Verify that the sets listed in (13.7e) are groups. Hint: See the proofs in (13.7f).

The matrices form a group, the set is closed under matrix multiplication, whether it is associative or as usual, and the existence of a unitary and inverse elements. It is known that matrix multiplication is associative. Matrices that belong to need to satisfy the following condition:

$$\begin{gathered} A{A}^T=A^{T}A=I \\ {A}^{-1}=A^T \end{gathered}$$

On the other hand, the associative attribute is concerned with the grouping of items in operation.

Let $A=I$, the equation holds, therefore unit element is included in $O\left(2\right)$group. The condition above is equivalent to the one that the absolute value of the determinant has to be equal to 1:

$\left|det\left(A\right)\right|=1$

Prove that the product of two orthogonal matrices has a unitary determinant, we will know that the resulting matrix is unitary as well. It is known that:

$det\left(AB\right)=det\left(A\right)·det\left(B\right)$

If A and B are orthogonal matrices, then their product has unitary determinants, therefore, belongs to the same group.

$A^{-1}=A^T$

And determinant of the transpose of the matrix is the same as that of the original matrix:

$det\left(A\right)=det\left(A^T\right)$

Orthogonal matrix:

$det\left(A\right)=det\left(A^{-1}\right)$

This means that the inverse matrix has the same determinant as the orthogonal matrix. The existence of an inverse element and $O\left(2\right)$is a group under matrix multiplication.