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?
Answer
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.
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