How do I prove that any abelian group $G$ contains divisible subgroup $H$, such that $G / H$ has no divisible subgroups other than $\{0\}$?


1) Using Zorn's lemma was suggested to me in the comments, so I started figuring out how to construct a partially ordered set that would allow me to extract maximal divisible subgroup.

2) A hint also provided a simple useful proposition for me to prove.

Edit: Based on the hints and my attempts in the comments, I have figured out the solution, which I have posted as an answer.

  • 1
    $\begingroup$ Have you heard of Zorn's lemma? $\endgroup$
    – Steve D
    Sep 21 '18 at 4:44
  • $\begingroup$ It was mentioned several times by our lecturer during some proofs. He never stated it though. I looked up its definition on Wikipedia, but have never used it. How does it apply to my question? $\endgroup$
    – Drinkwater
    Sep 21 '18 at 4:53
  • 2
    $\begingroup$ It helps ensure you can get a maximal divisble subgroup. Here's another hint: suppose you have an abelian group $K$ and a subgroup $L$, such that both $L$ and $K/L$ are divisible. Can you show $K$ is divisible? $\endgroup$
    – Steve D
    Sep 21 '18 at 4:57
  • 1
    $\begingroup$ Let's suppose you use Zorn's lemma to find a maximal divisible subgroup $H$. Then suppose $G/H$ has a divisible subgroup $K/H$. Then by what you've shown $K$ is divisble, contradicting the maximality of $H$. $\endgroup$
    – Steve D
    Sep 21 '18 at 7:54
  • 1
    $\begingroup$ In many algebraic cases, you take the union of a chain. Show this is a subgroup, and divisible. $\endgroup$
    – Steve D
    Sep 21 '18 at 8:13

Zorn's lemma is not needed.

I'll use additive notation for the abelian group $G$.

The sum of any family of divisible subgroups is divisible.

Indeed, let $(H_\alpha)$ be a family of divisible subgroups and $H=\sum_\alpha H_\alpha$. Let $x\in H$; then $x=\sum_{i=1}^n x_{\alpha_i}$, for $x_{\alpha_i}\in H_{\alpha_i}$. If $m>0$ is an integer, then for each $i$, there is $y_i\in H_{\alpha_i}$ with $$ my_i=x_{\alpha_i} $$ Set $y=\sum_{i=1}^ny_i$; then $y\in H$ and $my=x$.

Thus you can consider the sum $D$ of all divisible subgroups of $G$, which is a divisible subgroup and the larges such.

Let's prove that $G/D$ has no divisible subgroup except $\{0\}=D/D$.

To this end we show that if $A$ is a divisible subgroup of $B$ and $B/A$ is divisible, then also $B$ is divisible. Let $x\in B$ and $m>0$. By assumption there is $y\in B$ such that $$ x+A=my+A $$ (since $B/A$ is divisible). This means that $x-my\in A$, which is divisible as well, so there is $z\in A$ with $mz=x-my$. Hence $x=m(y+z)$ as desired.


I thank user Steve D for helping me figure this one out.

1 General structure of the proof

1) Consider a set $P$ of all divisible subgroups of $G$, endowed with the partial ordering $\le$: $A \le B \iff A \text{ is a subgroup of } B$.

2) Apply Zorn's lemma to extract $D$, the maximal element of $P$.

3) Show that existence of any non-trivial divisible subgroup of $G / D$ implies that $D$ is not maximal, which yields the contradiction that concludes the proof.

2 Some helpful details for expanding the proof


Zorn's lemma is applicable in step 2 because $P$ at the very least contains $\{0\}$ and each chain in $P$ has the upper bound: a union of all chain members. It is indeed a divisible group, which is easily proven.


Lemma Consider an abelian group $K$ and its subgroup $L$, such that both $L$ and $K/L$ are divisible.
Then $K$ is divisible as well.

$$(\forall l \in L) \ (\forall n \in \mathbb{N}) \ (\exists l_n \in L) \ (n l_n = l)$$ $$(\forall (k + L) \in K / L) \ (\forall n \in \mathbb{N}) \ (\exists (k'_n + L) \in K / L) \ (n (k'_n + L) = k + L)$$

Consider arbitrary fixed $k \in K$ and $n \in \mathbb{N}$. From the above statement: $$(\exists (k'_n + L) \in K / L) \ (n (k'_n + L) = k + L)$$

It follows that $l := (n k'_n - k) \in L$.

Also, from the above: $$(\exists l_n \in L) \ (n l_n = l)$$

Therefore $$ n k'_n - k = n l_n \Rightarrow k = n(k'_n - l_n)$$ Hence $k_n := (k'_n - l_n) \in K$ has the desired property $n k_n = k$, thus $K$ is divisible. $\square$


Using above lemma, if we now suppose (step 3) that $G / D$ contains a non-trivial divisible subgroup $K / D$, we automatically conclude $K$ is divisible. Combined with $D$ being a subgroup of $K$ (and thus $D \le K$, non-triviality ensures $K \ne D$) it follows that $D$ is not maximal. Contradiction.

2.4 Zorn's lemma

For the reference, Zorn's lemma states that every non-empty partially ordered set $(P, \le)$, with the property that every chain has an upper bound in $P$, contains a maximal element.

Maximal element of $(P, \le)$ is defined as the element which does not precede any other element of $P$.

