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.


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 :)

  • 1
    $\begingroup$ 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

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 $f$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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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