Largest divisible subgroup of an abelian group

Question

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

Attempts:

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.

Answer 1

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

2.1

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.

2.2

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.

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

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. $\square$

2.3

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.

Answer 2

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.

Answer 3

$\newcommand{\inv}{^{-1}}$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 $$D = \{a\in A: \text{$a$ is totally divisible}\}.$$

Proof.

$D$ 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. $\blacksquare$

Aside

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 $\phi\inv(H)$.

Proof.

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 $D$.