Let $M$ be a module over $\mathbb{Z}$. Prove that:
a.$M$ is a cyclic module if and only if it's a cyclic group.
b.$M$ is simple if and only if it's cyclic of prime order.
I tried solving it in the following way:
About a: If $M=\langle a\rangle$ where $a\in\mathbb{Z}$, I can verify it's a module very easily. On the other direction, I take $M=\{nx\mid n\in \mathbb{Z}\}$. Why this group is cyclic?
About b: Suppose the order is not prime, we can find subgroup of prime order which is a module. On the other direction, take a cyclic group of prime order. Why is it a simple module?
a) You just have to realize that a unital $\Bbb Z$ module is the same thing as an Abelian group. Of course, any module is an Abelian group (as part of the definition) and the $\Bbb Z$ module action matches the Abelian group operation. On the other hand, given an Abelian group, the action combining $n$ and $x$ into $nx$ in the Abelian group is a $\Bbb Z$ module structure.
As a special case, $M=x\Bbb Z$ for some element x of M iff M is a cyclic Abelian group.
b) a simple group is of course cyclic, and basic group theory says there is a subgroup for each divisor of $|G|$. There will be trivial subgroups iff the order is prime.