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 :)