$H^1(G, \mathbb{Z}/p \mathbb{Z})$ and linearly independent elements in open subgroups. 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.
 A: 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 :)
