The following is taken from Hungerford's undergraduate Abstract Algebra An Introduction text
Lemma 1: Let $M$ and $N$ be normal subgroup of a group $G$ such that $M\cap N=\langle e\rangle$. If $a\in M$ and $b\in M,$ then $ab=ba.$
Theorem 2: Let $N_1,N_2,\ldots,N_k$ be normal subgroups of a group $G$ such that every element in $G$ can be written uniquely in the form $a_1a_2\cdots a_k,$ with $a_i\in N_i.$ Then $G$ is isomorphic to the direct product $N_1\times N_2\times\cdots N_k.$
(3) 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 $$(a_1,a_2,\ldots)(b_1,b_2,\ldots)=(a_1b_1,a_2a_2,\ldots)$$
(Assumed 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 $G_i.$
Exercise 2: Let $G$ be a group and assume that for each positive integer $i, N_i$ is a normal subgroup of $G.$ If every element of $G$ can be written uniquely in the form $n_{i_1}\cdot n_{i_2}\cdots n_{i_k}$ with $i_1<i_2<\cdots <i_k$ and $n_{i_j}\in N_{i_j}.$ Prove that $G\equiv \sum_{i\in I}N_i.$ [Hint: Adapt the proof of Theorem 2 by defining $f(a_1,a_2,\ldots)$ to be the product of those $a_i$ that are not the identity element.]
Questions:
I want to proof exercise 2. But I am suppose to adapt the proof of Theorem 2. In the text, proof of theorem 2 make use of Lemma 1. I have make a post on its proof for the infinite direct sum of groups version. I would like some clarification for the notation used for part 2 of exercise 2. I am not sure how it differs from exercise 2 in the context of having to assume the notations from (Assumed Exercise 1). Here are other things i don't understand about the notations.
(i) I don't understand why the indices $i_k$-s are used. I mean how is it different than how Theorem 2 is formulated. Am I to assume that (a) the elements of $a_k$ will stand in for $n_{i_k}$ and (b) only finitely many of the $n_{i_k}$ will not equal to $e_k$ where each $e_k$ is the identity element of $G_i.$
(ii) Is the map I have to defined suppose to be:
$f:\sum_{i\in I}G_i\rightarrow G$ is $f(a_1,a_2,\ldots e_{k}, e_{k+1},\ldots)=a_1\cdot a_2,\cdots a_k$ where $a_k=n_{i_k},$ because I am not sure if I am suppose to interpret the meaning of $\sum_{i\in I}N_i$ to be the same as $\sum_{i\in I}G_i,$ with each $N_i$ being normal subgroups of $G_i.$
Thank you in advance
(i) If it was not guaranteed that $i_1 < i_2 < i_3 < ...$, then any every element in $G$ could be written as $n_{i_1}n_{i_2}n_{i_3}...$ or $n_{i_2}n_{i_1}n_{i_3}...$ or $n_{i_3}n_{i_2}n_{i_1}...$ or any other permutation of the $n's$ because all the $n's$ can commute with each other.
Imagine that $G$ was the set of positive integers and $n_1, n_2, n_3, ...$ are all the prime numbers. Of course, any positive integer $g \in G$ except for $1$ can be written unique product of prime powers $p_1^{k_1}$, $p_2^{k_2}$, $p_3^{k_3}$, where $p_1 < p_2 < p_3$. If I were to drop the ordering requirement, then it would make little sense to say that $30 = 2\cdot 3 \cdot 5$ is a unique factorization since $30 = 3\cdot 2 \cdot 5 = 5\cdot 2\cdot 3 = ...$.
The reason you don't see this ordering in theorem 2 is that the "indices" are 1, 2, 3, ... and their order is implied.
(ii) Yes, you are supposed to interpret $\sum$ the same way (for the rest of the book) unless otherwise specified.
Does this answer all your questions?