0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ A subgroup $N$ of ${\mathbb Z} \times {\mathbb Z}$ is pure if and only if $G/N$ is torsion-free. $\endgroup$ – Derek Holt Oct 22 '12 at 17:54
  • $\begingroup$ 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? $\endgroup$ – Conrad Oct 22 '12 at 17:56
  • $\begingroup$ @BabakSorouh An element $a\in G$ a group is $n$-divisible if $a=b^n$ for some $b\in G$. $\endgroup$ – Conrad Oct 22 '12 at 18:27
  • $\begingroup$ @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 $x=ns$. $\endgroup$ – mrs Oct 22 '12 at 18:32
  • 1
    $\begingroup$ @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. $\endgroup$ – Derek Holt Oct 24 '12 at 7:58
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Nice! I hope you wake up refreshed from a nice visit to heaven! $\endgroup$ – amWhy Apr 3 '13 at 0:57
  • $\begingroup$ Yes, soon I shall be in slumber...it's very late for me, but I've gotten "sucked in" here...at Math.SE! I hope you slept well? ;-) $\endgroup$ – amWhy Apr 3 '13 at 5:50
  • $\begingroup$ 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. $\endgroup$ – mrs Apr 3 '13 at 5:54
  • 1
    $\begingroup$ Very well, I'll leave the driving to you! ;-) $\endgroup$ – amWhy Apr 3 '13 at 5:59
  • $\begingroup$ @amWhy: I wonder why this prime question has not got any attention, Amy. $\endgroup$ – mrs Nov 1 '13 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.