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