Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I assume in this question that if $G$ is a group extension of $N$ by $K$ then $N$ is the normal subgroup in $G$.

Let $G$ be a group extension of $N$ by $K$ where $N$ is the direct limit of a system of groups $N_i$. Can this be considered as the direct limit of group extensions of the $N_i$ ?

What happens for example with $BS(1,2) \simeq \mathbb{Z}[\frac{1}{2}] \rtimes \mathbb{Z}$ ?

share|cite|improve this question
up vote 2 down vote accepted

Your example is likely to make one problem very clear: the $N_i$ need to be normalized by $K$ in order for the smaller extensions to make sense.

Consider the similar case of $\mathbb{Q}^2 \rtimes \mathbb{Q}$ where $\mathbb{Q}^2$ is the direct limit of some $\tfrac1n\mathbb{Z}^2$. The $N_i$ admit only a trivial action of $\mathbb{Q}$ (since $M_2(\mathbb{Z})$ contains no non-identity divisible abelian subgroups), so the direct limit of the extensions will be abelian. However, the original extension can be a Heisenberg group, so not abelian. The direct limit of abelian groups is abelian, so the direct limit of the extensions cannot be an extension of the direct limit.

share|cite|improve this answer
Just to be clear : $\mathbb{Q}^2=\mathbb{Q} \times \mathbb{Q}$ and $M_2(\mathbb{Z})$ is the group of 2x2 matrices with coefficients in $\mathbb{Z}$ ? – JeanThiviers Sep 4 '12 at 13:44
Yes. Though, $M_2(\mathbb{Z})$ should be replaced by the determinant ±1 guys $\operatorname{GL}(2,\mathbb{Z})$. The group $\mathbb{Q}^2 \rtimes \mathbb{Q}$ could be $\begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix}$. – Jack Schmidt Sep 4 '12 at 13:48
(The main point is that the $N_i$ may not support an action of $K$.) – Jack Schmidt Sep 4 '12 at 13:48
Thanks for your answer ! – JeanThiviers Sep 4 '12 at 16:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.