Group cohomology

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 01 February 2012

Group cohomology

Let G be a group and let M be an abelian group with an action of G by automorphisms, i.e. let M be a G-module. Define abelian groups C0GM =M and CnGM = { f:G××GM }, for   n >0, where the operation in CnGM is given by (f1+f2) = f1 + f2 . Define 0 C0GM 1 C1M 2 C2M 3 such that n+1 n=0, by defining (nf) = g1f + i=1n -1 i f g1 ... gi-1 gigi+1 gi+2 ... gn + -1 n f g1... gn-1 , for all f Cn-1 GM and G.

Let 𝔼 be a field and let G be a finite subgroup of Aut(𝔼).

  1. H1 G𝔼 is trivial.
  2. H1 G𝔼* is trivial.

b. Let f:GE* be a cocycle so that f(g1g2) = fg1 (g1 fg2) , for all   g1, g2G. By Dedekind's lemma we may choose x𝔼 such that b= gG fg gx is nonzero. If hG then hb= gG (hfg) hgx = gG f h - 1 f(hg) hgx= f h - 1 b. So fh = b/hb and f is a coboundary.

(Dedekind's Lemma.) Let 𝔽 be a subfield of 𝔼.

  1. Distinct embeddings of 𝔽 into 𝔼 are linearly independent.
  2. Distinct characters χ:G𝔼* are linearly independent.
  3. Distinct elements of Aut(𝔼) are linearly independent in End𝔽(𝔼).

Let 𝔼 be a finite extension of 𝔽. Let gi: A𝔼,   1in be algebra homomorphisms. Let 𝔼 be such that i=1n cigi =0. For any x,yA, 0= i=1n cigi (xy) -gn(x) i=1n cigi (y) = i=1 n-1 ci( gi(x) -gn(x) ) gi(y). So, by induction, ci( gi(x)- gn(x)) =0, for all   1in-1 ,   and all   xA. For each 1in-1 , we may choose x such that gi(x) gn(x), and conclude that li=0 for all   1in-1. So i=1n ligi = lngn =0. Thus 0=ln gn(1) =ln.

Notes and References

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