Let $G_i, f_{ij}$ be an inverse system of topological groups where each $G_i$ is finite in the discrete topology. A profinite group is defined to be an inverse limit of such an inverse system. However, my professor seemed to assume that the $f_{ij}$ needed to be surjective (so a projective system). Is it necessary to have this assumption? Are there any advantages to assuming surjectivity?

In any case one has that if $G = \lim\limits_{\leftarrow}G_i$, then $G$ is isomorphic to the inverse limit of all quotients $G/N$, where $N$ runs through the open normal subgroups of $G$. Therefore any profinite group will be the inverse limit of a system where the maps are surjective, so I guess we can assume this property holds without loss of generality.

  • $\begingroup$ Well, a projective system is nicer is if you want to explicitly describe the group. In this case if you take a "projective subsystem $H_i$", then the inverse limit $H$ of the $H_i$'s has a surjection from $G$. On the other hand, if you don't require the maps $f_{ij}$ to be surjective, then you can't necessarily tell from looking at a finite subsystem whether the inverse limit is the trivial group or not. However I don't have any idea if this is related to why your professor made this assumption. $\endgroup$ – Kimball Jan 25 '15 at 16:14

It is not necessary for the maps to be surjective in the definition. However, any profinite group is isomorphic to a limit of a projective system in which the maps are surjective. Indeed, suppose $(G_i, f_{ij})$ is a projective system. Define $G_i' = \cap_j f_{ij} (G_j)$, where the intersection is taken over all $j \to i$. Remark that if $G_j \to G_{k} \to G_i$, we have $f_{ij}(G_j) = f_{ik}(f_{kj}(G_j)) \subseteq f_{ik}(G_{k})$, so the $f_{ij}(G_j)$ get smaller as $j$ moves up the system. I'll let you prove that $(G_i', f'_{ij})$ forms a projective system, where $f'_{ij}$ are the restrictions of the $f_{ij}$'s, and that the $f'_{ij}$'s are surjective. I'll let you prove also that the inclusion $(G'_i, f_{ij}') \to (G_i, f_{ij})$ induces an isomorphism on the limits.

  • $\begingroup$ Why are the $f_{ij}'$ surjective? It's clear that $f_{ij}$ will map $G_j$ onto $G_i'$, but why should it map $G_j'$ onto this group? I've tried for two hours to prove this. $\endgroup$ – D_S Feb 16 '15 at 2:20
  • $\begingroup$ Nevermind I got it. That was harder than I thought it would be. $\endgroup$ – D_S Feb 16 '15 at 7:09

A detailed explanation of Bruno Joyal's answer:

I. For $k \geq i$, the image of the restriction $f_{ik}' = f_{ik} \mid G_k'$ is contained in $G_i'$.

Let $$g \in G_k' = \bigcap\limits_{j \geq k} f_{kj}(G_j)$$ In order to show $f_{ik}(g) \in G_i'$, we must show that for any $j \geq i$, there exists an $h \in G_j$ such that $f_{ij}(h) = f_{ik}(g)$. Since we are in a direct system, we may choose an index $s$ which is $\geq j$ and $k$. Since $g \in G_k'$, there exists a $y \in G_s$ with $f_{ks}(y) = g$. Setting $h = f_{js}(y)$, we have $$f_{ij}(h) = f_{ij}(f_{js}(y)) = f_{is}(y) = f_{ik}(f_{ks}(y)) = f_{ik}(g)$$

II. The mapping $f_{ik}':G_k' \rightarrow G_i'$ is surjective.

Let $x \in G_i'$. Then for any $j \geq i$, there exists an $x_j \in G_j$ with $f_{ij}(x_j) = x$. In particular, $f_{ik}(x_k) = x$. However, we do not know that $x_k$ actually lies in $G_k'$. If it does not, then nevertheless it suffices to find a $g \in f_{ik}^{-1}\{x\}$ which does lie in $G_k'$.

Suppose there is no such $g$. Then for every $g \in f_{ik}^{-1}\{x\}$, there exists an index $k_g \geq k$ for which $f_{kk_g}^{-1}\{g\} = \emptyset$. Since $f_{ik}^{-1}\{x\}$ is finite (we are dealing with finite groups), we can find an index $j$ which is $\geq$ all the indices $k_g$. We then have $$x = f_{ij}(x_j) = f_{ik}(f_{kj}(x_j))$$ So $f_{kj}(x_j) \in f_{ik}^{-1}\{x\}$, hence $f_{kj}(x_j)$ is equal to some $g$. But then $$g = f_{kj}(x_j) = f_{kk_g}(f_{k_gj}(x_j))$$ so $f_{k_gj}(x_j) \in f_{kk_g}^{-1}\{g\} = \emptyset$, a contradiction.

III. $(G_i', f_{ij}')$ forms an inverse system, with $\lim\limits_{\leftarrow} G_i' \cong \lim\limits_{\leftarrow} G_i$.

It's clear that $(G_i', f_{ij}')$ forms an inverse system. For its inverse limit, we use the canonical construction, namely the subset of the product group (in the product topology) $$ \mathcal G' \subseteq \prod\limits_i G_i'$$ consisting of all $(x_i)$ such that $x_i = f_{ij}'(x_j)$ whenever $i \leq j$. On the other hand, $\lim\limits_{\leftarrow} G_i$ can be taken as the product $$\mathcal G \subseteq \prod\limits_i G_i$$ again consisting of all sequences $(x_i)$ where $x_i = f_{ij}(x_j)$ for all $i \leq j$. For any such $(x_i)$, it follows that the $x_i$ actually lie in $G_i'$, hence $$\lim\limits_{\leftarrow} G_i =\mathcal G = \mathcal G' = \lim\limits_{\leftarrow} G_i'$$

  • 1
    $\begingroup$ Thanks for adding all of the details! I think the accepted answer should have gone to you. +1 from me anyways! $\endgroup$ – Bruno Joyal Feb 18 '15 at 22:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.