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Math Help - group of motions

  1. #1
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    group of motions

    Prove that O is not a normal subgroup of M
    where O is the group of of orthogonal operators
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  2. #2
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    Quote Originally Posted by jin_nzzang View Post
    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.
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