Finding all irreducible $\mathbb{Z}$-modules Module $M$ is irreducible means it doesn't contain any submodules except of $0$ and $M$, also I have proven that $M$ is irreducible if and only if it is cyclic and has any nonzero element as generator.
Now I have to find all irreducible $\mathbb{Z}$-modules. It is clear that $\mathbb{Z}/p\mathbb{Z}$ is irreducible by Lagrange's theorem. Why there are no another irreducible $\mathbb{Z}$-modules?
 A: In any $\Bbb Z$-module $M$ and any $a \in M$, we can look at the submodule generated by $a$, denoted $(a)$.  We can define this to be the smallest submodule containing $a$, or we can more directly define it as
$$
(a) = \{na : n \in \Bbb Z\}
$$
Now, if $a$ is non-zero, then $(a) \neq 0$.  So, if $M$ is irreducible, then $M = (a)$.  However, this means that $M$ is a cyclic (abelian) group.
Now, which cyclic groups have no proper, non-trivial subgroups?
Note that in $M = \Bbb Z/n\Bbb Z$ where $n = pq$, we have the subgroup $(p) = \{0,p,2p,\dots,(q-1)p\} \subsetneq M$.
A: Let $M$ be an irreducible $Z$-module, $m\neq 0, m\in M$. Denote by $Ann(m)=\{x\in Z, x.m=0\}$. 


*

*Suppose that $Ann(m)=0$. We have $2Z m=\{ 2x.m, x\in Z\}$ is a non trivial submodule thus it is equal to $M$, we have $m=x2m$ implies that $(2x-1)m=0$ and $2x=1$. Contradiction.


$Ann(m)=nZ$, suppose that $n=ab$, $a,b>1$, $bZm=\{xbm, x\in Z\}$ is a submodule, it is not trivial since $bm\neq 0$, we have $bZm=M$, we deduce that $xbm=m$ this implies tha $xabm=am=0$, contradiction since $am\neq 0$, so $n$ is a prime.
A: Another, very similar approach could be something like this. Let $M$ be irreducible $\Bbb Z$-module. As Abelian group, it is finite, otherwise there would be an element $a\in M$ with infinite order, and thus $(a)\cong\Bbb Z$ is a submodule of $M$. By irreducibility, $M = (a)$, i.e. $M\cong \Bbb Z$. But $\Bbb Z$ is obviously not irreducible. Now, let $p\mid|M|$ be a prime. By Cauchy's theorem, there is an element $a\in M$ with order $p$, and thus $(a)\neq \{e\}$ is a submodule. By irreducibility, $M=(a)$, and thus $|M| = p$, i.e. $M\cong\Bbb Z/p\Bbb Z$ and it is indeed irreducible, since any non-trivial submodule $N$ would have to satisfy $|N|\mid p$ by Lagrange's theorem, and therefore $|N|\in\{1,p\}$, which is contradiction with assumption that $N$ is non-trivial.
We have avoided case $\Bbb Z/n\Bbb Z$ this way for non-prime $n$, but it is easy to argue why this case doesn't work with same argument as above. If there is proper divisor $d$ of $n$, then there is element $a$ of order $d$ by Cauchy's theorem and $(a)$ is non-trivial submodule of $\Bbb Z/n\Bbb Z$.
