I'm trying to prove normal subgroups of the group $E(2)$, but I haven't been given, what the group $E(2)$ of isometries of $\mathbb{R}^2$ is like.
What is it like?
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1In general: $${\rm E}(n) = { T \circ A \mid A \in {\rm O}(n) \text{ and }T \text{ is a translation}}.$$ – Ivo Terek May 05 '16 at 18:30
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@IvoTerek What's $O(n)$? – mavavilj May 05 '16 at 18:32
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It is the group of orthogonal matrices: the ones satisfying $AA^T = {\rm Id}$. – Ivo Terek May 05 '16 at 18:35
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The isometry group $E(n)$ of $\mathbb{R}^n$ is the following matrix group $$ E(n)={\rm Isom} ( \mathbb{R}^n) =\left\{\begin{pmatrix} A & v \\ 0 & 1 \end{pmatrix} \mid A\in O_n(\mathbb{R}), v\in \mathbb{R}^n\right\}. $$ The multiplication is given by $$ \begin{pmatrix} A & v \\ 0 & 1 \end{pmatrix} \begin{pmatrix} B & w \\ 0 & 1 \end{pmatrix}= \begin{pmatrix} AB & Aw+v \\ 0 & 1 \end{pmatrix}. $$ A normal subgroup consists of translations, given by $$ T(n)=\left\{\begin{pmatrix} I_n & v \\ 0 & 1 \end{pmatrix} \mid v\in \mathbb{R}^n\right\}. $$
Dietrich Burde
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