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$\DeclareMathOperator{\Res}{Res} \DeclareMathOperator{\Cor}{Cor}$ This question was asked in MSE. It got no answers or comments, and so I post it here.

Let $H$ be a subgroup of a finite group $G$, and let $M$ be a $G$-module, that is, an abelian group on which $G$ acts. Write $n=[G:H]$. Consider the restriction and corestriction homomorphisms in Tate cohomology: \begin{align*} \Res\colon\, &H^{-1}(G,M) \to H^{-1}(H,M),\\ \Cor\colon\, &H^{-1}(H,M) \to H^{-1}(G,M). \end{align*} Since $$ \Cor \circ \Res =n,$$ we know that for $\xi \in H^{-1}(G, M)$, $$\text{if}\ \, \Res \xi=0,\ \text{then}\ \, n\xi=0.$$

Question: What is an example of $(G,H,M,\xi)$ such that $$n\xi=0,\ \ \text{but}\ \ \Res \xi \neq 0\ ?$$

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    $\begingroup$ What happens for $\mathbb Z/p\mathbb Z\subset \mathbb Z/p\mathbb Z\times\mathbb Z/p\mathbb Z$ with $M=\mathbb Z/p\mathbb Z$? Without computing I'd guess Res is identity (on $H^{-1}\subseteq H_0$), and exponents of these (sub)groups are p. $\endgroup$
    – Chris Gerig
    Commented Sep 19, 2022 at 17:04
  • $\begingroup$ @ChrisGerig: By Proposition 7(i) in Section 6 of the Chapter by Atiyah and Wall in the book "Algebraic Number Theory" edited by Cassels and Frohlich, the restriction map $${\rm Res}\colon H^{-1}(G,M)\to H^{-1}(H,M)$$ is induced by the operator $N'_{G/H}\colon M_G\to M_H$, where $$N'_{G/H}(a)=\sum_i s_i^{-1} s$$ and $s_i$ is a system of coset representatives of $G/H$. $\endgroup$
    – Mikhail Borovoi
    Commented Sep 19, 2022 at 17:48
  • $\begingroup$ @ChrisGerig: In your case, $N'_{G/H}$ is multiplication by $p$, and hence 0. Thus ${\rm Res}\,\xi=0$ for all $\xi\in H^{-1}(G,M)$ in your case. $\endgroup$
    – Mikhail Borovoi
    Commented Sep 19, 2022 at 17:52
  • $\begingroup$ @ChrisGerig: In the second displayed formula in my first comment, I meant $\sum_i s_i^{-1}a$. $\endgroup$
    – Mikhail Borovoi
    Commented Sep 19, 2022 at 18:10

1 Answer 1

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Let $G = \Bbb Z/4$, acting on the Gaussian integers $M = \Bbb Z[i]$ via multiplication by $i$. The transfer $M_G \to M^G$ is given by multiplication by $1 + i + (-1) + (-i) = 0$, so $$H^{-1}(G,M) = M_G = \Bbb Z[i] / (1-i) \cong \Bbb Z/2.$$ In particular, all elements are 2-torsion.

Similarly, if $H$ is the index-2 subgroup $\Bbb Z/2$, then $$H^{-1} = M_H = \Bbb Z[i] / (2).$$ Under this identification, the restriction $\Bbb Z[i]/(1-i) \to \Bbb Z[i]/(2)$ is multiplication by $(1+i)$. The restriction is therefore injective.

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  • $\begingroup$ Excellent! Many thanks! $\endgroup$
    – Mikhail Borovoi
    Commented Sep 19, 2022 at 18:07

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