Relations bet. Groups, from Relations between Resp. Presentations.

  • Thread starter Bacle
  • Start date
  • #1
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Hi, All:

I am given two groups G,G', and their respective presentations:

G=<g1,..,gn| R1,..,Rm> ;

G'=<g1,..,gn| R1,..,Rm, R_(m+1),...,Rj >

i.e., every relation in G is a relation in G', and they both have the same generating

set.

Does this relation (as a S.E.Sequence) between G,G' follow:

0---> Gp{ R_(m+1),...,Rj }--->G'--->G-->0 ,

where Gp{R_(m+1),...,Rj} is the group generated by the relations (more precisely, by

elements defining the relations) ?

Thanks.
 

Answers and Replies

  • #2
22,129
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I believe this is true, but you'll have to switch G and G' around in you short exact sequence.

The reason is, if you have a group G and a normal subgroup N, then there is a short exact sequence

[tex]0\rightarrow N\rightarrow G\rightarrow G/N\rightarrow 0[/tex]

What you do in your case is work with the free group. A quotient of the free group is exactly a presentation of the group.
 

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