Let Z denote the set of all integers, and let G be a finite cyclic Group.
For every ZG-module A, and n=1,2,3...
Show that:
Hn(G,A) is isomorphic to Hn+1(G,A).
I think that you can use the article
FROM HOMOLOGICAL ALGEBRA TO GROUP COHOMOLOGY
Semester Project By
Maximilien Holmberg-Péroux
Responsible Professor
prof. Jacques Thévenaz
Supervisor
Rosalie Chevalley
Academic year : 2013-2014
Spring Semester
You can calculate the cohomology/homology of cyclic groups G in different ways, e.g. by a 2-periodic ZG-free resolution of Z.:
.... --> ZG --> ZG --> ZG --> ZG --> Z
The last homomorphism is augmentation. The other homomorphisms are Multiplikation by t-1 and 1+t+t^2+ ...+t^{n-1}, if G is cyclic of order n and t a generator.
The calculation of the 2-periodic group homology/cohomology of G with coefficients A (by applying $Hom(.,A)$ and $\otimes A$ to this resolution) is straightforward and establishes the (co-)homology in terms of the G-Operation on A. The isomorphism ensues.
(For the construction of a periodic resolution see e.g. Brown: Cohomology of Groups.
For a more principled type of argument see chapter 7 of Hilton/Stammbach: Homological algebra.)
This is actually a question on the periodicity of the cohomology of a finite cyclic group, see Serre, chap. VIII. For a finite group G, define the modified Tate cohomology groups as follows : define the norm element N of Z[G] by N = sum of all elements g of G, and for any G-module A, H0^(G,A)=AG/NA, H-1^ (G,A)=Ker N /IGA , Hn^(G,A) = Hn(G,A) if n>=1, Hn^(G,A) = Hn-1(G,A) if n>=2. Tate showed that this deftermines a cohomological functor in all dimensions, and if G is cyclic, the modified Hn^(G,A) depend only on the parity of n. If a generator g of the cyclic G is chosen, then for all n in Z, one has isomorphisms Hn^(G,A) = Hn+2^(G,A) induced by the cup-product with a distinguish element of H2^(G,A). The result to keep in memory is that Hn^(G,A) = AG/NA if n is even, Ker N/(g-1)A if n is odd, but these iso. are not canonical. Needless to say, none of Tate's theorems is elementary, so your question is not either.
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