Is there any counter example for the following statement?


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.



Finite / uniquely divisible abelian groups
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. 


