The Weak Direct Product of an Arbitrary Collection of Groups

# The Weak Direct Product of an Arbitrary Collection of Groups

Definition: Let $\{ G_i : i \in I \}$ be an arbitrary collection of products. Define $\displaystyle{\prod_{i \in I}^{\mathrm{weak}} G_i = \left \{ f : I \to \bigcup_{i \in I} G_i \: \biggr | \: f(i) \in G_i \: \forall i \in I, \: f(i) = e_{G_i} \: \mathrm{for \: all \: but \: finitely \: many \:} i \right \}}$. The Weak Direct Product of the groups $\{ G_i : i \in I \}$ is the set $\prod_{i \in I}^{\mathrm{weak}} G_i$ with the operation of compontentwise multiplicative defined for all $f, g \in \prod_{i \in I}^{\mathrm{weak}} G_i$ by $(fg)(i) = f(i)g(i)$ for all $i \in I$. |

*Note that $\prod_{i \in I}^{\mathrm{weak}} G_i \subseteq \prod_{i \in I} G_i$ and $\prod_{i \in I}^{\mathrm{weak}} G_i$ consists of all functions $f : I \to \bigcup_{i \in I} G_i$ that are "almost" the function $e$ defined by $e(i) = e_{G_i}$.*

*Also note that if $I$ is a finite set then the weak direct product of $\{ G_i : i \in I \}$ is the same thing as the direct product of $\{ G_i : i \in I \}$.*

We now look at some basic properties of the weak direct product of groups.

Proposition 1: Let $\{ G_i : i \in I \}$ be an arbitrary collection of groups. Then:a) $\prod_{i \in I}^{\mathrm{weak}} G_i$ is a subgroup of $\prod_{i \in I} G_i$.b) $\prod_{i \in I}^{\mathrm{weak}} G_i$ is normal in $\prod_{i \in I} G_i$. |

*Recall from the Criteria for a Subgroup to be Normal page that a subgroup $H$ of $G$ is a normal subgroup of $G$ if and only if $gHg^{-1} \subseteq H$ for all $g \in G$. We use this in the proof of (b) below.*

**Proof of a)**Let $f, g \in \prod_{i \in I}^{\mathrm{weak}} G_i$. Let $\tilde{F} = \{ i \in I : f(i) \neq e_{G_i} \}$ and let $\tilde{G} = \{ i \in I : g(i) \neq e_{G_i} \}$. By definition $\tilde{F}$ and $\tilde{G}$ are finite sets. Observe that if $i \not \in \tilde{F} \cup \tilde{G}$ then $(fg)(i) = e_{G_i}$. Thus if $\tilde{FG} = \{ i \in I : (fg)(i) \neq e_{G_i} \}$ we have that $|\tilde{FG}| \leq |\tilde{F}| + |\tilde{G}| < \infty$. So $fg \in \prod_{i \in I}^{\mathrm{weak}} G_i$, which shows that $\prod_{i \in I}^{\mathrm{weak}} G_i$ is closed under the operation of componentwise multiplication.

- Now clearly $e : I \to \bigcup_{i \in I} G_i$ defined for all $i \in I$ by $e(i) = e_{G_i}$ is such that $e \in \prod_{i \in I}^{\mathrm{weak}} G_i$.

- Lastly, if $f \in \prod_{i \in I}^{\mathrm{weak}} f(i)$ then $f(i) \neq e_{G_i}$ for all but finitely many $i \in I$. So $f^{-1}(i) \neq e_{G_i}$ if and only if $f(i) \neq e_{G_i}$. That is, $f^{-1}(i) \neq e_{G_i}$ for all but finitely many $i \in I$. So $f^{-1} \in \prod_{i \in I}^{\mathrm{weak}} G_i$.

- Thus $\prod_{i \in I}^{\mathrm{weak}} G_i$ is a subgroup of $\prod_{i \in I} G_i$. $\blacksquare$

**Proof of b)**Let $G = \prod_{i \in I} G_i$ and let $H = \prod_{i \in I}^{\mathrm{weak}} G_i$. We aim to show that for all $g \in G$ that $gHg^{-1} \subseteq H$.

- Let $g \in G$. Let $h \in H$. Then for each $i \in I$ we have that $(ghg^{-1})(i) = g(i)h(i)g^{-1}(i)$.

- Suppose that $h(i) = e_{G_i}$. Then:

\begin{align} \quad (ghg^{-1})(i) = g(i)h(i)g^{-1}(i) = g(i)e_{G_i}g^{-1}(i) = g(i)g^{-1}(i) = e_{G_i} \end{align}

- Since $h(i) = e_{G_i}$ for all but finitely many $i \in I$ we see that $(ghg^{-1})(i) = e_{G_i}$ for all but finitely many $i \in I$ from above. Therefore $ghg^{-1} \in H$. This shows us that $gHg^{-1} \subseteq H$ and so $H = \prod_{i \in I}^{\mathrm{weak}} G_i$ is a normal subgroup of $G = \prod_{i \in I} G_i$. $\blacksquare$