TITLE: Showing that the group of Unitary matrices $(U_n)$ is non-abelian for $n \geq 2$
QUESTION [5 upvotes]: I know I could show this by counter-example, finding two unitary square matrices of size $2 \times 2$ at least, and conclude $U_n$ is non-abelian. The problem with this, is I think it's somewhat time consuming trying to work out two unitary matrices and showing they don't commute, so I'm hoping there's a more concise, and clever, way of doing it. 
I've tried looking at contradiction, assuming for two unitary matrices $A$ and $B$ we have
$AB = BA$
$A^{-1}ABB^{-1} = A^{-1}BAB^{-1}$
$I_n = \bar{A}^T BA \bar{B}^T$
Then maybe trying to show
$(I_n)_{11} = 1 = \left(\bar{A}^T BA \bar{B}^T \right)_{11}$
Doesn't hold for all unitary matrices $A$ and $B$, but short of actually finding $A$ and $B$ to disprove this I'm unsure what could be done. 
Any ideas greatly appreciated.

REPLY [3 votes]: Since this question has been answered explicitly, let me suggest a general fact that it useful here (and elsewhere), of which julien's answer is one case. If we take an $n \times n$ matrix (unitary or not), which is diagonal, with $n$ distinct entries on its main diagonal, it will only commute with other diagonal matrices. This is easy to check, and I omit the details. Therefore when $n >1,$ any time you can dream up two unitary $n \times n$ matrices, one diagonal with distinct entries on its diagonal, and the other not diagonal, they will not commute.