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Let $G$ be a profinite group and $p$ a prime number, and consider the following condition on $G$:
For every open normal subgroup $U$ of $G$ and any integer $N \geq 0,$ there are $N$ elements $$z_1, \ldots z_N \in H^1(U,\mathbb{Z}/p\mathbb{Z})$$ such that the elements $z_1, \ldots , z_N$ are linearly independent over $\mathbb{Z}/p\mathbb{Z}.$

Now, Serre, in his book on Galois cohomology claims that it is enough to verify that this property holds for all sufficiently small open subgroups. To be clear with what I believe he means, the claim is that if we for every open normal $U$ can find an open $U’$ for which this condition holds, then it holds for all open normal subgroups.

I am having a hard time seeing why this is so. Would anyone mind to give some help, comments or thoughts? Complete solutions are OK, but just thoughts work too.
Update
This is exercise 1, I.§3.4 in Serre’s book on Galois cohomology. If you do not see how to solve the exercise as it stands now, take the cohomological dimension of $G$ to be $1.$ maybe there is an assumption that Serre forgets to add.
Update 2*
I misquoted Serre, believing one could drop something which I now believe one can not drop.
The (correct) stronger assumption to make is that for every open normal subgroup$U$ of $G$ and any integer $N \geq 0,$ there are $N$ elements $$z_1, \ldots, z_N \in H^1(U,\mathbb{Z}/p\mathbb{Z})$$ such that the elements $s(z_i),$ where $s \in G/U, i = 1 , \ldots , n$ are linearly independent.

With this assumption, the statement holds.

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  • $\begingroup$ Is there some reason why H1(U) \to H1(U') is surjective, i.e. you can lift z_i's? I know that the ker of this map is H1(U/U') (a finite group), assuming that the action on Z/pZ is trivial. Do you know anything on the coker? $\endgroup$ – Andrea Marino Feb 12 '18 at 23:44
  • $\begingroup$ Do you have any other hp on G? $\endgroup$ – Andrea Marino Feb 12 '18 at 23:48
  • $\begingroup$ @AndreaMarino In general, as you probably know, the map $H^1(U) \to H^1(U’)$ is not surjective. If $H^2(U/U’)=0,$ then the map is surjective. If the map is surjective, then we are of course done, but I see no reason for it beomg surjective $\endgroup$ – Dedalus Feb 13 '18 at 8:30
  • $\begingroup$ Isn't the H^2 defined just for abelian groups? Furthermore, if the action of G on Z/pZ is trivial, H1(U,Z/pZ)= Hom(U,Z/pZ), right? In this case you can factor out the commutator of G and reduce to the abelian case. I am not fresh of group cohomology, do you confirm? $\endgroup$ – Andrea Marino Feb 13 '18 at 9:32
  • $\begingroup$ @AndreaMarino $H^2(G,A)$ is defined for $G$ arbitrary, but $A$ should be abelian (unless you want non-abeliqn cohomology). You are correct that if the action is trivial, first cohomology coincides with Hom. $\endgroup$ – Dedalus Feb 13 '18 at 18:55
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I hope this is right :)

As I mentioned in comments, the $H^1$ groups are just $Hom$ because the action is trivial. $Z/pZ$ is abelian, thus we can factor out the commutators of $G$ and suppose $G$ abelian. Furthermore, $Hom(Z/q^rZ, Z/pZ)=0$, thus we can suppose that $G$ is a p-group.

Take $U' \subset U$ and consider the exact sequence

$$ H^1(U/U', Z/pZ) \to H^1(U, Z/pZ) \to H^1(U', Z/pZ) \to H^2(U/U', Z/pZ)$$ Now we want to investigate the first and the last term. Firstly, note that $U/U'$ is finite, because $U,U'$ are open subgroups of a compact group, thus they have finite index. It is an abelian p-group, so it writes as a product of $Z/p^rZ$. But these factors all have cohomological dimension 0 over the ring of coefficients $Z/pZ$. Infact

$$Z/pZ[ Z/p^rZ] \simeq Z/pZ[x]/(x-1)^{p^r}$$

Has krull dimension 0, which is the projective dimension of the category of $Z/pZ[Z/p^rZ]$, which in turn is the definition of cohomological dimension.

This concludes, because cohomology of $U/U'$ vanishes and the exact sequence yields $H^1(U,Z/pZ) \simeq H^1(U', Z/pZ)$.

I suspect that this is wrong because it seems a too strong result... I wait for your opinions :)

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  • $\begingroup$ Thank you for your answer. There are some mistakes, and one of them I believe makes the argument invalid. 1. First, the exact sequence is not precisely as you claimed, but you need to take fixed points, see en.wikipedia.org/wiki/Inflation-restriction_exact_sequence (so the third term should be $H^1(U',\mathbb{Z}/p\mathbb{Z}).$ 2. Group cohomology is, as you seem to be aware of, the derived functor of $Hom_{\mathbb{Z}[G]}(\mathbb{Z},-).$ In our case, $G= \mathbb{Z}/p^r\mathbb{Z}.$ To calculate $H^i(G,A),$if $A$ is killed by $p,$ we can calculate the derived functors of (cont...) $\endgroup$ – Dedalus Feb 14 '18 at 11:53
  • $\begingroup$ $Hom_{\mathbb{Z}/p\mathbb{Z}[\mathbb{Z}/p^r\mathbb{Z}]}(\mathbb{Z}/p\mathbb{Z},-).$ You then write that $\mathbb{Z}/p \mathbb{Z}/[\mathbb{Z}/p^r\mathbb{Z}] \cong (\mathbb{Z}/p\mathbb{Z})^{p^r}.$ What kind of isomorphism is this? It can not be an isomorphism of rings, since the left hand side is not reduced, but the right hand side is. it is not enough for this to be an isomorphism of vector spaces over $\mathbb{Z}/p\mathbb{Z}$ to make your argument work. $\endgroup$ – Dedalus Feb 14 '18 at 11:53
  • $\begingroup$ In fact, you can show that if $U/U' = \mathbb{Z}/p^r \mathbb{Z},$ then the cohomology ring $H^*(\mathbb{Z}/p^r\mathbb{Z},\mathbb{Z}/p\mathbb{Z})$ has non-zero cohomology groups in arbitarily high degrees. $\endgroup$ – Dedalus Feb 14 '18 at 11:53
  • $\begingroup$ Yes, you are right. My isomorphism was wrong! But fortunately the krull dimension remains 0, because you factored a PID by a non-zero ideal. Why do you argue that Z/p^rZ has infinite cohomological dimension over Z/pZ? If we have to take Z in place of Z/pZ, the cohomological dimension remains 1, which is enough for our purpose. About the first point, I actually took the fixed points, but the action is trivial. It seems to me that you wrote H1(U', Z/pZ) which is my third term. Isn't it? Or are you referring to the H^2 term? $\endgroup$ – Andrea Marino Feb 14 '18 at 12:11
  • $\begingroup$ Consider the simplest case, when $G= \mathbb{Z}/2\mathbb{Z}.$ Then the group cohomology $H^*(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z})$ coincides with the cohomology of $\mathbb{R}P^\infty.$ The cohomology ring $H^*(\mathbb{R}P^\infty,\mathbb{Z}/2\mathbb{Z})$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}[x].$ Thus, the group cohomology of $\mathbb{Z}/2\mathbb{Z}$ has terms in infinitely high degree. Here is an online reference: groupprops.subwiki.org/wiki/… $\endgroup$ – Dedalus Feb 14 '18 at 12:17

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