Inflation and Restriction

Suppose $H$ is a subgroup of a finite group $G$ and $A$ is a $G$-module. For each $n\geq 0$, there is a natural map

$$\res _H : \H^n(G,A) \to \H^n(H,A)$$

called restriction. Elements of $\H^n(G,A)$ can be viewed as classes of $n$-cocycles, which are certain maps $G \times \cdots \times G \to A$, and the restriction maps restricts these cocycles to $H \times \cdots \times H$.

If $H$ is a normal subgroup of $G$, there is also an inflation map

$${\mathrm{inf}}_H: \H^n(G/H, A^H) \to \H^n(G,A),$$

given by taking a cocycle $f : G/H \times \cdots \times G/H \to A^H$ and precomposing with the quotient map $G\to G/H$ to obtain a cocycle for $G$.

Proposition 11.3.1   Suppose $H$ is a normal subgroup of $G$. Then there is an exact sequence

$$0 \to \H^1(G/H, A^H)\xrightarrow{{\mathrm{inf}}_H} \H^1(G,A)\xrightarrow{\res _H} \H^1(H,A).$$

Proof. Our proof follows [Ser79, pg. 117] closely.

We see that $\res \circ {\mathrm{inf}}= 0$ by looking at cochains. It remains to prove that ${\mathrm{inf}}_H$ is injective and that the image of ${\mathrm{inf}}_H$ is the kernel of $\res _H$.

1. That ${\mathrm{inf}}_H$ is injective: Suppose $f:G/H\to A^H$ is a cocycle whose image in $\H^1(G,A)$ is equivalent to 0 modulo coboundaries. Then there is an $a\in A$ such that $f(\sigma) = \sigma a - a$, where we identify $f$ with the map $G\to A$ that is constant on the costs of $H$. But $f$ depends only on the costs of $\sigma$ modulo $H$, so $\sigma a - a = \sigma \tau a - a$ for all $\tau \in H$, i.e., $\tau a = a$ (as we see by adding $a$ to both sides and multiplying by $\sigma^{-1}$).Thus $a\in A^H$, so $f$ is equivalent to 0 in $\H^1(H,A^H)$.

2. The image of ${\mathrm{inf}}_H$ contains the kernel of $\res _H$: Suppose $f:G\to A$ is a cocycle whose restriction to $H$ is a coboundary, i.e., there is $a\in A$ such that $f(\tau) = \tau a - a$ for all $\tau \in H$. Subtracting the coboundary $g(\sigma) = \sigma a - a$ for $\sigma\in G$ from $f$, we may assume $f(\tau) = 0$ for all $\tau \in H$. Examing the equation $f(\sigma\tau) = f(\sigma) + \sigma f(\tau)$ with $\tau \in H$ shows that $f$ is constant on the cosets of $H$. Again using this formula, but with $\sigma\in H$ and $\tau\in G$, we see that
   $$f(\tau) = f(\sigma \tau) = f(\sigma) + \sigma f(\tau) = \sigma f(\tau),$$
   so the image of $f$ is contained in $A^H$. Thus $f$ defines a cocycle $G/H \to A^H$, i.e., is in the image of ${\mathrm{inf}}_H$.

$\qedsymbol$

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 $H=A_3 \subset S_3$, and let $S_3$ act trivially on the cyclic group $C = \mathbf{Z}/3\mathbf{Z}$. Using the $\Hom$ interpretation of $\H^1$, we see that $\H^1(S_3/A_3, C) = \H^1(S_3, C) = 0$, but $\H^1(A_3, C)$ has order $3$.

Remark 11.3.3   On generalization of Proposition 11.3.1 is to a more complicated exact sequence involving the ``transgression map'' tr:

$$0 \to \H^1(G/H, A^H)\xrightarrow{{\mathrm{inf}}_H} \H^1(G,A)\xrig... ...w{\res _H} \H^1(H,A)^{G/H} \xrightarrow{{\rm tr}} \H^2(G/H,A^H) \to \H^2(G,A).$$

Another generalization of Proposition 11.3.1 is that if $\H^m(H,A) = 0$ for $1\leq m < n$, then there is an exact sequence

$$0 \to \H^n(G/H, A^H)\xrightarrow{{\mathrm{inf}}_H} \H^n(G,A)\xrightarrow{\res _H} \H^n(H,A).$$

Remark 11.3.4   If $H$ is a not-necessarily-normal subgroup of $G$, there are also maps

$$\cores _H: \H^n(H,A) \to \H^n(G,A)$$

for each $n$. For $n=0$ this is the trace map $a\mapsto \sum_{\sigma \in G/H} \sigma a$, but the definition for $n\geq 1$ is more involved. One has $\cores _H \circ \res _H = [\char93 (G/H)]$. Taking $H=1$ we see that for each $n\geq 1$ the group $\H^n(G,A)$ is annihilated by $\char93 G$.