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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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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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. |
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