If we let H be a subgroup of (X, x, T) where x is the binary operation. Show that
(i)if H is normal, then H* is normal
H* is the closure of H
Of course this is supposed to be a topological group, right?
So, suppose that is a topological group with multiplication and inversion maps given by respectively and . It is easy to prove that (to get an idea look here where I prove the inversion part.) Now, to prove it's normal notice that the conjugation map
is continuous since since
So, using continuity we may conclude that
. Since was arbitrary it follows that