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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 $2$.

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  • $\begingroup$ Your suggested answer is not correct $\endgroup$
    – MJD
    Nov 23, 2018 at 15:48
  • $\begingroup$ @MJD: In that case I assume that there are more than $3$. Is that right? $\endgroup$
    – John Doe
    Nov 23, 2018 at 15:49
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    $\begingroup$ 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 $\mathbb{Z}_p$. $\endgroup$
    – Kapil
    Nov 23, 2018 at 17:30

1 Answer 1

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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

$$\mathop{\mathrm{Aut}} (\mathbb{Z}_2\oplus\mathbb{Z}_2) \cong S_3.$$

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  • $\begingroup$ What happens when you keep taking $\text{Aut}(\cdot)$? Is it periodic? $\endgroup$
    – J. Moeller
    Nov 23, 2018 at 15:57
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    $\begingroup$ I'm a numpty and wrote the wrong thing, give me a second. $\endgroup$ Nov 23, 2018 at 15:58
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    $\begingroup$ @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 $n$. $\endgroup$ Nov 23, 2018 at 16:02
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    $\begingroup$ @Prototank That is known as the automorphism tower problem. There is a lot of literature on it. $\endgroup$
    – Derek Holt
    Nov 24, 2018 at 8:55

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