Solution: Any element of $GL(2, \mathbb{Z} _2)$ will be \[ \left\{ \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}: ad - bc \neq 0 \text{ and } a,b,c,d \in \mathbb{Z} _2. \right\} \] Since \( \mathbb{Z} _2 = \{0,1\} \), there are \( 2^4 = 16 \) possible matrices. However, only those matrices for which \( ad - bc \neq 0 \) are in \( GL(2, \mathbb{Z} _2) \). The six elements of \( GL(2, \mathbb{Z} _2) \) are \[ \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}, \quad \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix}, \quad \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ \end{bmatrix}, \quad \begin{bmatrix} 1 & 1 \\ 1 & 0 \\ \end{bmatrix}, \quad \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}, \quad \begin{bmatrix} 0 & 1 \\ 1 & 1 \\ \end{bmatrix}. \]

To show that \( GL(2, \mathbb{Z} _2) \) is non-Abelian, we will find two elements that do not commute. Consider the matrices \[ A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \\ \end{bmatrix}. \] Then \[ AB = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix} \quad \text{and} \quad BA = \begin{bmatrix} 1 & 1 \\ 1 & 0 \\ \end{bmatrix}. \] Since \( AB \neq BA \), we conclude that \( GL(2, \mathbb{Z} _2) \) is non-abelian.