Show $\operatorname{Aut}(C_2 \times C_2)$ is isomorphic to $S_3=D_6$

Solution 1:

Notice that $C_2\times C_2$ is a vector space over the field $Z_2$. Hence $Aut(Z_2\times Z_2)=Gl(2,2)$. It is easy to see that $Gl(2,2)\cong S_3 $.

Solution 2:

There are $3$ non-trivial elements of $C_2 \times C_2$. Show that an automorphism must permute them and that any such permutation does in fact give rise to an automorphism. The group $D_6$ does the same thing, except to vertices of a triangle. So you can imagine writing the $3$ non-trivial elements on the vertices of a triangle to get the isomorphism explicitly !

Solution 3:

$\mathbb{Z}_2\times \mathbb{Z}_2=\{(0,0), (1,1), (1,0), (0,1)\}$. Any permutation of the three non-zero elements determines an automorphism, since the three non-zero elements are "almost same" in the following sense:

1) square of any non-zero element is identity

2) product of two non-zero (distinct) elements gives the third element.

The permutation group on $3$ letters is isomorphic to dihedral group of order $6$