Exercise 26, § 2

Let $(G_i)_{i \in I}$ be a family of topological groups, and for each $i \in I$ let $K_i$ be an open normal subgroup of $G_i$. Let $\mathfrak{B}$ denote the neighbourhood filter of the identity element in the product topological group $K = \prod_{i \in I} K_i$. Show that in the product group $G = \prod_{i \in I} G_i$, $\mathfrak{B}$ is a base of the neighbourhood filter of $e$ for a topology compatible with the group structure of $G$. The group $G$, endowed with this topology, is called the local product of the $G_i$ (relative to the $K_i$); $K$ is then an open normal subgroup of $G$, and $G/K$ is isomorphic to the (discrete) product of the groups $G_i/K_i$. If, for each $i \in I$, $K'_i$ is another open normal subgroup of $G_i$, then the local product topologies on $G$, defined by the families $(K_i)$ and $(K'_i)$, coincide if and only if $K'_i = K_i$ for all but a finite number of indices.

For each $x \in I$, let $G'_k$ be the normal subgroup of $G$ consisting of all $x = (x_i)$ such that $x_i = e_i$ whenever $i \neq x$. Then $G'_x$, with the topology induced by a local product topology on $G$, is canonically isomorphic to $G_x$. The closure in $G$ of the subgroup generated by the $G'_i$ is the normal subgroup $G_0$ of $G$ which consists of all $x = (x_i)$ such that $x_i \in K_i$ for all but a finite number of indices. $G_0$ is said to be the local direct product of the $G_i$ (relative to the $K'$); $G_0/K$ is then a discrete group isomorphic to the direct product of the $G_i/K_i$.