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Show that the additive group $\mathbb{Z}$ acts on itself by $z \cdot a = z+a$ for all $z,a \in \mathbb{Z}$.
Solution: Let $a \in \mathbb{Z}$. We have $0 \cdot a = 0 + a = a$. Now let $z_1, z_2 \in \mathbb{Z}$. Then $$z_1 \cdot (z_2 \cdot a) = z_1 \cdot (z_2 + a) = z_1 + (z_2 + a) = (z_1 + z_2) + a = (z_1 + z_2) \cdot a.$$ Thus left addition by group elements is in fact a group action.