Profinite groups, directed sets and $H^1$

Question

Usually whenever one reads the definition of profinite group, one starts with an ordered set $I$ which is directed, meaning that for every $i,j\in I$ there is some $k\in I$ such that $i\leq k$ and $j\leq k$. Then one has a family of finite groups $\{G_i\}_{i\in i}$, compatible maps between them and defines $\varprojlim_i G_i$ as the subset of $\prod_{i\in I}G_i$ formed by coherent sequences.

The thing that is not clear at all to me is why we require $I$ to be directed, because it seems that dropping that assumption basically nothing would change and the definition would remain the same. Moreover, one thing we would like is for example to consider $\prod_iG_i$ a profinite group. This can be seen easily if we take the trivial ordering on $I$. But with this ordering, $I$ is not a directed set. Is there a way to see $\prod_i G_i$ as an inverse limit of finite group over a directed set? And more in general, if $\{P_i\}_{i\in I}$ is a family of profinite group and $P=\prod_iP_i$ is their direct product with the product topology, is $P$ always a profinite group?

Now let assume that $\{P_i\}_{i\in I}$ is a family of profinite groups for which the question above has an affirmative answer. Let $A$ be a discrete abelian group over which $P_i$ acts trivially for all $i$'s. Is it true that

$$H^1\left(\prod_iP_i,A\right)\simeq \bigoplus_iH^1(P_i,A)$$ Here $H^1$ stands for the usual first cohomology group with coefficients in $A$.

Thanks a lot to anyone which is willing to give me a hint!

Answer 1

I hope this will clarify the situation:

I: Why is $I$ a directed set: In fact, there is no need for $I$ to be directed. One can define the inverse limit of a inverse system without the restriction that $I$ is directed. However, one should be aware that by doing so, certain propositions about profinite groups resp. spaces won't work anymore. For example the following proposition is not true anymore: Let $G=\varprojlim_{i\in I}G_i$ be a profinite group, where $I$ is a directed set and let $\varphi_i:G\rightarrow G_i$ be the projection homomorphisms. Then $\{S_i:S_i=\mathrm{ker}(\varphi_i)\}$ is a fundamental system of open neighborhoods of the identity element of $1$ in $G$.

II: Is a direct product of profinite groups $\prod_iP_i$ (with the product topology) again profinite: Yes. See Proposition 2.2.1 in 'Ribes, Zalesskii, Profinite groups'.

III: Does $H^1\left(\prod_iP_i,A\right)\cong \bigoplus_iH^1(P_i,A)$ hold: Yes if $\prod_iP_i$ acts trivially on $A$. Set $P=\prod_iP_i$. For every $P_i$ there is a collection of open-closed normal subgroups $\mathcal{N}_i$ such that $H^1(P_i,A)=\varinjlim_{N\in \mathcal{N}_i}H^1(P_i/N,A)$ (See Corollary 6.5.6'Ribes, Zalesskii, Profinite groups'). Furthermore, there is collection of open-closed normal subgroups $\mathcal{N}$ such that $H^1(P,A)=\varinjlim_{N\in \mathcal{N}}H^1(P/N,A)$. In fact, we can choose $\mathcal{N}\subseteq \prod_i\mathcal{N}_i$ to be the set of all $N\in \prod_i\mathcal{N}_i$ such that $M_{N}:=\left\{k:\mathrm{pr}_k(N)\neq P_k\right\}$ is finite. It is easy to see that we can embed each $\mathcal{N}_i$ into $\mathcal{N}$ such that $\mathcal{N}_i\cap \mathcal{N}_j=\{P\}$ if $i\neq j$. We can rewrite $\varinjlim_{N\in \mathcal{N}}H^1(P/N,A)$ as $$\varinjlim_{N\in \mathcal{N}}H^1(P/N,A)=\left(\bigoplus_{N\in \mathcal{N}}H^1(P/N,A)\right)/\mathcal{I}.$$ Each non-zero element of the subgroup $\bigoplus_{i}H^1(P_i,A)$ of $\bigoplus_{N\in \mathcal{N}}H^1(P/N,A)$ is not contained in the subgroup $\mathcal{I}$, since $\mathcal{N}_i\cap \mathcal{N}_j=\{P\}$ for $i\neq j$. Hence $\bigoplus_{i}H^1(P_i,A)\subseteq H^1(P,A)$. It remains to verify $\bigoplus_{i}H^1(P_i,A)\supseteq H^1(P,A)$. Note that every $N\in \mathcal{N}$ can be written as a finite intersection $\bigcap N_i$ of elements in $N_i\in \mathcal{N}_i$. Set $\tilde{\mathcal{N}}=\mathcal{N}\backslash \bigcup_i\mathcal{N}_i$. Then $H^1(P/\tilde{N},A)\cong \bigoplus_{k\in M_{\tilde{N}}}H^1(P_k/\mathrm{pr}_k(\tilde{N}),A)$. Therefore, it is easy to see that $H^1(P/\tilde{N},A)$ is contained in $\mathcal{I}$. Thus $\bigoplus_{i}H^1(P_i,A)=H^1(P,A)$.