# Help with Proof: $Aut_{Grp}(\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}) \cong S_3$ [duplicate]

I need help improving my proof. It's a bit lacking in details, and I don't know how to phrase my reasoning.

For any homeomorphism, identities must map to each other. Therefore, any automorphism of $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ must map $(0,0) \mapsto (0,0)$.

This leaves three other elements, label them $1: (0,1), 2: (1,0), 3: (1,1)$.

Since all of these elements are order two, and the group action of any two will give the third, they can be interchanged with each other freely. So, any bijection of these three elements will result in an automorphism. Hence, there is a function between the automorphisms of $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ and the symmetric group $S_3$

To prove this is a homeomorphism, we show $\mu(g\circ h) = \mu(g)\cdot\mu(h)$. This is true because $g$ and $h$ are automorphisms, and the bijection above supports this property, and this is circular reasoning that I can't seem to articulate more clearly.

To prove this is a bijection, like before, the function maps the pairs to their labels. It's injective, because if two labels are equal, then their points are equal, and so it is with the functions inputs and outputs. It's also surjective because of labeling of inputs and outputs carrying over to functions.

Is "labeling" enough to prove bijection? I don't want to just enumerate all the functions, there has to be a better way. The reasoning for homeomorphism and bijection are both really lame, but I don't know what sort of reasoning I can employ to make this more concise.

This is in the "basic" Group Theory section of Paolo Aluffi's "Algebra: Chapter 0". I have not yet gotten to Rings, Fields, Modules, or any Linear Algebra.

Let $G=Aut(\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z})$.
• Prove that $G$ has over $6$. See this question.
• Prove that $G$ is not abelian. You just need to find two elements that do not commute.
The given group $G$ is isomorphic with the abelian group underlying in the vector space structure of $\Bbb F_2^2$ of dimension $2$ over the field with two elements $\Bbb F_2$. It has a canonical basis, the two basis elements $e_1=(1,0)$, $e_2=(0,1)$ were named already in the question. There is also an element $v=e_1+e_2=(1,1)$. The proposed solution should also show/check that moving $e_1,e_2$ to two different elements among the above three elements $e_1,e_2,v$ indeed leads to a permutation, i.e. the third element $v$ is moved to the not mapped element among them! (This is of couse so, since $e_1+e_2+v=0$, and $0$ is mapped to $0$ by a group/vectorspace automorphism.)
Alternatively: We are searching for the structure of $G=\operatorname{GL}(2, \Bbb F_2)$, an element $g$ of it is a $2\times 2$ matrix of determinant $1$, its first column (the image of $e_1$ by $g$) is any non-zero vector, there are $3=2^2-1$ such vectors, its second column (the image of $e_2$ by $g$) is any non-zero vector different from (i.e. independent of) the first column, there are $2=2^2-2^1$ such vectors, so $G$ has six elements.