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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 $6$.

  1. Is this correct?

  2. 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 $\mathbb{Z}_{720}$?


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up vote 2 down vote accepted

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 $12$.

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Sir could you please explain why (3rd last line) a,b,c are the multiplicative units and how you are proceeding afterwards? – akansha Mar 8 '15 at 10:51

If $\gcd(n,m)=1$, note that $$\mathbb Z_n\oplus\mathbb Z_m\cong\mathbb Z_{nm}.$$ Therefore $$\mathbb Z_4\oplus\mathbb Z_6\cong \mathbb Z_4\oplus\mathbb Z_2\oplus\mathbb Z_3\cong \mathbb Z_2\oplus\color{red}{\mathbb Z_{12}}.$$

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Once you have found the structure $\def\Z{\Bbb Z}\Z_4\oplus\Z_2$, you can apply the structure theorem for fintely generated Abelian groups (or more elementarily a few applications ot the Chinese remainder theorem) to conclude that $(720)\cong\Z_2\oplus\Z_2\oplus\Z_4\oplus\Z_{12}$. But if you only need to find the order of the largest cyclic subgroup, it is easier to use that the order of an element $g$ in a direct product of groups $G_i$ is the least common multiple of the orders of its components $g_i\in G_i$. In particular if the groups $G_i$ are Abelian (so that the set of occurring orders is closed under least common multiples) the maximal possible order is the least common multiple of the maximal orders in the groups $G_i$. For $U(720)$ the latter number is $\operatorname{lcm}(4,6,4)=12$.

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