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Let $H$ be a subgroup of a finite group $G$, and let $M$ be a $G$-module. Are there any simple conditions on $H,G$ and $M$ which would ensure that the transfer map $H^p(H,M)\to H^p(G,M)$ is the zero map? In my particular situation, $M$ is a trivial $[G:H]$-torsion module, does it buy me anything?

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If it's the zero map, then it's the zero map for all conjugacy classes $gHg^{-1}$ (transfer/restriction maps are "invariant under conjugation").

The composition of transfer map and restriction map is $z\mapsto|G:H|z$, so the transfer map can't be zero unless $H^\ast(G,M)$ is annihilated by $|G:H|$, which you shouldn't expect for general $M$. In other words, you need the exponent of $H^\ast(G,M)$ to divide into $|G:H|$, and in general the exponent divides into $|G|$, and not much is known from the literature besides upper bounds.

So $M=\mathbb{Z}/|G:H|\mathbb{Z}$ allows us to continue. In this case, the transfer map is trivial on the image of the restriction map, so we look for when the restriction map is surjective. This was asked here, Conditions for the restriction $H^i(G,A)\to H^i(H,A)$ being surjective (to which I replied), but we have to be careful due to the assumption on $M$. For example, if $|G:H|=p$ then the restriction map can't be an isomorphism (for then $H$ would be a Sylow $p$-subgroup).

One example I know of that works: $H\hookrightarrow G:=H\times\mathbb{Z}/p\mathbb{Z}$ for any prime $p$, with $M=\mathbb{Z}/p\mathbb{Z}$.

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  • $\begingroup$ Actually, a simp;e example where $H^p(G,M)\to H^p(H,M)$ is NOT surjective, and therefore the transfer map is not zero, would still be very helpful. Especially, for p=1 and p=2. But at least I am now convinced such examples do exist. (There is a fairly complicated p=1 example in the answer to the question that you cited). $\endgroup$ Commented May 2, 2016 at 5:27
  • $\begingroup$ For a simple example in degree two take the cyclic groups $G=C_4, H=C_2$ of order 4 resp. 2. The integral cohomology rings are $H^\ast(G)=\mathbb{Z}[x]/(4x)$ and $\mathbb{Z}[y]/(2y)$ where $x,y$ have degree two. $x$ restricts to $y$ and by for formular in Chris' answer, $tr^G_H(y)=2x \neq 0$. $\endgroup$ Commented May 2, 2016 at 17:14

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