Let $1 \to K \to G \to H \to 1$ be a short exact sequence in the category of groups(interested in non-Abelian groups). My question is the following:

Where does the obstruction to the splitting of the above short exact sequence lie?

Any help will be appreciated.

Thank you so much.


The main problem arising when you deal with anabelian groups sequences is that the you don't have an unique notion of splitness (recall that groups don't form an abelian category).

There are two "canonically split" exact sequences: the direct product exact sequence: $$1\longrightarrow G\longrightarrow G\times H\longrightarrow H\longrightarrow 1$$ and the semi-direct product exact sequence: $$1\longrightarrow G\longrightarrow G\rtimes H\longrightarrow H\longrightarrow 1$$

Splitness is usually understood by means of semi-direct product, since this notion is more general and flexible. The obstructions look like the same ones in the abelian case (existence of a section or a retract) but there are some subleties because you need to distinguish when an exact sequence is of "direct product" type or "semi-direct product" type. Here the main result:

Theorem. Let $1\to H\to G\to K\to 1$ a short exact sequence. The following are equivalent:

  • there exists a morphism $G\to H$ which is a left inverse for $H\to G$;
  • there exists an isomorphism $G\to H\times K$ which fits in the following commutative diagram: enter image description here

Similarly, the following are equivalent:

  • there exists a morphism $K\to G$ which is a right inverse for $G\to K$;
  • there exists an isomorphism $G\to H\rtimes K$ which fits in the following commutative diagram: enter image description here

For more detailed informations, you can read this nice introduction: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/splittinggp.pdf

| cite | improve this answer | |
  • $\begingroup$ The obstruction class for spliting should lie in some group cohomology of $H.$ $\endgroup$ – Math Dec 31 '16 at 11:56
  • $\begingroup$ That's exactly the meaning of my post, but in a more down-to-earth way. Splitness is related to group extensions, which are related to Ext^1 and then group cohomology. I wanted to warn you that the usual homological algebra tools may not suffice as you don't have an abelian structure in the group category. $\endgroup$ – Caligula Jan 6 '17 at 13:17
  • $\begingroup$ Yes. I get that the obstruction class should lies in the second cohomology of H with coefficient in center of K. $\endgroup$ – Math Jan 6 '17 at 14:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.