Group of Unitary Matrices under Multiplication is not Abelian

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Theorem

Let $n > 1$ be a natural number.

Let $\map U n$ denote the unitary group.


Then $\map U n$ is not abelian.


Proof

Proof by Counterexample

Recall the definition of Unitary Group:

Let $n$ be a positive integer.


The unitary group $\operatorname U_n$ is the group of all $n \times n$ unitary matrices under (conventional) matrix multiplication.


The matrices:

$A = \begin {pmatrix} 1 & 0 \\ 0 & -1 \end {pmatrix}$

and

$B = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end {pmatrix}$

are both real and symmetric.

Hence, by Real Symmetric Matrix is Hermitian, both are Hermitian.

Furthermore, both are their own inverse.

Hence they satisfy the definition of a unitary matrix.


However:

$AB = \begin {pmatrix} 0 & 1 \\ -1 & 0\end {pmatrix}$

and

$BA = \begin {pmatrix} 0 & -1 \\ 1 & 0 \end {pmatrix}$

Hence their product is not commutative.


Hence, by definition, $\map U n$ is not an abelian group.

$\blacksquare$


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