$n$-divisible group

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 $n=p_1^{m_1}{p}_2^{m_2}\mathrm{\cdots }{p}_k^{m_k}$ where each $p_i$ is a prime number (distinct from one another) and $m_i$ a positive integer.

Proposition 1.

If $x$ is divisible by $n$, then $x$ is divisible by $p_{\mathrm{1}}\mathrm{,}{p}_{\mathrm{2}}\mathrm{,}\mathrm{\dots }\mathrm{,}{p}_k$.

Proof.

If $x$ is divisible by $n$, write $x=ny$, where $y\in G$. Since $p_i$ divides $n$, write $n=p_i{t}_i$ where $t_i$ is a positive integer. Then $x=p_i{t}_i(y)=p_i(t_{i}y)$. Since $t_{i}y\in G$, $x$ 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:

Proposition 2.

If $G$ is $n$-divisible, it is also $n^s$-divisible for every non-negative integer $s$.

Proposition 3.

Suppose $p$ and $q$ are coprime, then $G$ is $p$-divisible and $q$-divisible iff it is $p\mathit{}q$-divisible.

Proof.

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

Proposition 4.

$G$ is $n$-divisible iff $G$ is $p$-divisible for every prime $p$ dividing $n$.

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 $n=p_1^{m_1}{p}_2^{m_2}\mathrm{\cdots }{p}_k^{m_k}$. Then if $G$ is $p_i^{m_i}$-divisible for every $i=1,\mathrm{\dots },k$. 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 $p$.