Basic question about automorphisms of structure $\mathbb{Z}_{adgp}$

Question

So, I've started reading "An invitation to model theory" by Jonathan Kirby. I got stuck at the very beginning already.

In Example 1.12 there is:

The only automorphisms of $\mathbb{Z}_{adgp}$ are the identity map and the map $x\mapsto -x$. To see this, note that if $\pi : \mathbb{Z} \rightarrow \mathbb{Z}$ is an $L_{adgp}$-embedding, then $\pi(0)=0$ ...

Why is it that $\pi(0) = 0$?

Here:

$L_{adgp}=\left\langle +, -, 0\right\rangle$ is the language of groups written additively.

Answer 1

If $\mathcal{M}$ and $\mathcal{N}$ are any structures (in any language $L$), and $\alpha : \mathcal{M} \to \mathcal{N}$ is an embedding, then for each constant symbol $c \in L$ we must have $\alpha(c^{\mathcal{M}}) = c^{\mathcal{N}}$ (this is part of the definition of embedding).

In particular, if $\mathcal{M} = \mathcal{N}$, as in your case, then any embedding of $\mathcal{M}$ to itself must fix the interpretations of all constant symbols. Since $0$ is in your language, we have $\pi(0) = 0$.

Answer 2

This follows immediately from the definition of embedding. Embeddings preserve the interpretations of the symbols in the language. In particular, any embedding must send the element named by the constant symbol $0$ to the element named by the constant symbol $0$.