How to prove something like $\mathopen\mid{\rm Aut}(\mathbb{Z}_2\times\mathbb{Z}_2)\mathclose\mid=6$

Question

Doing this by trying all possible cases is not hard for a group like this but I've been wondering if there is a more systematic method, applicable to groups of higher order, that doesn't rely on visualising the "symmetries" of the group. In general, finding the symmetries of an object seems to me to be hard to formalise and to require just trying a number of possibilities.

Answer 1

$\DeclareMathOperator{\Aut}{Aut}$In general, this is not an easy problem.

In your particular case you may note that if $a, b, c$ are the three non-zero elements of $\mathbb{Z}_2\times\mathbb{Z}_2$ (I am using additive notation),then \begin{equation*}a + a = b + b = c + c = 0,\end{equation*} the group is commutative, and \begin{equation*}a + b = c, b + c = a, c + a = b.\end{equation*} To see the latter, just note that $a + b \ne 0$ (as $a \ne b$), and $a + b \ne a, b$ (as $a, b \ne 0$).

Then all permutations of $a, b, c$ yield an automorphism, so $\Aut(\mathbb{Z}_2\times\mathbb{Z}_2) \cong S_{3}$.