# What is $\operatorname{Aut}(\mathbb{Z}_2\oplus \mathbb{Z}_2)$?

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 $$2$$.

– MJD
Nov 23, 2018 at 15:48
• @MJD: In that case I assume that there are more than $3$. Is that right? Nov 23, 2018 at 15:49
• For any prime $p$ and any $n$, the automorphisms of $\mathbb{Z}_p^n$ are all invertible linear maps of this group considered as a vector space over $\mathbb{Z}_p$. Nov 23, 2018 at 17:30

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).
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
$$\mathop{\mathrm{Aut}} (\mathbb{Z}_2\oplus\mathbb{Z}_2) \cong S_3.$$
• What happens when you keep taking $\text{Aut}(\cdot)$? Is it periodic? Nov 23, 2018 at 15:57
• @Prototank No, not at all. In particular, $\mathop{\mathrm{Aut}}(S_3)=S_3$, so $\mathop{\mathrm{Aut}}^n(\mathbb{Z}_2\oplus\mathbb{Z_2})\neq \mathbb{Z}_2\oplus\mathbb{Z_2}$ for any $n$. Nov 23, 2018 at 16:02