Describe the second cohomology group $H^2(Z_n times Z

Friday, 25 January 2008

I would take the standard cyclic resolution of $G = Z/nZ$ :

$dots stackrel{1-t}to Z[G] stackrel{sum t^i}to Z[G] stackrel{1-t}to Z[G] to Z to 0,$
where

$t$ is the generator of

$G$ , and then take the tensor square of two such --- this would give a resolution

$dots to Z[G_1times G_2]^3 stackrel{d_2}to Z[G_1times G_2]^2 stackrel{d_1}to Z[G_1times G_2] to Z to 0,$
where

$G_1 = G_2 = Z/nZ$ and the maps are given by

$d_1 = (1-t_1,1-t_2), qquad d_2 = left(begin{array}{ccc} sum t_1^i & 1-t_2 & 0 cr 0 & 1-t_1 & sum t_2^i end{array}right)$
(

$t_1$ and

$t_2$ are the generators of

$G_1$ and

$G_2$ respectively).
I think you can use this for the calculations.

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