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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.

Shaun
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Vladimir
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  • The converse (assuming abelian) is at http://math.stackexchange.com/questions/193273/when-abelian-group-is-divisible – Jack Schmidt Sep 24 '12 at 03:29

2 Answers2

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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 $x \notin N$.

Shaun
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Nicky Hekster
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  • Why "$\overline{x} = \overline{y^m} = \overline {1}$"? – Vladimir Sep 24 '12 at 12:37
  • 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 $\overline{y^m}=\overline{y}^m=\overline{1}$. – Nicky Hekster Sep 24 '12 at 19:04
  • 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. – Nicky Hekster Sep 24 '12 at 19:39
  • 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? – Vladimir Sep 24 '12 at 19:39
  • Can I say $K=$ in the general case you provided? – Vladimir Sep 24 '12 at 19:41
  • No $G/N$ would be the $K$. Look at divisibility in the sense that you can extract any m-power root of any element and for any $m$. The order of the element $y$ equals the order of the subgroup $$. – Nicky Hekster Sep 24 '12 at 19:47
  • I still don't quite understand... I wonder how the Lagrange's theorem can be used. Could you explain it in details? – Vladimir Sep 24 '12 at 19:53
  • Yeah, but since the order of $y$ equals to the order of $G/N$, can't I say $G/N=$? – Vladimir Sep 24 '12 at 19:55
  • let us continue this discussion in chat – Nicky Hekster Sep 24 '12 at 20:01
  • Sorry, Nicky. My iPad can't direct you to that chat... – Vladimir Sep 24 '12 at 20:07
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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.

M Turgeon
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