1
$\begingroup$

On my homework today, we had to show that Tr(n,R), the translational group on R^n, is a normal subgroup of M(n,R), the group all Isometries on R^n. In fact, the quotient group M(n,R)/Tr(n,R) is isomorphic to O(n,R), the real orthogonal group.

What homomorphism can have Tr(n,R) as its kernel?

This is easily undrestandble when n=1, since the Index or Tr(1,R) in M(1,R) is two.

Thanks :)

1 Answers 1

5

Every element in $M(n,R)$ is of the form $\phi_{A,b}:x\in\mathbb R^n\mapsto Ax+b\in\mathbb R^n$ for some $A\in O(n,R)$ and some $b\in \mathbb R^n$. You should have little trouble checking that the map $\phi_{A,b}\in M(n,\mathbb R)\mapsto A\in O(n,\mathbb R)$ is an homomorphism of groups with $Tr(n,\mathbb R)$ as kernel.

  • 0
    Thanks Mariano. What about the map Phi:M(n,R)->O(n,R) , Phi(T)=T-I-T(0)? Trouble is still remained.2011-03-28