Let $n \in \mathbb{N}$ be odd. Show that: $$\Aut(\mathbb{Z}/{n\mathbb{Z}}) \cong \Aut(\mathbb{Z}/{2n\mathbb{Z}})$$

$\DeclareMathOperator{\Aut}{Aut}$ My attempt:

An automorphism $f \in \Aut(\mathbb{Z}/{n\mathbb{Z}})$ is uniquely represented by $f(1)$ since $1$ generates $\mathbb{Z}/{n\mathbb{Z}}$. $f(1)$ has to be a generator of $\mathbb{Z}/{2n\mathbb{Z}}$, which means $2n$ and $f(1)$ are relatively prime.

Thus, $\Aut(\mathbb{Z}/{n\mathbb{Z}})$ and $(\mathbb{Z}/{n\mathbb{Z}})^\times$ are actually isomorphic via the isomorphism $i \mapsto f_i$, where $f_i$ is an automorphism of $\mathbb{Z}/{n\mathbb{Z}}$ having $f_i(1) = i$.

Since we can similarly deduce that $\Aut(\mathbb{Z}/{2n\mathbb{Z}}) \cong (\mathbb{Z}/{2n\mathbb{Z}})^\times$, we have reduced the problem to showing that $$(\mathbb{Z}/{n\mathbb{Z}})^\times = (\mathbb{Z}/{2n\mathbb{Z}})^\times$$

Since $n$ and $2$ are relatively prime, we have:

$$\phi(2n) = \phi(2)\phi(n) = \phi(n)$$

Hence, both groups are of the same order, $\phi(n)$.

Now, I'm aware of the fact that $(\mathbb{Z}/{p\mathbb{Z}})^\times$ is a cyclic group if $p$ is prime, but that is certainly not the case for $2n$, so we cannot easily conclude that they are isomorphic.

The only thing that comes to mind is to try to find what the elementary divisors for an Abelian group of order $\phi(n)$ could be. For example, $\phi(n)$ is even for $n \ge 3$ so there exists a unique element of order $2$ in both groups, which is $-1$. So the isomorphism would send $-1$ to $-1$.

How should I proceed here?


If $n$ is odd, then $\newcommand{\Z}{\Bbb Z}\Z/2n\Z$ is isomorphic to $\Z/2\Z\times\Z/n\Z$. As $n$ is odd, any automorphism of $\Z/2\Z\times\Z/n\Z$ preserves this factorization, so comes from an automorphism of $\Z/2\Z$ and one of $\Z/n\Z$. But $\Z/2$ has only the trivial automorphism.

  • $\begingroup$ So, basically the argument goes like this$\DeclareMathOperator{\Aut}{Aut}$: $$\Aut(\mathbb{Z}/{2n\mathbb{Z}}) \cong \Aut(\mathbb{Z}/{2\mathbb{Z}} \times \mathbb{Z}/{n\mathbb{Z}}) \cong \Aut(\mathbb{Z}/{2\mathbb{Z}}) \times\Aut(\mathbb{Z}/{n\mathbb{Z}}) = \{0\} \times \Aut(\mathbb{Z}/{n\mathbb{Z}}) \cong \Aut(\mathbb{Z}/{n\mathbb{Z}}) $$ I was not aware of the fact that $\Aut(H \times K) = \Aut H \times \Aut K$ when $|H|$ and $|K|$ are relatively prime. $\endgroup$ – mechanodroid Nov 11 '17 at 15:53
  • $\begingroup$ @mechanodroid That's because any element of order dividing $|H|$ must be mapped into $H$ by any automorphism of $H\times K$ etc. $\endgroup$ – Lord Shark the Unknown Nov 11 '17 at 15:57

Hint : by the CRT, $\mathbb{Z}/2n\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ are isomorphic as rings for odd $n$, so their invertible groups are isomorphic. For rings $R_1, R_2$, what's $(R_1\times R_2)^\times$ ?

  • $\begingroup$ Ok, so basically this: $$(\mathbb{Z}/{2n\mathbb{Z}})^\times \cong (\mathbb{Z}/{2\mathbb{Z}} \times \mathbb{Z}/{n\mathbb{Z}})^\times \cong (\mathbb{Z}/{2\mathbb{Z}})^\times \times (\mathbb{Z}/{n\mathbb{Z}})^\times = \{1\} \times (\mathbb{Z}/{n\mathbb{Z}})^\times \cong (\mathbb{Z}/{n\mathbb{Z}})^\times$$ $\endgroup$ – mechanodroid Nov 11 '17 at 16:02
  • $\begingroup$ Yep that's exactly it ! $\endgroup$ – Max Nov 11 '17 at 17:01

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