Take the 2-minute tour ×

Is there any counter example for the following statement?


Let $0 \to F \to A \to Q \to 0$ be a short exact sequence of abelian groups. Assume that $F$ is a finite group, and $Q$ is a uniquely divisible abelian group. Then, this exact sequence splits.

Please give me any advice.

share|improve this question

3 Answers 3

up vote 6 down vote accepted

By "uniquely divisible", I guess you mean torsion-free and divisible, i.e., $Q$ is a $\mathbb{Q}$-vector space. If so, the answer to your question is that every such exact sequence splits.

It is enough to show $\mathrm{Ext}(Q, \mathbb{Z}/n) = 0$. We have an exact sequence

$$\hom(Q, \mathbb{Z}/n) \to \mathrm{Ext}(Q, \mathbb{Z}) \stackrel{ \cdot n}{\to} \mathrm{Ext}(Q, \mathbb{Z}) \to \mathrm{Ext}(Q, \mathbb{Z}/n) \to \mathrm{Ext}^2(Q, \mathbb{Z})$$

where the first and last terms are zero. From Hilton-Stammbach, A Course in Homological Algebra exercise 6.1 (p. 109), we have that $\mathrm{Ext}(Q, \mathbb{Z})$ is divisible if $Q$ is torsionfree, and $\mathrm{Ext}(Q, \mathbb{Z})$ is torsionfree if $Q$ is divisible. Hence in our case $\mathrm{Ext}(Q, \mathbb{Z})$ is torsionfree and divisible, hence a $\mathbb{Q}$-vector space, and so multiplication by $n$ is an isomorphism and the claim follows.

share|improve this answer
Of course, as Simone points out, this really only uses that $Q$ is torsionfree (making $\mathrm{Ext}(Q, \mathbb{Z})$ divisible, so that the multiplication by $n$ map above is surjective, etc.). –  Todd Trimble Dec 15 '12 at 18:35

I couldn't help thinking that there must be a lower-power proof of this. Here's what I came up with; I'd be interested if there are shorter proofs that are as elementary.

First, as both of the previous answers pointed out, unique divisibility makes $Q$ torsion-free; if there were non-zero $x\in Q$ and $n\in\mathbb N$ with $nx=0$ then division of 0 by $n$ wouldn't be unique.

Consider elements $a\in A$ that are divisible in $A$, meaning that $nx=a$ has solutions $x\in A$ for all non-zero integers $n$. I claim that each coset $C$ of $F$ contains a divisible element. Proof by contradiction: Suppose that each $c\in C$ failed to be divisible by some non-zero integer $n_c$. Let $m=\prod_{c\in C}n_c$, and use the assumption that the quotient $Q$ is divisible to get some $a\in A$ with $ma\in C$. But then this $ma$ is an element $c\in C$ divisible (in $A$) by $m$ and thus by $n_c$, a contradiction.

Next, I claim that each coset $C$ of $F$ contains only one divisible element. To see this, suppose there were two, say $c$ and $c'$. Then $c-c'$ is an element of $F$ that is divisible in $A$. In particular, $c-c'=|F|x$ for some $x\in A$. (Here $|F|$ denotes the order of $F$.) Projecting to $Q$ and using the fact that $c$ and $c'$ were in the same coset $C$, we get $0=|F|\bar x$ in $Q$, where $\bar x$ is the projection of $x$ into $Q$. But $Q$ is torsion-free, so $\bar x=0$, which means $x\in F$. But then $|F|x=0$, so $c=c'$.

So every coset of $F$ in $A$ contains a unique divisible element. The sum and the difference of two divisible elements are obviously divisible. So the divisible elements of $A$ constitute a subgroup of $A$ complementary to $F$, and we have the desired splitting.

share|improve this answer
A nice argument from scratch, Andreas. –  Todd Trimble Dec 15 '12 at 23:48

well... as the Prüfer groups are not uniquely divisible, by the structure of divisible abelian groups you get that $Q$ is a rational vector space. In particular, $Q$ is flat (i.e., torsion-free as $\mathbb Z$ is a PID). This means that $F$ is pure in $A$. Now, it is well-known that finite pure subgroups split.

share|improve this answer

Your Answer


