Theorem
Prove that \(\operatorname{Aut}(\mathbb{Z}_n, +)\) is isomorphic with \(U_n\) (the group of units modulo \(n\)).
Proof
Step 1
Let us define the mapping
\[
f: \operatorname{Aut}(\mathbb{Z}_n, +) \to U_n \quad \text{by} \quad f(\varphi) = \varphi(1).
\]
where \(\varphi \in \operatorname{Aut}(\mathbb{Z}_n) \) for every automorphism \(\varphi\) of the cyclic group \((\mathbb{Z}_n, +)\) is determined by its action on the generator \(1\).
Step 2
To prove \(f\) is Well-Defined
We must verify that for any automorphism \(\varphi \in \operatorname{Aut}(\mathbb{Z}_n, +)\), the element \(\varphi(1)\) is a unit in \(\mathbb{Z}_n\). Since \(1\) is a generator of \(\mathbb{Z}_n\), its image \(\varphi(1)\) must also be a generator of \(\mathbb{Z}_n\). An element \(a \in \mathbb{Z}_n\) is a generator if and only if \(\gcd(a, n) = 1\); that is, \(a \in U_n\). Therefore,
\[
f(\varphi) = \varphi(1) \in U_n.
\]
This confirms that \(f\) is well-defined.
Step 3
To prove \(f\) is a homomorphism
For any \(\varphi, \psi \in \operatorname{Aut}(\mathbb{Z}_n, +)\), we need to show that \(f(\varphi \circ \psi) = f(\varphi) f(\psi)\). Recall that \(f(\varphi) = \varphi(1)\), and since \(1\) is the generator of \(\mathbb{Z}_n\), the behavior of \(\varphi\) and \(\psi\) is completely determined by their action on \(1\).
\begin{align}
f(\varphi \circ \psi) &= (\varphi \circ \psi)(1) \nonumber\\
&= \varphi\big(\psi(1)\big) \quad \nonumber\\
&= \varphi(1)\,\psi(1) \quad \text{(since \(\varphi\) acts linearly on generators)} \nonumber\\
&= f(\varphi)\,f(\psi) \quad \nonumber
\end{align}
This confirms that \(f\) preserves the group operation, so it is a homomorphism.
Step 4
To prove \(f\) is Injective
Suppose \(f(\varphi) = f(\psi)\), that is, \(\varphi(1) = \psi(1)\). Since every element \(k \in \mathbb{Z}_n\) can be written as
\[
k = \underbrace{1 + 1 + \cdots + 1}_{k \text{ times}},
\]
we have:
\begin{align}
\varphi(k) &= k\,\varphi(1) \nonumber\\
&= k\,\psi(1) \nonumber\\
&= \psi(k)\nonumber
\end{align}
Hence, \(\varphi = \psi\), and \(f\) is injective.
Step 5
To prove \(f\) is Surjective
Let \(a \in U_n\). Since \(a\) is a unit modulo \(n\) (i.e., \(\gcd(a, n) = 1\)), consider the map
\[
\varphi_a: \mathbb{Z}_n \to \mathbb{Z}_n \quad \text{defined by} \quad \varphi_a(k) = ak.
\]
Well-Defined and Homomorphism:
First, \(\varphi_a\) is well-defined because multiplication by \(a\) respects the equivalence classes in \(\mathbb{Z}_n\); that is, if \(k_1 \equiv k_2 \pmod{n}\), then \(ak_1 \equiv ak_2 \pmod{n}\). Also, for any \(k, l \in \mathbb{Z}_n\),
\begin{align}
\varphi_a(k+l) &= a(k+l) \nonumber\\[5mm]
&= ak + al \nonumber\\[5mm]
&= \varphi_a(k) + \varphi_a(l), \nonumber
\end{align}
so \(\varphi_a\) is a group homomorphism.
Invertibility:
Since \(a \in U_n\), there exists an inverse \(a^{-1} \in \mathbb{Z}_n\) such that \(aa^{-1} \equiv 1 \pmod{n}\). Define
\[
\varphi_{a^{-1}}(k) = a^{-1}k.
\]
Then, for any \(k \in \mathbb{Z}_n\),
\begin{align}
(\varphi_a \circ \varphi_{a^{-1}})(k) &= \varphi_a(a^{-1}k) = a(a^{-1}k) = k \nonumber\\[5mm]
(\varphi_{a^{-1}} \circ \varphi_a)(k) &= \varphi_{a^{-1}}(ak) = a^{-1}(ak) = k \nonumber.
\end{align}
Hence, \(\varphi_a\) is invertible and is an automorphism of \((\mathbb{Z}_n, +)\).
Clearly,
\[
f(\varphi_a) = \varphi_a(1) = a.
\]
Therefore, every element \(a \in U_n\) is in the image of \(f\), proving that \(f\) is surjective.
