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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 $BS(1,2) \simeq \mathbb{Z}[\frac{1}{2}] \rtimes \mathbb{Z}$ ?

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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.

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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
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