= Solution
The three nonzero <vectors> in $\mathbb F_2^2$ are $e_1$, $e_2$ and $e_1+e_2$. An invertible <linear map> permutes this three-element set, so part (b) gives a <group homomorphism>
$$
\rho:GL_2(\mathbb F_2)\longrightarrow S_3.
$$
If a matrix acts trivially on all three vectors, it fixes both basis vectors $e_1,e_2$, so it is the identity matrix. Thus the <kernel of a group homomorphism> is trivial and $\rho$ is injective.
To count the domain, its first column may be any of the three nonzero vectors. The second column must be outside the two-element span of the first, leaving two choices. Hence
$$
|GL_2(\mathbb F_2)|=3\cdot2=6=|S_3|.
$$
An injection between these finite sets of equal size is surjective, proving
$$
\boxed{GL_2(\mathbb F_2)\cong S_3.}
$$
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