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Show $\operatorname{Aut}(C_2 \times C_2)$ is isomorphic to $D_6$ (the group with $x^3=1$, $y^2=1$ and $xy=yx^2$).

I'm not really sure how to express the elements of $\operatorname{Aut}(C_2 \times C_2)$. Would it be sufficient to show the elements of $\operatorname{Aut}(C_2 \times C_2)$, find their order and show they bijectively map to every element of $D_6$ and satisfy $xy=yx^2$?

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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 !

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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 $.

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$\mathbb{Z}_2\times \mathbb{Z}_2=\{(0,0), (1,1), (1,0), (0,1)\}$. 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 $6$

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