Problem
Find \(\operatorname{Aut}(\mathbb{Z})\).
Solution
Step 1
To determine the form of an automorphism
Since \(\mathbb{Z}\) is a cyclic group generated by \(1\), every automorphism \(\varphi: \mathbb{Z} \to \mathbb{Z}\) is completely determined by the image of \(1\). That is, for any \(n \in \mathbb{Z}\),
\[
\varphi(n) = n \cdot \varphi(1).
\]
Step 2
To identify the Possible Values of \(\varphi(1)\)
For \(\varphi\) to be an automorphism, it must be bijective and a homomorphism. The image \(\varphi(1)\) must be a generator of \(\mathbb{Z}\). The only generators of \(\mathbb{Z}\) are \(1\) and \(-1\). Therefore,
\[
\varphi(1) \in \{1, -1\}.
\]
Step 3
Define the isomorphism
Define the mapping
\[
f: \operatorname{Aut}(\mathbb{Z}) \to \{1, -1\} \quad \text{by} \quad f(\varphi) = \varphi(1).
\]
We now check that \(f\) is an isomorphism.
Step 4
To verify \(f\) is a homomorphism and bijective
Let \(\varphi, \psi \in \operatorname{Aut}(\mathbb{Z})\). Then, for the composition \(\varphi \circ \psi\):
\begin{align}
f(\varphi \circ \psi) &= (\varphi \circ \psi)(1) \nonumber\\[5mm]
&= \varphi\big(\psi(1)\big) \nonumber\\[5mm]
&= \varphi(1)\,\psi(1) \nonumber\\[5mm]
&= f(\varphi) \, f(\psi). \tag{1}
\end{align}
This shows that \(f\) is a homomorphism.
Since the only possible values for \(\varphi(1)\) are \(1\) and \(-1\), \(f\) is clearly surjective. Moreover, if \(f(\varphi) = f(\psi)\), then \(\varphi(1)=\psi(1)\) and hence \(\varphi=\psi\) (because automorphisms of \(\mathbb{Z}\) are determined by the image of \(1\)). Thus, \(f\) is injective.
