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I am thinking of the following problem $^*$:

Given an example in which the subgroups generated by two pure subgroups ia not pure. (Hint: Look within a free abelian group of rank $2$.).

So, as Rotman hinted, I consider $\mathbb Z\times\mathbb Z$, a free abelian group of rank 2. What I want to be guided about it is "What a pure of $\mathbb Z\times\mathbb Z$ looks like, until I am able to work on this problem? Thanks.

$*$ An Introduction to the theory of groups.

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    The subgroup $\langle (1,2),(3,2)\rangle$ contains $(2,0)$. It does not contain $(1,0)$ since if it did there would exist $n,m\in\mathbb{N}$ such that $1=n+3m$ and $0=2n+2m$. But this would mean $1=2m$. You don't need any tools to do this question.2012-11-01

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This answer has been arranged with the help of @Derek and @Conrad.

Definition: $S is pure in $G$ if and only if $x=ng$ for $n\in\mathbb N$ and $y\in G$ then we can always find $s\in S$ such that $x=ns$.

So a subgroup $S$ of $G$ is pure if all elements of $S$ which are $n$-divisible in $G$ are $n$-divisible in $S$.

Theorem: A subgroup $N$ of a torsion-free group $G$ is pure if and only if $G/N$ is torsion-free.


We consider abelian torssion-free group $G=\mathbb Z\times\mathbb Z$, subgroups $H=\langle (3,2)\rangle$ and $K=\langle (1,2)\rangle$. $G/H$ and $G/K$ both are infinite and isomorphic to $\mathbb Z$, so since $\mathbb Z$ is torsion-free then these subgroups are pure in $G$.

Now we consider $S=\langle (1,2),(3,2)\rangle=\{(k+3k’,2k+2k’)|\exists k,k’\in\mathbb Z \}\leq \mathbb Z\times\mathbb Z$. $S$ is not pure in $G$. In fact $(2,0), (1,0)\in G\;\; \text{and}\;\;(2,0)=2(1,0)$ but these two ordered pairs are not not connected as $(2,0)=n(1,0)$ in $S$ because $(2,0)\in S$ and $(1,0)\notin S$. Hence, $H\leq_{pure} G, \;K\leq_{pure} G$ but $\langle H,K\rangle$ is not pure in $G$.

Thanks for step by step hitting me.

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    @amWhy: I wonder why this prime question has not got any attention, Amy.2013-11-01