Simple and semi-simple over $\mathbb{Z}$ What is a necessary and sufficient condition on an integer $n$ for $\mathbb{Z}/n \mathbb{Z}$ to be simple as a module over $\mathbb{Z}$? Semisimple? In the case of simple I think that because it is a cycle group it will be simple iff $n$ is prime. But for semisimple I have no clue. 
Thanks
 A: First, let us remind ourselves that $\mathbb Z$-modules are exactly the same thing as abelian groups. So we're dealing with abelian groups here.
To avoid lots of cumbersome notation, we write
$$
\mathbb Z/n\mathbb Z =: C_n
$$
in the following.
Regarding simplicity, you're right. $C_n$ is simple iff $n$ is prime. A somewhat more detailed explanation could be as follows.
According to the definition, $C_n$ is simple iff it has no non-trivial proper subgroups. Now, the correspondence between subgroups of $C_n$ and divisors of $n$ is well known. From this, we see that indeed $C_n$ is simple iff $n$ is prime.
On to semisimplicity. Let's assume that $C_n$ is semisimple. Then we get from our characterization of simplicity that
$$
C_n \simeq C_{r_1}\oplus\cdots\oplus C_{r_k} =: D
$$
with $k$ appropriate primes $r_j.$ Consider the (unique) prime factorization of $n,$
$$
N = p_1^{e_1}\cdots p_m^{e_m}
$$
with pairwise different primes $p_j$ and exponents $e_j\in\mathbb N.$ Comparing sizes of $C$ and $D$ (and using the uniqueness of prime factorization in $\mathbb Z$), we find that we must have
$$
D = (C_{p_1})^{e_1}\oplus\cdots\oplus(C_{p_m})^{e_m}.
$$
Otherwise, $C$ and $D$ won't have the same size. From this representation of $D$, we see that the order of any $x\in D$ is at most
$$
p_1\cdots p_m =: h
$$
(indeed, there are elements of order exactly $h$).
Now, assume that $n$ is not square-free, i.e.
$$
\exists i\in\{1,\ldots,m\}:e_i > 1.
$$
Then we have
$$
h < n.
$$
Now, since $C_n,$ being cyclic, contains an element of order $n,$ but $D$ does not, we see that $C_n \simeq D$ can't hold, and $C_n$ isn't semisimple. We have reached a contradiction. So we have shown
$$
n\ not\ square-free \Rightarrow C_n\ not\ semisimple.
$$

Next, assume that $n$ is square-free, i.e.
$$
n = p_1\cdots p_m
$$
(all exponents are $1$). Then we get from the fundamental theorem of finite abelian groups, or alternatively from the Chinese remainder theorem,
$$
C_n \simeq C_{p_1}\oplus\cdots\oplus C_{p_m}.
$$
Here, the r.h.s. is semisimple according to our characterization of simplicity. So we have shown
$$
n\ square-free \Rightarrow C_n\ semisimple.
$$

All in all, we have shown
$$
n\ square-free \Leftrightarrow C_n\ semisimple,
$$
which is a nice characterization of semisimplicity.
