Let $n$ be a positive integer and $G$ an abelian group. An element $x\in G$ is said to be divisible by $n$ if there is $y\in G$ such that $x=ny$

By the unique factorization of $\mathbb{Z}$ write $n=p_1^{m_1}p_2^{m_2}\cdots p_k^{m_k}$ where each $p_i$ is a prime number (distinct from one another) and $m_i$ a positive integer.

Proof. If $x$ is divisible by $n$ write $x=ny$ where $y\in G$ Since $p_i$ divides $n$ write $n=p_it_i$ where $t_i$ is a positive integer. Then $x=p_it_i(y)=p_i(t_iy)$ Since $t_iy\in G$ $x$ is divisible by $p_i$

Definition. An abelian group $G$ such that every element is divisible by $n$ is called an $n$ divisible group. Clearly, every group is $1$ divisible.

For example, the subset $D\subseteq \mathbb{Q}$ of all decimal fractions is $10$ divisible. $D$ is also $2$ and $5$ divisible. In general, we have the following:

Proof. This follows from proposition 1 and the fact that if $p|n$ $q|n$ and $\gcd(p,q)=1$ then $pq|n$

Proof. Suppose $G$ is $n$ divisible. By proposition 1, every element $x\in G$ is divisible by $p$ so that $G$ is $p$ divisible. Conversely, suppose $G$ is $p$ divisible for every $p|n$ Write $n=p_1^{m_1}p_2^{m_2}\cdots p_k^{m_k}$ Then if $G$ is $p_i^{m_i}$ divisible for every $i=1,\ldots, k$ Since $p_i^{m_i}$ and $p_j^{m_j}$ are coprime, $G$ is $n$ divisible by induction and proposition 3.

Remark. $G$ is a divisible group iff $G$ is $p$ divisible for every prime $p$