Is $U(1)$ a normal subgroup of $U(2)$ and does the question even make sense?

509 Views Asked by At

I have been wondering whether $U(1)$, defined as the group of complex phases (edit for clarity: I mean complex numbers of unit absolute value, such as $e^{i\alpha}$ with $\alpha \in \mathbb{R}$) with multiplication as group operation, is a normal subgroup of $U(2)$, defined as the group of $2\times 2$ unitary matrices. With these definitions I would say that $U(1)$ isn't even a subgroup of $U(2)$, let alone normal. However, it is isomorphic to many subgroups of $U(2)$ such as:

  • $A=\begin{bmatrix} e^{i\alpha} & 0 \\ 0 & e^{i\alpha} \\ \end{bmatrix}$

  • $B=\begin{bmatrix} 1 & 0 \\ 0 & e^{i\alpha} \\ \end{bmatrix}$

Now $A$ is normal in $U(2)$ while $B$ is not. I am not particularly surprised that normality isn't preserved by isomorphism, but then I can only answer to questions like "Is $A$ (or $B$) normal in $U(2)$?".

What happens to the quotient space? Since $B$ is not normal then I think $U(2)/B$ is not a group, however $U(2)/A$ can be. Can you help me to find the manifold underlying these quotient spaces/quotient groups? Shouldn't the quotient be the same if I change a group to an isomorphic one?