The following is taken from Hungerford's undergraduate Abstract Algebra An Introduction text
(1) Let $I$ be the set of positive integers and assume that for each $i\in I, G_i$ is a group. The infinite direct product of the $G_i$ is denoted $\prod_{i\in I}G_i$ and consists of all sequences $(a_1,a_2,\ldots)$ with $a_i\in G_i.$ Prove that $\prod_{i\in I}G_i$ is a group under the coordinatewise operation $$(a_1,a_2,\ldots)(b_1,b_2,\ldots)=(a_1b_1,a_2a_2,\ldots)$$
Exercise 1: With the notation as in (1), let $\sum_{i\in I}G_i$ denote the subset of $\prod_{i\in I}G_i$ consisting of all sequences $(c_1,c_2,\ldots)$ such that there are at most a finite number of coordinates with $c_j\neq e_j,$ where $e_j$ is the identity element of $G_j.$ Prove that $\sum_{i\in I}G_i$ is a normal subgroup of $\prod_{i\in I}G_i.$ $\sum_{i\in I}G_i$ is called the infinite direct sum of the $G_i.$
Proof for Exercise 1:
Proof: We assumed the notation of $\prod_{i\in I}G_i$ in (1). For any $c\in \sum_{i\in I}G_i,$ let $c=(c_i)_{i\in I}=(c_1,c_2,\ldots)$ such that only finitely many of the $c_j\neq e_j$ where $e_j$ is an identity element of each group $G_j$. For any $a\in \prod_{i\in I}G_i,$ where $a=(a_i)_{i\in I}=(a_1,a_2,\ldots).$ since each $G_i$ is a group, and if $a_i\in G_i,$ ${a_i}^{-1}\in G_i,$ and $a^{-1}=({a_i}^{-1})_{i\in I}=({a_1}^{-1},{a_2}^{-1},\ldots)\in \prod_{i\in I}G_i.$ Finally, if $b\in \sum_{i\in I}G_i,$ then $b\in \prod_{i\in I}G_i,$ since $\sum_{i\in I}G_i\subset \prod_{i\in I}G_i.$ After having defined our notations, we proceed with our proof. So suppose for any $a, a^{-1}\in \prod_{i\in I}G_i,$ and any $c\in \sum_{i\in I}G_i,$ then $aca^{-1}=(a_ic_i{a_i}^{-1})_{i\in I}.$ Since only at most finitely many of the $c_j\neq e_j,$ then for the coordinates of each of $a_ic_i{a_i}^{-1}$ in $aca^{-1},$ $a_ic_i{a_i}^{-1}=e_i,$ if $c_i=e_i,$, else $a_ic_i{a_i}^{-1}\in \sum_{i\in I}G_i$. Hence $a_ic_i{a_i}^{-1}\in a\sum_{i\in I}G_ia^{-1},$ and $\sum_{i\in I}G_i$ is a normal subgroup of $\prod_{i\in I}G_i.$
Can someone check if my proof is correct please.
Thank you in advance.
For $a\in\prod$, call: $$J(a):=\{i\in I\mid a_i\ne e_i\}$$ Both subgroup and normality verifications for $\sum$ proceed from noting that: $J(a^{-1})=J(a)$, $J(ab)\subseteq J(a)\cup J(b)$ and $\sum:=\{a\in\prod\mid J(a)\text{ is finite}\}$.