Investigate whether a modified approach to calculating profinite completion of a direct sum of cyclic groups produces an isomorphic result | Step-by-Step Solution

Understanding Profinite Completions of Direct Sums

What We're Solving: We want to explore whether there's a "shortcut" method for computing the profinite completion of a direct sum of cyclic groups by using a smaller index set in our inverse limit construction, and determine if this modified approach gives us the same mathematical object as the standard method.

The Approach: This is a question that gets to the heart of how inverse limits behave with respect to direct sums! We're asking: "Can we be clever about our construction and still get the right answer?" To tackle this, we need to:

• 1. Understand what the profinite completion actually measures
• 2. Set up both the "standard" and "modified" constructions carefully
• 3. Use the universal property of inverse limits to compare them
• 4. Think about what "smaller index set" might mean in this context

Step-by-Step Solution:

Step 1: Set up the standard construction Let's say we have a direct sum $G = \bigoplus_{i \in I} C_i$ where each $C_i$ is cyclic. The profinite completion $\hat{G}$ is constructed as: $$\hat{G} = \lim_{\leftarrow} G/N$$ where the inverse limit runs over all finite index normal subgroups $N$ of $G$.

Step 2: Understand what "smaller index set" could mean There are several interpretations, but the most natural one is: instead of considering ALL finite index subgroups, we consider only those in some smaller collection that still captures the essential information. For example, we might only look at subgroups that interact "nicely" with the direct sum structure.

Step 3: Use the key insight about direct sums and profinite completions Here's the crucial theorem that will guide us: For a direct sum of cyclic groups, we have: $$\widehat{\bigoplus_{i \in I} C_i} \cong \prod_{i \in I} \hat{C_i}$$

This tells us the profinite completion of a direct sum equals the direct product of the individual completions!

Step 4: Analyze when the shortcut works The modified construction will work if and only if our "smaller collection" of subgroups still generates the same inverse limit. This happens when:

• Every finite index subgroup can be "approximated" by subgroups in our smaller collection
• The smaller collection is still "cofinal" in the directed system

Step 5: Consider specific cases For cyclic groups, we could potentially use only the subgroups of the form $\prod_{i \in F} (C_i/H_i) \times \prod_{i \notin F} C_i$ where $F$ is finite and each $H_i$ is a finite index subgroup of $C_i$.

The Answer: The modified approach CAN produce the same result, but only under specific conditions! The key is that your smaller index set must be "cofinal" - meaning every subgroup in the original construction can be "refined" by one from your smaller collection.

When this cofinality condition holds, the universal property of inverse limits guarantees the constructions are isomorphic. However, if your smaller collection misses important "directions" of approximation, you'll get a different (usually smaller) completion.

Memory Tip: Think of it like taking photos of a sculpture - if you only take pictures from the front and back (smaller collection), you might miss important details that side views would capture. But if your "smaller collection" of viewpoints still captures all the essential features, you'll reconstruct the same sculpture! The mathematical version is: "cofinal subcollections give the same inverse limit."

The beauty of this problem is that it shows how the rigid-seeming definition of profinite completion actually has some flexibility in its construction, as long as we preserve the essential limiting behavior!