Does the additive group modulo $n$ have proper subgroups if and only if $n$ is not prime? After checking a few examples is seems as if it is true and I would have thought the elements of the group would be the multiples of the divisors of $n$ up until $n$, and how would I attempt to prove this?
 A: You have that $(\mathbb{Z}_n,+)$ is cyclic, generated by $1$ (modulo $n$, course), so $1$ has order $n$ in $(\mathbb{Z}_n,+)$, then you can show that for every $d|n$, $d \cdot 1$ has order $n/d$, and the subgroups generated by $d \cdot 1, d | n$ are the only subgroups of $(\mathbb{Z}_n,+)$.
You can show that $d \cdot 1$ has order $n/d$ in the following way: let $o$ be the order of $d \cdot 1$. You have that $\displaystyle \frac{n}{d} \cdot \left(d  \cdot 1\right) = 0,$ so $o | \displaystyle \frac{n}{d}.$
On the other hand, you also have that $\displaystyle \left(o \cdot d \right)\cdot 1 = d \cdot (o \cdot 1) = 0,$ so $n | (o \cdot d) \implies \left( \displaystyle d \cdot \frac{n}{d} \right) | (o \cdot d) \implies \displaystyle \frac{n}{d} | o,$ so $o = \displaystyle \frac{n}{d}$.
If $n$ is not prime, you have that there is some $d \in \mathbb{N}, d|n$ such that the subgroup generetaed by $d \cdot 1$ is proper.
If $(\mathbb{Z}_n, +)$ has proper subgroups, then assume that $n$ is prime. Then, from Lagrange's theorem, you would have that the order of every subgroup would be a divisor of $n$, but since $n$ is prime, you would have no proper subgroups, which is a contradiction.