Chain in $(P, \le)$ is defined as a subset of $P$ where each pair of elements is $\le$-comparable.

  • 2
    $\begingroup$ Minor comment on Step 2.1: the union works for all but the empty chain; you need to deal with the empty chain separately, which can be done by exhibiting an element of $P$ (which is an upper bound for the empty chain). $\endgroup$ Sep 21 '18 at 18:04
  • $\begingroup$ Interesting insight. Easy to look over, but ultimately trivial, as long as there's at least one element in $P$. $\endgroup$
    – Drinkwater
    Sep 22 '18 at 9:10
  • 1
    $\begingroup$ Yes, but it’s an important point of the formulation of Zorn’s Lemma you have given. Other (equivalent) formulations require $P$ to be nonempty, and then restrict to “nonempty chains”. That’s why you often see applications of Zorn’s Lemma begin by saying “the set is nonempty, since blah is in there”. That’s dealing with the empty chain. $\endgroup$ Sep 22 '18 at 17:01
  • $\begingroup$ I believe I have now accounted for those formalities. $\endgroup$
    – Drinkwater
    Sep 23 '18 at 8:31

$\newcommand{\inv}{^{-1}}$While there are already two good answers, in this case there is a concrete description of the maximal divisible subgroup, which doesn't seem to be included in either answer.

Def: Let $A$ be abelian. Call an element $a\in A$ totally divisible if you can find a family of elements $(b_n)_{n\in\Bbb{N}^+}\subseteq A$ such that

  1. $b_1=a$
  2. for all $n$, $m$, $mb_{nm}=b_n$.

The maximal divisible subgroup $D$ is $$D = \{a\in A: \text{$a$ is totally divisible}\}.$$


$D$ is a subgroup. First we check that this is a subgroup of $A$.

  1. $0\in D$. The family $b_n=0$ for all $n$ shows $0$ is totally divisible.
  2. If $a\in D$, and $(b_n)$ is a family showing $a$ is totally divisible, $(-b_n)$ will be a family showing $-a$ is totally divisible, so $-a\in D$.
  3. Similarly, if $a,c\in A$, and $(b_n)$, $(d_n)$ are families showing $a$ and $c$ are totally divisible, then $(b_n+d_n)$ is a family showing $a+c$ is totally divisible. Thus $a+c\in D$.

$D$ is divisible. Now we check that $D$ is divisible. Suppose $a\in A$, $(b_n)$ is a family showing $a$ is totally divisible. Then for any $n\in\Bbb{N}^+$, $nb_n=b_1=a$, and the family $(b_{nm})_{m\in\Bbb{N}^+}$ shows $b_n$ is totally divisible. Thus $b_n\in D$. Hence $D$ is divisible.

$D$ is maximal. Now we check that $D$ is the maximal divisible subgroup. Let $H$ be a divisible subgroup of $A$. Let $h\in H$. Since $H$ is divisible, inductively define $b_{n!}$ by choosing $b_{n!}$ to be any solution to $nb_{n!}=b_{(n-1)!}$, with $b_{1!}=h$. Then for any other integer $m$, there is some least integer $n$ so that $m\mid n!$. Define $$b_m = \frac{n!}{m}b_{n!}.$$

Now we check that the $(b_m)$ satisfy the properties for a family showing that $h$ is totally divisible. Certainly $b_1=h$, so we just need to check that for all integers $n$ and $m$, $mb_{nm}=b_n$.

Let $j$ be the least integer large enough that $n\mid j!$. Let $k$ be the least integer large enough that $nm \mid (j+k)!$. Then $$m b_{nm} = m \frac{(j+k)!}{nm} b_{(j+k)!} = \frac{j!}{n} \frac{(j+k)!}{j!}b_{(j+k)!} = \frac{j!}{n} b_{j!} = b_n.$$ The equality $\frac{(j+k)!}{j!}b_{(j+k)!} = b_{j!}$ comes from the inductive definition of the $b_{j!}$.

This shows that every element of a divisible subgroup is totally divisible. Hence $H\subseteq D$. Thus $D$ is the maximal divisible subgroup. $\blacksquare$


Because the original question asks us to show that there is a divisible subgroup $H$, such that $A/H$ has no nonzero divisible subgroups, we'll show that the maximal divisible subgroup $D$ has this property.

To this end, we'll prove the following lemma: If $A$ is an abelian group, $K$ is a divisible subgroup, $\phi :A \to A/K$ is the quotient map, then if $H\subseteq A/K$ is divisible, so is $\phi\inv(H)$.


Let $b\in \phi\inv(H)$, let $n\in\Bbb{N}^+$. We want to construct $c\in \phi\inv(H)$ with $nc=b$. $\phi(b)\in H$, which is divisible, so choose $b' \in H$ with $nb'=\phi(b)$. Since $\phi$ is surjective, there is $c'\in A$ with $\phi(c')=b'$.

Now $\phi(b-nc')=0$, so $b-nc' \in K$. Choose $d \in K$ with $nd = b-nc'$. Then $b=n(d+c')$. $d+c'\in \phi\inv(H)$. Thus $\phi\inv(H)$ is divisible. $\blacksquare$

This implies that $A/D$ has no nonzero divisible subgroups, since if had a nonzero divisible subgroup $H$, then $\phi\inv(H)$ would be a divisible subgroup of $A$ strictly larger than $D$, which is impossible by maximality of $D$.


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