# Thread: group of motions

1. ## group of motions

Prove that O is not a normal subgroup of M
where O is the group of of orthogonal operators

2. Originally Posted by jin_nzzang
Prove that O is not a normal subgroup of M
where O is the group of of orthogonal operators
Some definitions and facts.

1. A rigid motion is a map $m:\mathbb{Re}^n \longrightarrow \mathbb{Re}^n$ such that if X, Y are points of $\mathbb{Re}^n$, then the distance from X to Y is equal to distance from m(X) to m(Y), i.e, |m(X) - m(Y)| = |X-Y|. The rigid motion of $\mathbb{Re}^n$ forms a group $M_n$.
2. The orthogonal $n \times n$ matrices from a subgroup of $GL_n(\mathbb{Re})$ forms a subgroup of $M_n$ denoted by $O_n = \{A \in GL_n(\mathbb{Re}) |A^TA = I\}$.
3. A subgroup $T_n$ of $M_n$ whose elements do not fix the origin is the translation. A translation by p is the map $t_p(X) = X + p = \left [ \begin{matrix} x_1 +p_1 \\ \vdots \\x_n + p_n \end{matrix} \right ]$.

Now going back to your problem, suppose to the contrary that $O_n$ is a normal subgroup of $M_n$. This implies that for all $m \in M_n$, $mO_nm^{-1} \subset O_n$. However, the set of motions fixing p involves the conjugate subgroup $O_n' = t_pO_nt_p^{-1}$, which does not necessarily fix an origin (Remember that every orthogonal operator fixes an origin), contradicting that $O_n$ is a normal subgroup of $M_n$. Thus $O_n$ is not a normal subgroup of $M_n$. Q.E.D.

On the contrary, $T_n$ is a normal subgroup of $M_n$. You can verify this by defining a map $\phi:M_n \rightarrow O_n$, given by $\phi(t_a \rho_\theta)=\rho_\theta$ and $\phi(t_a \rho_\theta r)=\rho_\theta r$, where $\rho_\theta$ denotes origin-fixing rotations and r is a reflection in $\mathbb{Re}^n$. Note that every element of m in $M_n$ can have forms either $m=t_a \rho_\theta$ or else $m = t_a \rho_\theta r$. You can see that the kernel of $\phi$ is $T_n$, which implies that $T_n$ is a normal subgroup of $M_n$.