I assume that is the unitary group of one by one matrices? Clearly (not to insult your intelligence) is an isomorphism (where is the circle group) but it's also clear that since the map is a homomorphism with kernel (just use the FIT). This is sort of close to what you want? It's also true that (the special orthogonal group) where I'm sure it's clear to you that you just send .

So, the answer to your question is no. Is there a particular reason you think that? You know that since is an element for