I need to find all free list, generating list and bases of $\mathbb Z/6\mathbb Z$ as $\mathbb Z/6\mathbb Z-$module and as $\mathbb Z-$module.
As $\mathbb Z/6\mathbb Z-$module : $\{1\}$ and $\{5\}$ are bases. For generating, all set that contain $1$ or $5$ are generating. For free list, only basis a are free list.
As $\mathbb Z$ module, for me it's the same. But I guess there is a problem since $\mathbb Z\to \mathbb Z/6\mathbb Z$ is surjective but no injective, so it's not free... So I don't really understand the thing here. Because $1$ looks to be a basis.
The module $\mathbb{Z}/6\mathbb{Z}$ is not free: indeed, there is no linearly independent subset, because $6x=0$ for every $x\in\mathbb{Z}/6\mathbb{Z}$.
You can also argue that there exists $\mathbb{Z}\to\mathbb{Z}/6\mathbb{Z}$ which is surjective but cannot split, because $\mathbb{Z}$ is indecomposable.
The generating sets of $\mathbb{Z}/6\mathbb{Z}$ are those that contain either $1$ or $5=-1$ (the only elements having order $6$). The generating sets with minimal cardinality are $\{1\}$ and $\{5\}$. On the other hand, $\{2,3\}$ is a generating set which is minimal in the sense that no proper subset thereof is a generating set.