TITLE: Is $G/pG$ is a $p$-group?
QUESTION [6 upvotes]: Jack is trying to prove:

Let $G$ be an abelian group, and $n\in\Bbb Z$. Denote $nG = \{ng \mid g\in G\}$.
(1) Show that $nG$ is a subgroup in $G$.
(2) Show that if $G$ is a finitely generated abelian group, and $p$ is prime,
   then $G/pG$ is a $p$-group (a group whose order is a power of $p$).

I think $G/pG$ is a $p$-group because it is a direct sum of cyclic groups of order $p$.
But I cannot give a detailed proof.

REPLY [8 votes]: Following my comment:
$$∀g∈G,pg∈pG⟹p(g+pG)=pG⟹ $$
 the element $\,p(g+pG)\,$  is zero in the quotient $\,G/pG\,$  and from here that all the elements in this quotient have order a power of $\,p\,$ ,  which is precisely the definition of $\,p$-group, no matter if it is finitely generated or not.