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show that the group $GL(2,\mathbb{Z}_2)$ has order 6 by listing all its elements

I think that $GL(2,Z_2)$ is 2x2 matrices with $Z_2$ elements that is

$$\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} ,\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} $$

$$ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} , \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} $$ $$ \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$

$$ \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} $$

I got that there are 15 elements so that would make it order of 15. but there are suppose to be 6 right?

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    How many of those are invertible? – 2017-01-27
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    ok did not know that thanks got the 6 – 2017-01-27
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    You've listed the $16$ elements of $M(2,\mathbb Z_2)$, not $GL(2,\mathbb Z_2)$. – 2018-03-16

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By definition, elements in $GL(2,Z_2)$ have nonzero determinants, so you have to delete those matrix with determinants equals to 0.