# Square Matrices with +1 or -1 Determinant under Multiplication forms Group

## Theorem

Let $n \in \Z_{>0}$ be a strictly positive integer.

Let $S$ be the set of square matrices of order $n$ of real numbers whose determinant is either $1$ or $-1$.

Let $\struct {S, \times}$ denote the algebraic structure formed by $S$ whose operation is (conventional) matrix multiplication.

Then $\struct {S, \times}$ is a group.

## Proof

Taking the group axioms in turn:

### Group Axiom $\text G 0$: Closure

Let $\mathbf A, \mathbf B \in S$.

By definition of matrix product, $\mathbf {A B}$ is a square matrix of order $n$.

From Determinant of Matrix Product:

- $\det \mathbf A \det \mathbf B = \map \det {\mathbf {A B} }$

Hence the determinant of $\mathbf {A B}$ is either $1$ or $-1$.

Thus $\mathbf {A B} \in S$ and so $\struct {S, \times}$ is closed.

$\Box$

### Group Axiom $\text G 1$: Associativity

We have that Matrix Multiplication is Associative.

Thus $\times$ is associative on $\struct {S, \times}$.

$\Box$

### Group Axiom $\text G 2$: Existence of Identity Element

From Determinant of Unit Matrix, the unit matrix $\mathbf I$ is in $S$.

From Unit Matrix is Unity of Ring of Square Matrices, the unit matrix $\mathbf I$ serves as the identity element of $\struct {S, \times}$.

$\Box$

### Group Axiom $\text G 3$: Existence of Inverse Element

Because the determinants of the elements of $S$ are $1$ or $-1$, they are by definition invertible.

We have that $\mathbf I$ is the identity element of $\struct {\R, \circ}$.

From the definition of invertible matrix, the inverse of any invertible matrix $\mathbf A$ is $\mathbf A^{-1}$.

$\Box$

All the group axioms are thus seen to be fulfilled, and so $\struct {S, \times}$ is a group.

$\blacksquare$

## Sources

- 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $1$: Definitions and Examples: Exercise $1 \ \text{(d)}$