Exercise 1.7.2 ($\mathbb{Z}$ acts on itself by translation)

beginproof The action is defined by $z\cdot a=z+a$ for all $z,a\in \mathbb{Z}$.

The law of the group $\mathbb{Z}$ is an additive law, with $0$ as identity element.

(i)

For all $z,z^{\prime },a\in \mathbb{Z}$, $$(z+z^{\prime })\cdot a=(z+z^{\prime })+a=z+(z^{\prime }+a)=z\cdot (z^{\prime }\cdot a).$$

(ii)

For all $z\in \mathbb{Z}$, $$0\cdot z=0+z=z.$$

Therefore the group $(\mathbb{Z},+)$ acts on itself by $z\cdot a=z+a$.

(This is a particular case of left translation described in Example (5).)