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This is Exercise 4.3.3 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.

(NB: I have left out the tag for a reason: the tools available here are entirely group theoretic.)

The Details:

Since definitions vary, on page 15, ibid., paraphrased, it states that

A subgroup $N$ of $G$ is normal in $G$ if one of the following equivalent statements is satisfied:

(i) $xN=Nx$ for all $x\in G$.

(ii) $x^{-1}Nx=N$ for all $x\in G$.

(iii) $x^{-1}nx\in N$ for all $x\in G, n\in N$.

On page 94, ibid.,

An element $g$ of an abelian group $G$ is said to be divisible in $G$ by a positive integer $m$ if $g=mg_1$ for some $g_1$ in $G$. [. . .]

An abelian group $G$ is said to be divisible of each element is divisible by every positive integer.

On page 106, ibid.,

A subgroup $H$ of an abelian group $G$ is called pure if

$$nG\cap H=nH$$

for all integers $n\ge 0$; in words, $H$ is pure if every element of $H$ that is divisible by $n$ in $G$ is divisible by $n$ in $H$.

The Question:

The pure subgroups of a divisible abelian group are just the direct summands.

Thoughts:

This exercise feels as if it would follow from the relevant definitions rather easily, but I've got nowhere so far. At the risk of repeating myself, here's how I start . . .

Let $H\le G$ be a pure subgroup of a divisible abelian group $G$. Then, for every integer $n\ge 0$, we have

$$nG\cap H=nH$$

and for each $g\in G$ and each integer $m\ge 0$, there exists a $g_1\in G$ such that $g=mg_1$.


Observation: Using notation as above, we have for each such $m\in \Bbb Z$,

$$mG\cap H=mH$$


What do I do next?


Since $G$ is abelian, each of its subgroups is normal, so $H\unlhd G$.

I need to show that there exists a $K\unlhd G$ such that $H\cap K=\{e\}$ and $G=HK=\{ hk\in G\mid h\in H, k\in K\}$; that is, $G=H\oplus K$.


The question is trivial if $H$ is trivial or $H=G$.


I think I could answer this question myself if I had enough time; as such, a good hint is preferred over a full answer.

Please help :)

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    $\begingroup$ If you want to avoid using modules, one way forward is to prove that any pure subgroup $H$ is a summand of $n^{-1}H$ first. $\endgroup$ Aug 11 at 22:11
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Let $G$ be a divisible abelian group and $H \subset G$ be pure. Then for any $n \geq 1$, $nH=nG \cap H=G \cap H=H$, so $H$ is divisible.

Now, you can use Zorn’s lemma (or the well-known fact that $H$, being divisible, is an injective $\mathbb{Z}$-module) to show that the identity $H \rightarrow H$ extends to some map $f:G \rightarrow H$ and then $G=H \oplus K$, where $K$ is the kernel of $f$.

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  • $\begingroup$ Thank you. I'm not sure how $nG \cap H=G \cap H$ at first glance. Let me think about it . . . $\endgroup$
    – Shaun
    Aug 11 at 22:08
  • $\begingroup$ @Shaun Because $G$ is divisible. $\endgroup$ Aug 11 at 22:09
  • $\begingroup$ I figured as much, @DavidA.Craven. Thank you. How do I prove it though? $\endgroup$
    – Shaun
    Aug 11 at 22:11
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    $\begingroup$ @Shaun $g=ng'$, so $g\in nG$. Thus $G=nG$. $\endgroup$ Aug 11 at 22:12
  • $\begingroup$ Ah, of course! Thank you again, @DavidA.Craven. $\endgroup$
    – Shaun
    Aug 11 at 22:13

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