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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.2012-10-22
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    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?2012-10-22
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    @BabakSorouh An element $a\in G$ a group is $n$-divisible if $a=b^n$ for some $b\in G$.2012-10-22
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    @Conrad: $S 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 $x=ns$.2012-10-22
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    @BabakSorouh So perhaps you could first find some examples of elements that generate pure-subgroups.2012-10-22
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    @BabakSorouh So a subgroup $S$ of $G$ is pure if all elements of $S$ which are $n$-divisible in $G$ are $n$-divisible in $S$.2012-10-22
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    @Conrad: As Derek pointed, I can't choose subgroups as $m\mathbb Z\times n \mathbb Z$ to be pure in $Z\times Z$. Searching in Rose's book, I found another subgroup of $Z\times Z$ as $\{(k,k)|k\in Z\}$, so all of these kinds of subgroups distract my mind. I am thinking.2012-10-22
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6208/discussion-between-conrad-and-babak-sorouh)2012-10-22
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    @DerekHolt: Conrad and I had a discussion last night to have a plan for this problem. I have just a question: Does your claim at first comment remain valid if G be a mixed group? Thanks so much for lighting me.2012-10-23
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    @Babak: No. For finitely generated abelian groups, a subgroup is pure if and only if it is a direct summand. This is not true for general abelian groups. The torsion subgroup of an abelian group is always pure, but is not always a direct summand. (I believe it was that property that motivated the definition of pure subgroups by Prüfer.)2012-10-23
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    @Conrad: As we chatted before, we consider abelian t.f. 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 t.f. then these subgroups are pure in $G$ (Derek’s comment). Now we consider $\langle (1,2),(3,2)\rangle$ which is isomorphic to $S=\langle (4,4)\rangle$.2012-10-23
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    @Conrad: We go for checking that if $G/S$ is finite or if it is not t.f. I say this second claim because I want to use the fact Derek noted first again. $G/S$ is not finite since $(0,1)+S\in G/S$ is not of finite order. Clearly, $G/S$ is a mixed group and then it is not a torsion-free one, so do you think, according to Derek's last comment, we can conclude $S$ is not pure? Thanks again for your patient.2012-10-23
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    @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.2012-10-24
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    @DerekHolt: Yes. I had a bad mistake. For a while I forgot $\langle (1,2),(3,2)\rangle$ is abelian and not cyclic (Sorry). If fact, $$\langle (1,2),(3,2)\rangle=\{(k+3k’,2k+2k’)|k,k’\in\mathbb Z \}=\langle (1,2)(3,2)\rangle$$ Yet, I think of finding a way proving the qoutient group is finite.2012-10-24
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    Yes, or you could just prove that it is not pure. Tho' it is probably a good exercise to "guess" what the quotient is isomorphic to and then exhibit an isomorphism.2012-10-24
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    @Conrad: I assume $A=\langle (1,2),(3,2)\rangle$ be pure, so if $(x,y)=n(a,b)$ wherein $(a,b)\in \mathbb Z\times\mathbb Z$, $(x,y)\in A$ and arbitrary $n\in\mathbb N$ then the equation can be solved in $A$ by some $(c,d)$. So for $k,k'\in\mathbb Z$, $(c,d)\in A$ can be written as $$(c,d)=(k+3k',2k+2k')$$ and for $s,s'\in\mathbb Z$, we have $$(x,y)=(s+3s',2s+2s)$$ Therefore, $(x,y)=n(c,d)$ lead us to a system of equation: $$s+3s'=nk+3nk'\\ 2s+2s'=2nk+2nk'$$ or $s=nk$, $s'=nk'$. I think this cannot be true for an arbitrary $n$ especially when $k,k',s,s'$ be fixed in $A$. Right?2012-10-24
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    Is $(2,0)\in\langle (1,2),(3,2)\rangle$? Is $(1,0)\in \langle (1,2),(3,2)\rangle$?2012-10-24
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    @Conrad: 1. I wasn't noticed that if my approach above was correct, however, I knew that is so basic for this problem? 2. The answer of your last kind comment is no and I didn't get why you gave me it (sorry). And finally I have found a new problem as follows. I think this is the key which both of you were trying to indicate me that. I find this problem so interesting; however, it needs a proof independently.2012-10-26
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    >Let $F$ be free abelian of rank $n$ and let $H$ be a subgroup of the same rank. Let $\{x_1,x_2,...x_n\}$ be a basis of $F$, and $\{y_1,y_2,...y_n\}$ be a basis for $H$ such that $y_j=\sum m_{ij}x_i$. Then $$[F:H]=|\det[m_{ij}]|$$ So if I use this problem without proof, the quotient group above is isomorphic to a group of order $4$ and so $\langle (1,2),(3,2)\rangle$ is not pure.2012-10-26
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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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    Nice! I hope you wake up refreshed from a nice visit to heaven!2013-04-03
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    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? ;-)2013-04-03
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    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.2013-04-03
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    Very well, I'll leave the driving to you! ;-)2013-04-03
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    @amWhy: I wonder why this prime question has not got any attention, Amy.2013-11-01