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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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A subgroup $N$ of ${\mathbb Z} \times {\mathbb Z}$ is pure if and only if $G/N$ is torsion-free. – Derek Holt Oct 22 '12 at 17:54
Why don't you think about which elements of $\mathbb{Z}\times\mathbb{Z}$ are $n$-divisible. For instance, are $(1,2)$ or $(3,2)$ $2$-divisible? – Conrad Oct 22 '12 at 17:56
@BabakSorouh An element $a\in G$ a group is $n$-divisible if $a=b^n$ for some $b\in G$. – Conrad Oct 22 '12 at 18:27
@Conrad: $S<G$ is pure in $G$ iff $x=ng$ for $n$ in $\mathbb N$ and $y $ in $G$ then we can always find $s$ in $S$ such that $x=ns$. – Babak S. Oct 22 '12 at 18:32
@Babak: Think again: it is not true that $\langle (1,2),(3,2) \rangle \cong \langle (4,4) \rangle$. In fact $G/\langle (1,2), (3,2) \rangle$ is finite. – Derek Holt Oct 24 '12 at 7:58
up vote 1 down vote accepted

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

Definition: $S<G$ 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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Nice! I hope you wake up refreshed from a nice visit to heaven! – amWhy Apr 3 '13 at 0:57
Yes, soon I shall be in's very late for me, but I've gotten "sucked in" Math.SE! I hope you slept well? ;-) – amWhy Apr 3 '13 at 5:50
Yes! I had enough. I was exhausted yesterday. Now, let me derive at Math.S.E highway and you can take a deep sleep near me. However, I couldn't answer many problems as you have done. – Babak S. Apr 3 '13 at 5:54
Very well, I'll leave the driving to you! ;-) – amWhy Apr 3 '13 at 5:59
@amWhy: I wonder why this prime question has not got any attention, Amy. – Babak S. Nov 1 '13 at 18:07

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