Finding middle group in short exact sequence

[wodhas]

Hi,

Is there any method by which problems involve short exact sequences can be solved.

For example, if I have :

0->Z->A->Z_4->0

how can I decide what is the middle group without knowing any of the maps

Thanks!
S.

 

[quasar987]

Well, by the first isomorphism theorem, we have, modulo isomorphisms, that A/Z = Z_4. So if you can find a group having Z as a subgroup and such that when quotiented by Z gives Z_4, then A will be isomorphic to that...

 

[g_edgar]

0->Z->A->Z_4->0

Then A is called an extension of Z by Z_4 ... one example is the product Z x Z_4, but there could be others as well. The collection of all such extensions is sometimes given some structure and made into an algebraic object itself ... look up Ext and Tor I don't remember the details.

 

[wofsy]

Hi,

Is there any method by which problems involve short exact sequences can be solved.

For example, if I have :

0->Z->A->Z_4->0

how can I decide what is the middle group without knowing any of the maps

Thanks!
S.

In general you can not tell. But you can narrow down the possibilities. As other replies pointed out the middle group in your example could be Z or it could be ZxZ/4.

 

[Tzar]

In general there are infinitely many groups that can go in the middle of a short exact sequence. In fact they are in one to one correspondence with elements of Ext^1(Z_4,Z) (which is an Abelian group). Now if this Ext group is zero, then you know for sure, the middle term must be ZxZ/4, otherwise there are many more possibilities. To compute this Ext^1 group look at the long exact sequence of cohomology groups arising from your short exact sequence or try to construct some resolutions. I'm not sure how hard either one of those tasks are in practice.(If you aren't sure what that means I recommend looking it up Ext on wikipedia)

 

[wofsy]

In general there are infinitely many groups that can go in the middle of a short exact sequence. In fact they are in one to one correspondence with elements of Ext^1(Z_4,Z) (which is an Abelian group). Now if this Ext group is zero, then you know for sure, the middle term must be ZxZ/4, otherwise there are many more possibilities. To compute this Ext^1 group look at the long exact sequence of cohomology groups arising from your short exact sequence or try to construct some resolutions. I'm not sure how hard either one of those tasks are in practice.(If you aren't sure what that means I recommend looking it up Ext on wikipedia)

Can you give me five other possibilities?