External Weak Product is the Internal Weak Product of…

[Bashyboy]

Homework Statement

Let $\{G_i \mid i \in I\}$ be a family of groups, then $\prod^w G_i$, the external weak direct product, is the internal weak direct product of the subgroups $\{i_k(G_k) \mid k \in I\}$, where $i_k : G_k \to \prod G_i$ is the canonical embedding.

Homework Equations

The Attempt at a Solution

I have already shown that the $i_k(G_k)$ are normal in $\prod^w G_i$, and clearly $i_k(G_k) \cap \langle \bigcup_{k \in I} i_k(G_k) \rangle = \{e\}$ But I don't see how this is true. Specifically, isn't $\langle \bigcup_{k \in I} i_k(G_k) \rangle$ bigger than $\prod^w G_i$, as it contains elements whose every component is an nonidentity element? I could see $\prod G_i$, the 'regular' direct product, being the internal weak direct product of these subgroups, but not $\prod^w G_i$. What am I missing?

 

[mighty2000]

First of all, let's clarify the definitions of the external and internal weak direct products. The external weak direct product $\prod^w G_i$ is defined as the set of all functions $f : I \to \bigcup_{i \in I} G_i$ such that $f(i) \in G_i$ for all $i \in I$ and $f(i) = e$ for all but finitely many $i \in I$, with the operation being defined pointwise. The internal weak direct product of the subgroups $\{i_k(G_k) \mid k \in I\}$ is defined as the subgroup of $\prod^w G_i$ consisting of all functions $f : I \to \bigcup_{i \in I} G_i$ such that $f(i) \in i_k(G_k)$ for all $i \in I$ and $f(i) = e$ for all but finitely many $i \in I$.

Now, let's address your concern about $\langle \bigcup_{k \in I} i_k(G_k) \rangle$ being bigger than $\prod^w G_i$. Note that, by definition, $\langle \bigcup_{k \in I} i_k(G_k) \rangle$ is the smallest subgroup of $\prod^w G_i$ containing all the elements of $\bigcup_{k \in I} i_k(G_k)$. It may seem like this subgroup is bigger than $\prod^w G_i$, but in reality it is not. In fact, we can show that $\prod^w G_i$ is contained in $\langle \bigcup_{k \in I} i_k(G_k) \rangle$.

To see this, let $f \in \prod^w G_i$ be an element of the external weak direct product. Then, by definition, $f(i) = e$ for all but finitely many $i \in I$. This means that $f(i) \in i_k(G_k)$ for all but finitely many $i \in I$, which implies that $f \in \langle \bigcup_{k \in I} i_k(G_k) \rangle$. Therefore, $\prod^w G_i$ is indeed a subgroup