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