How to find $\left|\operatorname{Aut}(\mathbb Z_2\times\mathbb Z_2)\right|$

Question

How to find $\left|\operatorname{Aut}(\mathbb Z_2\times\mathbb Z_2)\right|=?$

Since $\mathbb Z_2\times\mathbb Z_2$ has 4 elements, $1 \leq \left|\operatorname{Aut}(\mathbb Z_2\times\mathbb Z_2)\right|\leq \left|S_4\right|$.

How to proceed further?

Thanks in advance...

Answer 1

Notice that in $(\mathbb Z /2\mathbb Z) \times (\mathbb Z /2\mathbb Z)$ all elements besides the neutral have order two (and each of them is just the product of the other two!), so every automorphisms "permutes" them and maps the neutral element to itself.

We have therefore shown something stronger: $\textrm{Aut} (\mathbb Z /2\mathbb Z) \cong S_3$.