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# $n$-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 $n=p_{1}^{{m_{1}}}p_{2}^{{m_{2}}}\cdots p_{k}^{{m_{k}}}$ 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 $p_{1},p_{2},\ldots,p_{k}$.

###### 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 $pq$-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 $n=p_{1}^{{m_{1}}}p_{2}^{{m_{2}}}\cdots p_{k}^{{m_{k}}}$. 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. ∎

## Mathematics Subject Classification

20K99*no label found*

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