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.


If $G$ is non-abelian, the statement still holds true. If $H$ is a subgroup of finite index $n$, then $G/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 $N=core_G(H) \subseteq H$ 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 $x \notin N$.

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  • $\begingroup$ Why "$\overline{x} = \overline{y^m} = \overline {1}$"? $\endgroup$ – Vladimir Sep 24 '12 at 12:37
  • $\begingroup$ In general, if $K$ is a finite group of order $m$ and $y \in K$, then $y^m=1$. This follows from the Theorem of Lagrange. In your case $\overline{y^m}=\overline{y}^m=\overline{1}$. $\endgroup$ – Nicky Hekster Sep 24 '12 at 19:04
  • $\begingroup$ You are welcome! By the way the space of orientation-preserving isometries of $\mathbb R^2$ is divisible. This is because each such isometry is either a translation or a rotation around a point, and in either case the ability to "divide by n" is obvious. This is an example of a non-abelian divisible group. $\endgroup$ – Nicky Hekster Sep 24 '12 at 19:39
  • $\begingroup$ I'm sorry I still feel confused about this... If so, it seems that every finite group is cyclic... $m$ is the index of $N$, why can we have $my=x$ with respect to the same m? $\endgroup$ – Vladimir Sep 24 '12 at 19:39
  • $\begingroup$ Can I say $K=<y>$ in the general case you provided? $\endgroup$ – Vladimir Sep 24 '12 at 19:41

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.

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