Hi people!
This my first question, here. I don't sure if it has a trivial answer, or not.
Let G a group, N normal subgroup in G. In which cases there is a subgroup in G isomorphic to G/N?
TIA
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$\begingroup$ A first example that comes to mind: let G be a finitely generated free abelian group. Then N is also finitely generated free. Thus, G/N is free of smaller rank. I don't know of any other examples (but that doesn't mean alot coming from me). $\endgroup$– Ariyan JavanpeykarCommented May 26, 2010 at 15:04
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7$\begingroup$ @Ariyan: If G=Z and N= 2Z, then G/N = Z/2Z which doesn't embed into Z. $\endgroup$– Jason DeVito - on hiatusCommented May 26, 2010 at 15:17
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6$\begingroup$ You should clarify whether you mean a subgroup of $G$ that is mapped isomorphically to $G/N$ by the natural quotient mapping $\pi:G \to G/N$, or simply a subgroup of $G$ that is "abstractly" isomorphic to $G/N$. The notions aren't the same. $\endgroup$– George McNinchCommented May 26, 2010 at 15:30
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1$\begingroup$ George: hmmmm... My original idea was asked for any subgroup in G that were isomorphic to G/N. I guess it's your meaning of "abstractly" isomorphic. It could be a subgroup "abstractly" isomorphic to G/N, but no subgroup mapped to G/N via the natural quotient mapping? $\endgroup$– user6334Commented May 26, 2010 at 17:12
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4$\begingroup$ The title of this question should be made more specific. $\endgroup$– j.c.Commented May 26, 2010 at 18:23
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3 Answers
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Assuming you're looking at the case where the isomorphism is induced by the quotient $G \to G/N$ (as per George McNinch's comment), then this should be if and only if the sequence $$ 0 \to N \to G \to G/N \to 0$$ splits. i.e. there is a section $\sigma : G/N \to G$. This is then seen to be equivalent to $G$ being isomorphic to the semidirect product $N \rtimes G/N$.
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4$\begingroup$ Nice! Thanks to your message, I just discovered the term "split" (I read it in some elementary intro to Category Theory). Interesting... I read a bit more at: en.wikipedia.org/wiki/Group_extension "Split extensions are very easy to classify..." The splitting lemma: en.wikipedia.org/wiki/Splitting_lemma Semidirect product en.wikipedia.org/wiki/Semidirect_product $\endgroup$– user6334Commented May 26, 2010 at 17:08
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$\begingroup$ But he said it's another meaning in the comment as well. $\endgroup$– awllowerCommented Feb 9, 2011 at 11:43
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You have to be careful! Of course in split extensions it is trivial that $G/N$ is isomorphic to a subgroup of $G$. On the other hand there are examples of extensions $$1\rightarrow N\rightarrow G\rightarrow G/N\rightarrow 1$$ that are not split but nevertheless there is a subgroup $H\le G$ with $H\cong G/N$.
An example would be the quaternion group that cannot be written as a nontrivial extension. But it contains a normal subgroup of index 2 and a subgroup of order 2.
Unfortunatly I don't see a solution to your problem in general.
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A good theorem related to your problem is Schur-Zassenhaus theorem. It states that when the normal subgroup N is a Hall subgroup, namely the order of N and the index of N are coprime, then there exists a complement of N, that is a subgroup H s.t. G=NH and N\cap H={identity}. So H is isomorphic to G/N.
