Group of unitary matrices is not commutative I am trying to show that for all $n \geq 2$, the group of $n \times n$ complex unitary matrices is not commutative. The "answer" at Showing that the group of Unitary matrices $(U_n)$ is non-abelian for $n \geq 2$ is not an answer as it only shows that $2 \times 2$ unitary matrices are not commutative, whereas I need to show it for all $n \geq 2$.
 A: If U is a 2×2 unitary, I is the (n-2) × (n-2) identity matrix, O is a 2× (n-2 ) zero matrix, then the block matrix
$$ \begin{bmatrix} U & O \\ O^T & I \end{bmatrix}$$
is an n×n unitary matrix. 
This way you can construct n×n unitaries from 2×2 unitaries. Now suppose $U$ and $V$ are two 2×2 unitaries such that $UV \neq VU$. Then corresponding n×n block matrices aren't commuting either. 
A: If you have a $2\times 2$ unitary matrix
$$U=\begin{pmatrix}
u_{11} & u_{12}\\
u_{21} & u_{22}
\end{pmatrix}$$
then it is easy to show that the $n\times n$ (with $n>2$) matrix
$$V = \begin{pmatrix}
u_{11} & u_{12} \\
u_{21} & u_{22} \\
       &        & 1 \\
       &        &   & \ddots \\
       &        &   &        & 1
\end{pmatrix}$$
is also unitary. Given two non-commuting unitary $2\times 2$ matrices $U_1$ and $U_2$, you now can easily construct two non-commuting $n\times n$ matrices $V_1$ and $V_2$.
Alternatively, you can use the easy to prove fact that permutation matrices are unitary, and use the non-commutativity for $n>2$ of $S_n$, the group of permutations of $n$ elements, to construct counterexamples to commutativity that do not correspond to $2\times 2$ matrices.
