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the title pretty much says it all... anyway... let G be a semi-direct product of N by Q, and let H be a subgroup of G
can one always find subgroups N1 and Q1, of N and Q respectively, such that H is the semi-direct product of these two groups?
if not in general, can one say anything about the following cases:
i) G = GL(n,K), i.e. the general linear group of dimension n over some field K
ii) G = Aff(E), i.e. the group of all affine motions of some (finite dimensional) linear space

thx in advance

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For case i) do you really want $G = \mathrm{GL}(n, K)$ or did you intend to write $Q = \mathrm{GL}(n, K)$? –  j.p. Jun 4 '11 at 11:38
    
Thanks for your input. I'm embarresed by the triviallity of the counterexamples. @Jug: No I don't, why (honestly I don't understand the question, I think...) Also: About the infinite cyclyc subgroups of Aff(E)...it supersedes my imagination, when I think of the homgenized version of Aff(E), i.e. the embedding into GL(n+1,E), how such a subgroup, i.e. one that is not of the form mentioned in my first post, can arise. Therefor I'd really appreciate it, if Jim Belk, or someone else, could elaborate on that. Lastly: I was thinking about GL(n,E) being the semi-direct product of SL(E) by K*. –  yoeh Jun 6 '11 at 21:08
    
(The previous comment was converted from a "comment as answer" posted by the OP, and was edited slightly for length. 10K users can see the original below.) –  Willie Wong Sep 5 '11 at 16:06
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3 Answers 3

In addition to what others have said, let me remark that it is not even true that any subgroup is isomorphic to a semi-direct product of subgroups (note that this weaker notion does take care of the counterexamples given, since the diagonal group is isomorphic to a semi-direct product of subgroups).

Here is an example: let $D$ be the dihedral group of order $2^{n+1}$: $$ D=\langle x,c | x^2=c^{2^n}=1, xcx=c^{-1}\rangle. $$ Let $H=\langle h| h^{2^n}=1\rangle$ act on $D$ via $x^h = cx$, $c^h=c$, form the semi-direct product $G=D\rtimes H$. Then, you can check that the element $xh$ has order $2^{n+1}$, so the cyclic group generated by it is not isomorphic to a semi-direct product of any subgroups of $D$ and/or $H$.

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Do there exist any such counter-examples in a direct product? –  user1729 Jun 7 '11 at 8:14
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The general answer to your question is no. For example, the Klein four-group is the direct product (and hence semidirect product) of two cyclic groups, but the (diagonal) subgroup $\{(0,0),(1,1)\}$ is not the product of subgroups.

Indeed, I would say that almost any example of a semidirect product will not have the property that you mention. It is certainly not the case for $\text{GL}(n,K)$ or $\text{Aff}(E)$, though I'm not sure exactly what semidirect product structure you have in mind for $\text{GL}(n,K)$. But I can tell you that $\text{Aff}(\mathbb{R})$ has plenty of infinite cyclic subgroups that are not of the form you mention.

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For the direct product $G \times G$ ($G$ being any nontrivial group), the diagonal subgroup is clearly not a semidirect product of subgroups of $G$.

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