I have the following question on Lie theory:
Prove that $U(n)$ has no nontrivial normal subgroup except $Z(U(n))$.
I know from Schreier's theorem that since $U(n)$ is path-connected, if it does have a discrete normal subgroup then it is contained in the centre of $U(n)$, this being $\left\{ \omega I: |\omega| = 1 \right\}$ (i.e. has to be the nth roots of unity). I also know the centre is the largest discrete normal subgroup. However, these don't appear to help much.
I have proved there is a 1-dimensional ideal of $\mathfrak{u}(n)$, the space spanned by $iI$, where $I$ is the $n \times n$ identity matrix. This is the tangent space of $Z(U(n))$. So I'm guessing the problem is reduced to showing that this is the only nontrivial ideal of $\mathfrak u(n)$, which I just can't seem to figure out how to do.
This is substantially different from the $SU(n)$ case (which is dealt with here: Normal Subgroups of $SU(n)$), since in that case $\mathfrak g$ is simple, thus has no nontrivial ideals. In this case, we have a nontrivial ideal, and the goal (I'm guessing) is to show it is the only one.
If anyone has any hints on how to do this, or some other way of approaching the problem, I'd be very thankful.