Back to easier problems for a bit…
I have been told that it is possible to find the order of the largest cyclic subgroup of $\mathrm{Aut}(\mathbb{Z}_{720})$ without considering automorphisms of $\mathbb{Z}_{720}$. Here's what I have:
We have that $\mathrm{Aut}(\mathbb{Z}_{720}) \cong U(720)$. Then since $720=2^4 \cdot 3^2 \cdot 5$ we have $U(720)=U(16) \oplus U(9) \oplus U(5)$. Some simple computations show that $U(16)$ is not cyclic, while $U(9)$ and $U(5)$ are, so $U(720) \cong U(16) \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_4$. Since $\gcd(6,4) \ne 1$, $\mathbb{Z}_6 \oplus \mathbb{Z}_4$ is not cyclic, so this, coupled with the fact that the noncyclic $U(16)$ of order 8 would have a cyclic subgroup of order 4 at the most, gives us that the largest cyclic subgroup of $\mathrm{Aut}(\mathbb{Z}_{720})$ is of order $6$.
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Is this correct?
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I didn't look at automorphisms of $\mathrm{Aut}(\mathbb{Z}_{720})$ per se, but I did consider elements of $\mathrm{Aut}(\mathbb{Z}_{720})$ up to isomorphism… Is there a way to derive this result just from looking at the structure of $\mathbb{Z}_{720}$?
Thanks.
Best Answer
If we find an element of maximal order in $Aut({\mathbb Z_{720}})$, then this element will be a generator of the maximal order cyclic subgroup of $Aut({\mathbb Z_{720}})$.
First, we use Thm: $Aut({\mathbb Z_n})\cong U(n)$. You had the first step:
$$Aut({\mathbb Z_{720}})\cong U(720)=U(16\cdot 9\cdot 5)\cong U(16)\oplus U(9)\oplus U(5)\cong {\mathbb Z_2}\oplus{\mathbb Z_4}\oplus{\mathbb Z_6}\oplus{\mathbb Z_4} $$
We see immediately that the orders of the elements of $U(720)$ can only be $1,2,3,4,6,$ and $12$, since an element from ${\mathbb Z_2}\oplus{\mathbb Z_4}\oplus{\mathbb Z_6}\oplus{\mathbb Z_4}$ has the form $(a,b,c,d)$, where $|a|=1$, $|b|\in \{1,2,4\}$, $|c|\in\{1,2,3,6\}$, and $|d|\in\{1,2,4\}$. Since the maximum order of an element in this form $(a,b,c,d)$ is the least common multiple of the orders of each element, it is easy to see that we see that $12$ is the maximum order of an element in $U(720)$. Hence the maximal order cyclic subgroup of $Aut({\mathbb Z_{720}})$ has order $12$.
We can arrive at the same conclusion considering the properties of $Aut({\mathbb Z_{720}})$ alone:
Let $\phi\in Aut({\mathbb Z_{720}})$. Then the mapping $\phi$ is completely determined by $\phi(1)$, which must be relatively prime to $720$. Note that ${\mathbb Z_{720}}\cong {\mathbb Z_{16}}\oplus {\mathbb Z_9}\oplus {\mathbb Z_5}$. Now the image under isomorphism $\phi(1)=(a,b,c)$, where $a,b$, and $c$ are multiplicative units of ${\mathbb Z_{16}}$, is a multiplicative unit of ${\mathbb Z_{16}}$, ${\mathbb Z_9}$, and ${\mathbb Z_5}$, respectively. But these groups of units are isomorphic to ${\mathbb Z_2}\oplus {\mathbb Z_4}, {\mathbb Z_6}$, and ${\mathbb Z_4}$ respectively. The order of $\phi$ is the least common multiple of $(|a|,|b|,|c|)$, and therefore is at most $12$.