I am pretty sure that this has been asked before, but I can't find it. My question is what
$$
\operatorname{Aut}(\mathbb{Z}_2\oplus \mathbb{Z}_2)
$$
is. (Here $\mathbb{Z}_2\oplus \mathbb{Z}_2$ is the (external) direct product).
My thinking is that this would be isomorphic to $\mathbb{Z}_3$ since $\mathbb{Z}_2\oplus \mathbb{Z}_2$ has three elements of order $2$, so there are three choices for where to send an element of order $2$.
Best Answer
Take a pair of generators $a,b$ for $\mathbb{Z}_2\oplus \mathbb{Z}_2$. Then any automorphism is determined by where it sends $a$ and $b$. There are three places to send $a$ (we can send it to $a$, $b$, or $ab$), and for each of those, there are two places to send $b$ (we can send it to either of those that we didn't send $a$ to). Thus, our group has 6 elements, so certainly isn't $\mathbb{Z}_3$: it's either $\mathbb{Z}_6$ or $S_3$ (these being the only two groups of order 6).
To distinguish them, we just need to check the orders of our elements:
Note that the map swapping $a$ and $b$ has order $2$, as do the maps sending $a$ to $ab$ and fixing $b$, and the map fixing $a$ and sending $b$ to $ab$. Thus, our automorphism group has at least three elements of order $2$, but $\mathbb{Z}_6$ has only one such element, so we must have
$$\mathop{\mathrm{Aut}} (\mathbb{Z}_2\oplus\mathbb{Z}_2) \cong S_3.$$