Given as exercise 2.5.1:
Show that the complete type realized by $1$ in $(\mathbb Z,+)$ is non-isolated. HINT Use the preceding exercise.
The previous exercise:
Let $\vec a$ and $\vec b $ be finite sequences from $\mathcal M.$ Prove that $\operatorname{tp}_{\mathcal M}(\vec a\vec b)$ is isolated iff $\operatorname{tp}_{\mathcal M}(\vec a/\vec b)$ and $\operatorname{tp}_{\mathcal M}(\vec b)$ are both isolated. Using this fact, show that when $\mathcal M$ is an atomic model and $\vec b\in M,$ then $\mathcal M$ is atomic over $\vec b.$ Conversely, if $\mathcal M$ is atomic over $\vec b$ and $\operatorname{tp}_{\mathcal M}(\vec b)$ is isolated, then $\mathcal M$ is atomic.
The only way I can see to "use the hint" here is to show $(\mathbb Z,+)$ is atomic over $1$ (which is clear since every element is definable from $1$) and then by the last part of the previous exercise, if $\operatorname{tp}(1)$ were isolated, $(\mathbb Z, +)$ would be atomic, hence prime, and there is a mention earlier on page 13, without proof, that $(\mathbb Z, +)$ is not prime.
When I was writing up solutions a while ago, I noted if indeed this was the intended solution, it was rather artificial, since thinking about it and looking at the reference given on page 13 (A 'natural' theory without a prime model, by Baldwin, Blass, Glass, and Kueker.), it seemed any reasonable way to show $(\mathbb Z,+)$ is not prime would basically amount to showing $\operatorname{tp}(1)$ is not isolated.
Then I saw this answer today which reminded me about this problem and reinforced my previous take on it.
My question is just if I am making an oversight here... is there something in the hint that's actually useful for seeing the big picture?
