Subgroups of finite index in divisible group Given a divisible group $G$, I wonder why $G$ has no nontrivial subgroup of finite index.
Suppose $H$ is a subgroup (of $G$) of finite index. Then there exists a normal subgroup $K$ of $G$ which is contained in $H$ and also has finite index. Given an element $g$ of $G$, I need to show that $g$ is in $K$. But I don't know how to continue... Could you explain it for me?
Thank you. 
 A: If $G$ is non-abelian, the statement still holds true. If $H$ is a subgroup of finite index $n$, then $G/{\rm core}_G(H) $ can be embedded in $S_n$ (look at the left multiplication action of $G$ on the $n$ left cosets of $H$), and hence $G$ has a normal subgroup $N={\rm core}_G(H) \subseteq H$ of finite index, say $m$. Now let $x \in G \backslash N$. Since $G$ is divisible, there exists an $y \in G$ with $y^m=x$. Hence in $G/N$ this implies $\overline{x} = \overline{y^m} = \overline {1}$. This contradicts the fact that $x \notin N$.
A: I will assume that you are dealing with abelian groups (which is usally the case when dealing with divisible groups).
First, prove that a quotient of a divisible group is divisible.
Hence, if you have a subgroup of finite index, the quotient is a finite divisible group, which is necessarilly trivial. To see this, note that if $n=|G|$, then the map $g\mapsto g^n$ is not surjective. Therefore, we can conclude that a divisible group has no non-trivial subgroup of finite index.
Now, if you are dealing with non-commutative divisible group, this argument clearly doesn't work in the case of non-normal subgroups. But I'm useful the notion of non-commutative divisible groups really is.
