TITLE: Subgroups of (Z,+) of order n
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: 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.)