Group of Unitary Matrices under Multiplication is not Abelian

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.

$\blacksquare$

Sources

• 1974: Robert Gilmore: Lie Groups, Lie Algebras and Some of their Applications ... (previous) ... (next): Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $2$. GROUP