I have to prove that the only automorphism of the structure $(\mathbb{Q}, \cdot, S)$ (where $S$ is a successor function) is the identity.
My question is: am I allowed to use that, since we have multiplication and we need neutral element, for any automorphism $h$, $h(1)=1$? I mean, is $(\mathbb{Q}, \cdot, S)$ really the same as $(\mathbb{Q}, 1, \cdot, S)$?
Let $h$ be an automorphism of the structure $(\Bbb{Q}, \cdot, S)$ (comprising the rational numbers, the multiplication function $x \cdot y = xy$ and the "successor" function $S(x) = x + 1$). Now $0$ is characterised by the property that $0 \cdot x = 0$ for every $x \in \Bbb{Q}$, but then:
$$h(0) \cdot x = h(0) \cdot h(h^{-1}(x)) = h(0 \cdot h^{-1}(x)) = h(0)$$ for every $x \in \Bbb{Q}$, so that $h(0)$ satisfies the characterising property of $0$ and hence $h(0) = 0$. Now using $h(S(0)) = S(h(0))$, you have, by induction, that $h(x) = x$ for every $x$ in $\Bbb{N}$. Using the fact that $x = y - 1$ iff $y = S(x)$ (which is equivalent to $h(y) = S(h(x))$) you get that $h(x) = x$ for every $x \in \Bbb{Z}$. Finally, assuming $y \in \Bbb{Z}$ and $z \in \Bbb{N}_{>0}$, you can use the fact that $x = \frac{y}{z}$ iff $y = x\cdot z$ (which is equivalent to $h(y) = h(x) \cdot h(z)$) to get that $h(x) = x$ for every $x \in \Bbb{Q}$.
In passing, you have also shown that an automorphism of $(\Bbb{Q}, \cdot, S)$, is also an automorphism of $(\Bbb{Q}, 1, \cdot, S)$. You are allowed to use this fact, but only after you have proved it $\ddot{\smile}$.
