Subgroups of (Z,+) of order n Studying groups and subgroups I find this question:

Are there subgroups of order $\mathbf 6 \mathbf 5$ in the additive group $(\Bbb Z$,$+)$?

I would answer no, because a subgroups of $(\Bbb Z,+)$ is the multiple of a Natural number $n $ and it has the form:
$n\Bbb Z$={$na|n \in \Bbb N, a \in \Bbb Z$} and they have no finite order.
But I'm not sure of this answer.
Could someone explain if I'm wrong and why?
Thanks.
$\mathbf {edit}:$
My question comes because discussing with a course mate he argued that exist Z/65Z in (Z,+), the group module 65 and it  has order 65 so I'm a bit confused. 
 A: Your thinking looks good! Try showing that any (non-trivial) subgroup of $(\mathbb{Z}, +)$ is infinite as follows:
Suppose there were a finite subgroup $G \leq (\mathbb{Z}, +)$. Since $G$ is finite and is a subgroup of the integers, $G$ must have a greatest element; call it $n$. If $n \neq 0$, by closure under addition and inverses we must have $n+n = 2n \in G$ and $-n \in G$, one of which (depending on whether $n < 0$ or $n > 0$) is a contradiction to the assumption that $n$ was the greatest element. If $n = 0$, take the least nonzero element $k \in G$ ($k$ must again exist since $G$ is finite) and note $-k \in G$, again a contradiction. So $G$ can't be finite. 
(This is really the same as the suggestions that no element in $\mathbb{Z}$ has finite order under addition, but I thought it might be nice to see a proof.)  
A: Your reasoning is perfectly correct, the infinite cyclic group $(\mathbb{Z}, +)$ has no elements of finite order, and so can have no finite cyclic subgroups. Can you make it into a proof now?
Your friends comment is, unfortunately, wrong, but this is a common error. It is not always true that factor groups are isomorphic to a subgroup of the group. Note that every subgroup of $\mathbb{Z}$ has finite index (why?), and so every proper quotient group is finite cyclic. Therefore, the factor group is a subgroup of $\mathbb{Z}$ if and only if the associated normal subgroup is the whole group or is the trivial group.
The error is perfectly understandable though! This is because it holds for finite cyclic groups. Can you see why?
