4

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

Anton Geraschenko
  • 23,718
user6334
  • 43
  • 3
  • 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). – Ariyan Javanpeykar May 26 '10 at 15:04
  • 7
    @Ariyan: If G=Z and N= 2Z, then G/N = Z/2Z which doesn't embed into Z. – Jason DeVito - on hiatus May 26 '10 at 15:17
  • 6
    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. – George McNinch May 26 '10 at 15:30
  • Whoops. I'll just leave the comment though. – Ariyan Javanpeykar May 26 '10 at 15:59
  • 1
    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?

     – user6334 May 26 '10 at 17:12
  • @Ariyan - If you'd like to delete it, I can also delete my comment (assuming I remember to come back here!) – Jason DeVito - on hiatus May 26 '10 at 17:14
  • 4
    The title of this question should be made more specific. – j.c. May 26 '10 at 18:23

3 Answers3

8

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 $N \rtimes G/N$.

Simon Rose
  • 6,250
  • 4
    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: http://en.wikipedia.org/wiki/Group_extension

    "Split extensions are very easy to classify..."

    The splitting lemma: http://en.wikipedia.org/wiki/Splitting_lemma

    Semidirect product http://en.wikipedia.org/wiki/Semidirect_product

     – user6334 May 26 '10 at 17:08
  • But he said it's another meaning in the comment as well. – awllower Feb 09 '11 at 11:43
8

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.

Tile
  • 435
3

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.

Jimmy
  • 545