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I know that $\mathrm{Aut}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} ) \cong S_3$, where $S_3$ is the symmetric group. I do not know how to prove that they are isomorphic, however.

What I tried was finding a specific $\phi:\mathrm{Aut}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})\rightarrow S_3$, but the part that confuses me is that the automorphism group is a group of isomorphisms itself, so I don't know how to send an isomorphism to $S_3$.

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  • $\begingroup$ One step that may take you a long way: can you find an element in $Aut(\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z})$ of order 3? $\endgroup$
    – Steven Stadnicki
    Feb 28, 2014 at 5:26
  • $\begingroup$ Let $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} = G$. If $\phi \in Aut(G)$ then $\phi(e_G)=e_G$ (where $e_G = (0,0))$. Now choose $\phi(a)=a, \phi(b)=b, \phi(c)=c$. $S_3 = (abc)$ (i.e. permutations of $\{a,b,c\}$ where $a,b,c$ have order $2$). This is as far as I got; I do not see intuitively the element of order 3. $\endgroup$
    – user113525
    Feb 28, 2014 at 5:37
  • $\begingroup$ What if $\phi(a) = b, \phi(b) = c, \phi(c) = a$? $\endgroup$
    – John Habert
    Feb 28, 2014 at 5:40
  • $\begingroup$ Then that wouldn't be a bijection? I'm not following. $\endgroup$
    – user113525
    Feb 28, 2014 at 5:41
  • $\begingroup$ Why wouldn't it be a bijection? It is one-to-one and onto. (We must always have $\phi(e) = e$ so I did not write it down.) $\endgroup$
    – John Habert
    Feb 28, 2014 at 5:43

1 Answer 1

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Notice that $\mathbb{Z}/ 2 \mathbb{Z} \times \mathbb{Z}/ 2 \mathbb{Z}$ is a two dimensional vector space over $\mathbb{Z}/ 2 \mathbb{Z}$. There are $3$ nonzero vectors $(1,0), (0,1), (1,1)$. Now, any automorphism $\phi$ of this vector space will be a linear transformation and so is determined by where it map $(1,0)$ and $(0,1)$. Moreover, we observe that if we are given the action of $\phi$ on one of these basis vectors, the image under $\phi$ of the remaining basis vector can be either of the other two nonzero vectors. Thus, we see that the automorphisms are precisely the permuations of the three nonzero vectors in this vector space.

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  • $\begingroup$ What remains to be seen is that an automorphism of the group $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ corresponds uniquely to an automorphism of the vector space $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, and conversely $\endgroup$
    – zcn
    Feb 28, 2014 at 6:44
  • $\begingroup$ Right. A vector space is an abelian group though, so any vector space automorphism is a group automorphism. But $\mathbb{Z}/2 \mathbb{Z}$ just has two elements, and multiplication by either of them clearly commutes with a group automorphism. $\endgroup$
    – Alexander
    Feb 28, 2014 at 6:48
  • $\begingroup$ It is certainly straightforward to do in this case, but still necessary to check, since the automorphism groups are in different categories (personally, I would stay in the category of groups to prove this particular fact, but the fact that the automorphism groups are the same (for groups or vector spaces) is also important) $\endgroup$
    – zcn
    Feb 28, 2014 at 6:55
  • $\begingroup$ That's a fair point. $\endgroup$
    – Alexander
    Feb 28, 2014 at 6:57

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