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So I believe I've finished the proof, but it all amounts to case-work and I really think (and hope) there's a better approach than this. Here's what I've done:

Attempt:

Within $\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ identities must be mapped to identities thus what defines each isomorphism is the bijective function between its last three elements: \begin{matrix} (0,0) & \mapsto & (0,0) \\ (0,1) & & (0,1) \\(1,0) & & (1,0) \\(1,1) & & (1,1) \\ \end{matrix} If every bijective function between these last three elements constitutes a homomorphism (including the first unchanged mapping of identities), then $\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \cong S_3$ in that isomorphisms in $\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ can be identified with permutations of three elements, which is exactly what $S_3$ is.

I then, essentially go case-by-case showing every one of the six different bijections is a homomorphism. For two of those bijections (which are just compositions of other bijections I've already checked) I apply the following lemma:

If $\phi :G \to H$ and $\psi : H \to K$ are bijective homomorphisms, then $(\psi \circ \phi)$ is a homomorphism. As proof, assume the premise. We then have $\forall a,b \in G:$ $$\psi (\phi (ab))= \psi(\phi(a) \phi(b))=\psi (\phi(a))\psi (\phi(b))$$

$\implies (\psi \circ \phi)$ is a homomorphism.

So my questions are:

  1. Is the proof that I have sufficient?
  2. Is there a better proof for this problem? (A hint towards it rather than complete give-away would be much more appreciated)

EDIT: I have indeed looked at this post already, but the approach carried out there uses vector spaces which I have not been introduced to or am supposed to use to solve this problem.

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  • $\begingroup$ Since you're requiring $(1,1)\mapsto (1,1),$ then it seems like you're thinking about ring automorphisms. A group automorphism would allow $(1,1)\mapsto (1,0)$ for example. Right now the map you defined explicitly is of order $2$ in the automorphism group. You need an element of order $3$ to even start thinking of $S_3$ as an option. $\endgroup$ – Chickenmancer Aug 14 '17 at 3:22
  • $\begingroup$ Every homomorphism of a group with a generating set $S$ is completely determined by its action on the elements of $S$. For instance, every homomorphism $\varphi\colon \mathbb Z \to G$ (for any group $G$) is completely determined by $\varphi(1)$. For, $\varphi(n) = (\varphi(1))^n, \forall n \in \mathbb Z$. $\endgroup$ – M. Vinay Aug 14 '17 at 3:23
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    $\begingroup$ You don't need group actions. Show that you have an automorphism $\phi,$ of order 3 and an automorphism $\psi$ of order 2, show they don't commute, and then use the fact that $S_3$ is the unique (up to isomorphism) group of order 6 which is non-commutative. $\endgroup$ – Chickenmancer Aug 14 '17 at 3:26
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    $\begingroup$ @AndrewTawfeek Oh, by "action", I did not mean "group action" — I meant the action of the homomorphism on the set of generators. To be precise, if $g_1, g_2, \ldots$ together form a set of generators of the group $G$, then any homomorphism $\varphi \colon G \to H$ (to any group $H$) is completely determined by the images (in $H$) of $\varphi(g_1), \varphi(g_2), \ldots$. This is because every element of $G$ is a product of the $g_i$'s and their inverses. $\endgroup$ – M. Vinay Aug 14 '17 at 3:30
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    $\begingroup$ I suppose it depends on what you've done in your course so far. To be totally rigorous, just show that $S_3$ is generated by any one of its elements of order 2 together with a 3-cycle, and then show that you have an isomorphism. You might want to try showing that $S_3$ is unique as another exercise. $\endgroup$ – Chickenmancer Aug 14 '17 at 3:38
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Consider a group automorphism $T : \mathbb Z_2\times \mathbb Z_2\to \mathbb Z_2\times \mathbb Z_2$ and write $e_1 = (1,0)$ and $e_2 = (0,1)$ for the group generators. This determines a matrix (just like with vector spaces, by looking at the images of $e_1$ and $e_2$), and it is invertible because $T$ is. Conversely, any $2\times 2$ invertible matrix with entries in $\mathbb Z_2$ determines an automorphism of your group, and this assignment is bijective.

One can list all possible such matrices by inspection (just look at those with determinant $1$):

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

Finally, note that this is not an abelian group, and recall that the only non-abelian group of order $6$, up to isomorphism, is $S_3$, generated by $s,t$ subject to $r^2,s^3$ and $(sr)^2$.

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Note that $\Bbb Z_ 2 \times\Bbb Z_2=<(0,1),(1,0)>=<(1,0),(1,1)>=<(1,1),(0,1)>$ and these are only only generators with minimum no. of elements(equivalent to basis of a vector space). Now f is a automorphism on $\Bbb Z_ 2 \times\Bbb Z_2$ iff $f$ sends any of these three basis-sets to another basis-sets.i.e. we have 3 choices of a basis-set to be image of f .Also for any of these choices we have 2 choices to pick a particular value of a element.i. e there are 6 automorphism(for example Consider $f({(0,1),(1,1)})={(0,1),(1,0)}$ . The rhs can be chosen 3 ways and each case can be done for 2 ways for $f((0,1))=(0,1)$ or $(1,0))$ So$ |Aut(\Bbb Z_ 2 \times\Bbb Z_2)|=6$ Also note that the maximum order of a element this group is 3 i.e $Aut(\Bbb Z_ 2 \times\Bbb Z_2) \cong S_3$

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