The set of all translations is a normal subgroup of $\text{I}(\mathbb{R}^n)$.

Question

I want to show that the set of all translations forms a normal subgroup of $\text{I}(\mathbb{R}^n)$.

Before we discuss it, I would like to share definitions of my textbook: a translation $\text{T}_a(x)=x+a$ for any $x,a \in \mathbb{R}^n$. And, the set $\text{I}(\mathbb{R}^n)$ is the set of all isometries of $\mathbb{R}^n$.

Here's my approach to show a contradiction.

Let $\text{T}=\left\{\text{T}_a | \text{T}_a:\mathbb{R}^n\to \mathbb{R}^n \; \text{s.t}\; x\mapsto x+a \; \text{with}\; a\in \mathbb{R}^n\right\}$

Assume that $\text{T}$ is not a normal subgroup of $\text{I}(\mathbb{R}^n)$.

Then, $\exists f\in \text{I}(\mathbb{R}^n) $ s.t $f\text{T} \neq \text{T}f$.

$\Rightarrow \exists a\in \mathbb{R}^n $ s.t $f\circ \text{T}_a \notin \text{T}f$.

Note that for any $x \in \mathbb{R}^n$, $f\circ \text{T}_a(x)=f(x+a)=A(x+a)+c$ for some $A \in \text{O}(n)$ and some $c \in \mathbb{R}^n$. (Here, $\text{O}(n)$ means the orthogonal group.)

And, $\forall b \in \mathbb{R}^n$ we must have $\text{T}_b \circ f \in \text{T}f$. So, for any $x \in \mathbb{R}^n$, $\text{T}_b \circ f(x) = \text{T}_b(Ax+c) = Ax +b +c$

But, as $f\circ \text{T}_a \notin \text{T}f$, $A(x+a)+c \neq Ax+b+c$

$\Leftrightarrow Aa \neq b $ for any $b \in \mathbb{R}^n$.

But, as $Aa \in \mathbb{R}^n$, this is a contradiction.

Hence, $\text{T}$ is a normal subgroup of $\text{I}(\mathbb{R}^n)$.

Does this answer make any sense?

More specifically, did I use the concept of cosets correctly?

Answer 1

For an orthogonal matrix $A\in O(n)$, let us denote (slight abuse) by $A$ the induced map $A:\Bbb R^n\to\Bbb R^n$. The translation by a vector $a\in\Bbb R^n$ will be denoted by $T_a$.

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The above solution seems to know that every isometry is a composition of an orthogonal map, and a translation. Let us rewrite it, so that the idea is transparent.

To show that the translations form a normal subgroup, it is enough to show that each conjugation of a translation is also a translation.

So let us fix a translation, $T_a$. Let $A\circ T_c$ be an arbitrary isometry. Conjugating a translation by $T_c$ is a simple process, the translation is invariated, since translations commute with each other. So it is enough to check the case of a purely "rotational" isometry $A$. Then testing on some vector $v\in\Bbb R^n$: $$ \begin{aligned} A\circ T_a\circ A^{-1}\ (v) &= A\circ T_a\ (A^{-1}v) \\ &= A\ (A^{-1}v + a) \\ &= AA^{-1}v + Aa \\ &= v + Aa \\ &=T_{Aa}\ v\ . \end{aligned} $$ So the conjugated translation is again a translation.

$\square$