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.
Best Answer
Or take $U=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $V=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. We have $UV=-VU\neq0$.