why your definition of $T(n)$ is equivalent to mine. – mavavilj May 06 '16 at 12:04
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    Identify $x$ with $(x,1)$ and let $A$ be a matrix in $T(n)$. Then $Ax=(Ix+v,1)=(x+v,1)$, which is identified with $x+v$. So it is the same map as your $t(x)=x+v$. – Dietrich Burde May 06 '16 at 12:49