# Inflation and Restriction

Suppose is a subgroup of a finite group and is a -module. For each , there is a natural map called restriction. Elements of can be viewed as classes of -cocycles, which are certain maps , and the restriction maps restricts these cocycles to .

If is a normal subgroup of , there is also an inflation map given by taking a cocycle and precomposing with the quotient map to obtain a cocycle for .

Proposition 11.3.1   Suppose is a normal subgroup of . Then there is an exact sequence Proof. Our proof follows [Ser79, pg. 117] closely.

We see that by looking at cochains. It remains to prove that is injective and that the image of is the kernel of .

1. That is injective: Suppose is a cocycle whose image in is equivalent to 0 modulo coboundaries. Then there is an such that , where we identify with the map that is constant on the costs of . But depends only on the costs of modulo , so for all , i.e., (as we see by adding to both sides and multiplying by ).Thus , so is equivalent to 0 in .

2. The image of contains the kernel of : Suppose is a cocycle whose restriction to is a coboundary, i.e., there is such that for all . Subtracting the coboundary for from , we may assume for all . Examing the equation with shows that is constant on the cosets of . Again using this formula, but with and , we see that so the image of is contained in . Thus defines a cocycle , i.e., is in the image of . This proposition will be useful when proving the weak Mordell-Weil theorem.

Example 11.3.2   The sequence of Proposition 11.3.1 need not be surjective on the right. For example, suppose , and let act trivially on the cyclic group . Using the interpretation of , we see that , but has order .

Remark 11.3.3   On generalization of Proposition 11.3.1 is to a more complicated exact sequence involving the transgression map'' tr: Another generalization of Proposition 11.3.1 is that if for , then there is an exact sequence Remark 11.3.4   If is a not-necessarily-normal subgroup of , there are also maps for each . For this is the trace map , but the definition for is more involved. One has . Taking we see that for each the group is annihilated by .

William Stein 2012-09-24