Prove thatOis not a normal subgroup ofM

where O is the group of of orthogonal operators

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- Sep 24th 2009, 10:33 AMjin_nzzanggroup of motions
Prove that

**O**is not a normal subgroup of**M**

where O is the group of of orthogonal operators - Sep 26th 2009, 01:05 PMaliceinwonderland
Some definitions and facts.

1. A rigid motion is a map $\displaystyle m:\mathbb{Re}^n \longrightarrow \mathbb{Re}^n$ such that if X, Y are points of $\displaystyle \mathbb{Re}^n$, 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 $\displaystyle \mathbb{Re}^n$ forms a group $\displaystyle M_n$.

2. The orthogonal $\displaystyle n \times n$ matrices from a subgroup of $\displaystyle GL_n(\mathbb{Re})$ forms a subgroup of $\displaystyle M_n$ denoted by $\displaystyle O_n = \{A \in GL_n(\mathbb{Re}) |A^TA = I\}$.

3. A subgroup $\displaystyle T_n$ of $\displaystyle M_n$ whose elements do not fix the origin is the translation. A translation by p is the map $\displaystyle t_p(X) = X + p = \left [ \begin{matrix} x_1 +p_1 \\ \vdots \\x_n + p_n \end{matrix} \right ]$.

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

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