Problem
Prove that \(\operatorname{Aut}(\mathbb{Q},+)\) is isomorphic with \((\mathbb{Q}^*,\cdot)\).
Solution
Step 1
To determine the form of an automorphism
Since \((\mathbb{Q},+)\) is a cyclic group generated by \(1\), any automorphism \(\varphi: \mathbb{Q} \to \mathbb{Q}\) is completely determined by \(\varphi(1)\). In fact, for any \(r \in \mathbb{Q}\),
\[
\varphi(r) = r \cdot \varphi(1)
\]
Step 2
Let us define the mapping
\[
f: \operatorname{Aut}(\mathbb{Q},+) \to \mathbb{Q}^*
\]
by
\[
f(\varphi) = \varphi(1)
\]
Note that \(\mathbb{Q}^*\) denotes the set of nonzero rational numbers under multiplication.
Step 3
To prove \(f\) is well-defined
We must verify that for any automorphism \(\varphi \in \operatorname{Aut}(\mathbb{Q},+)\), the element \(\varphi(1)\) is nonzero, i.e., \(\varphi(1) \in \mathbb{Q}^*\). If \(\varphi(1) = 0\), then for every \(r \in \mathbb{Q}\) we would have
\[
\varphi(r) = r \cdot \varphi(1) = 0
\]
which contradicts the fact that \(\varphi\) is an isomorphism (since it would not be injective). Therefore, \(\varphi(1) \neq 0\) and \(f(\varphi) \in \mathbb{Q}^*\). This shows that \(f\) is well-defined.
Step 4
To prove \(f\) is a homomorphism
Let \(\varphi, \psi \in \operatorname{Aut}(\mathbb{Q},+)\). Then for any \(r \in \mathbb{Q}\),
\begin{align}
f(\varphi \circ \psi) &= (\varphi \circ \psi)(1) \nonumber\\
&= \varphi\big(\psi(1)\big) \nonumber\\
&= \varphi(1) \, \psi(1) \nonumber\\
&= f(\varphi) \, f(\psi)\nonumber
\end{align}
Step 5
To prove \(f\) is injective
Let \(f(\varphi) = f(\psi)\); that is, \(\varphi(1) = \psi(1)\). Then for any \(r \in \mathbb{Q}\),
\begin{align}
\varphi(r) &= r \cdot \varphi(1) \nonumber\\
&= r \cdot \psi(1) \nonumber\\
&= \psi(r) \nonumber
\end{align}
Hence, \(\varphi = \psi\) and \(f\) is injective.
Step 6
To prove \(f\) is surjective
For any \(q \in \mathbb{Q}^*\), define \(\varphi_q: \mathbb{Q} \to \mathbb{Q}\) by
\[
\varphi_q(r) = q \cdot r
\]
It is straightforward to check that \(\varphi_q\) is an automorphism of \((\mathbb{Q},+)\) and
\[
f(\varphi_q) = \varphi_q(1) = q
\]
Thus, \(f\) is surjective.
Step 7
To prove \(f\) is isomorphism
Since \(f\) is a bijective homomorphism, then
\[
\operatorname{Aut}(\mathbb{Q},+) \cong \mathbb{Q}^*
\]
