Clarifications needed for a question concerning ${\rm Aut}(\mathbb{Z}_{2}\times \mathbb{Z}_{2})\cong S_{3}$

I am trying to show ${\rm Aut}(\mathbb{Z}_{2}\times \mathbb{Z}_{2})\cong S_{3}$. So in the included image, example 8, it asks the reader to find all the automorphism group in the set $\mathbb{Z}_{2}\times \mathbb{Z}_{2}$. The solution states:

Every automorphism of $\mathbb{Z}_{2}\times \mathbb{Z}_{2}$ must leave the identity $(0,0)$ fixed. However, the three other elements are algebraically identical--for instance, they are each of order 2 and commute with the other elements. So any of the six permutations of them gives an automorphism. Thus ${\rm Aut}(\mathbb{Z}_{2}\times \mathbb{Z}_{2})\cong D_{3}$. We can also represent these automorphism as $2 \times 2$ matrices from the ring of matrices $M_{2}(\mathbb{Z}_{2})$, mapping the four vectors [0 0], [1 0], [0 1], [1 1] to themselves.

I am confused about what the sentence "we can also represent these automorphism as $2 \times 2$ matrices from the ring of matrices $M_{2}(\mathbb{Z}_{2})$, mapping the four vectors [0 0], [1 0], [0 1], [1 1] to themselves." means. I mean is it saying that the reader can create some sort of isomorphic linear transformations sending each one of those four vectors to a two by two matrix with a determinant equal to one? Or does it mean something else. I am also not clear in how to transcribe that sentence into mathematical notation. Thank you in advance.

Solution 1:

Every automorphism of $G:=(\Bbb{Z}/2\Bbb{Z})\times(\Bbb{Z}/2\Bbb{Z})$ is a bijection from $G$ to itself. Moreover any automorphism maps $(0,0)$ to $(0,0)$. From here it is not difficult to check that any such automorphism is in fact an $\Bbb{F}_2$-linear map, where $\Bbb{F}_2=\Bbb{Z}/2\Bbb{Z}$ is the field of two elements.

Every $\Bbb{F}_2$-linear map from $G=\Bbb{F}_2^2$ to itself can be represented by a $2\times2$-matrix with coefficients in $\Bbb{F}_2$. This is what the quoted sentence intends to convey; that there exists a map $$\operatorname{Aut}(G)\ \longrightarrow\ \operatorname{GL}_2(\Bbb{F}_2),$$ that maps automorphisms of $G$ to (invertible) $2\times2$-matrices over $\Bbb{F}_2$. Such a map is in fact an isomorphism, but the text you quote does not make a convincing argument that this is so, in my humble opinion.