Properties of Restricted Direct Products (5.1.17)

Dummit and Foote Abstract Algebra, section 5.1, exercise 17:

Let $I$ be any nonempty index set and $G_i$ be a group for each $i∈I$. The restricted direct product or direct sum $R$ of the groups $G_i$ is the set of elements of the direct product that are the identity in all but finitely many components, i.e. are the nonidentity in finitely many coordinates, which is to say, the set of all elements $\prod a_i \in \prod_{i \in I} G_i$ such that $a_i = 1_i$ for all but a finite number of $i \in I$.
(a) Prove that $R ≤ \prod_{i \in I} G_i$.
(b) Prove that $R \trianglelefteq \prod_{i \in I} G_i$.
(c) Let $I = \mathbb{Z}^+$ and let $p_i$ be the $i^{th}$ prime. Show that if $G_i = \mathbb{Z}/p_i\mathbb{Z}$ for all $i \in \mathbb{Z}^+$, then every element of $R$ of $\prod_{i \in I} G_i$ has finite order, but the unabridged group has elements of infinite order as well. In this case, show that $Tor(\prod_{i \in I} G_i) = R$ (cf. 2.1.6).

Proof: (a) We have $\prod a_i\prod b_i = \prod a_i b_i$ has a nonidentity element in the $j^{th}$ coordinate only if either $\prod a_i$ or $\prod b_i$ has a nonidentity element in that coordinate, which limits the potential coordinates of their product for nonidentity elements to occur down to a finite number, and thus the product is within $R$. Furthermore, inverses clearly don't change nonidentity coordinates, the identity element is within $R$, and the associativity axiom is evident.

(b) For any $\prod c_i \in \prod_{i \in I} G_i$ and $\prod a_i \in R$, we have the $j^{th}$ coordinate of $\prod c_i\prod a_i\prod c_i^{-1}$ is equal to $1$ if $a_j = 1$. This implies that there is only a finite number of coordinates for nonidentity coordinates to occur in, and therefore $\prod c_i\prod a_i\prod c_i^{-1} \in R$ and $\prod c_i R \prod c_i^{-1} = R$ so that $R \trianglelefteq \prod_{i \in I} G_i$.

(c) Let $\prod a_i$ be the element such that $a_i = 1$ for all $i$. If $| \prod a_i | = n < \infty$, then observe the $(n+1)^{th}$ coordinate; we should have $a_{n+1} n ≡ n ≡ 0 \mod p_{n+1}$, even though $0 < n < p_{n+1}$. Therefore $| \prod a_i | = \infty $.

For any $\prod a_i \in R$, set $n = \prod_{i \in K} p_i$, where $K$ is the subset of $I$ such that $a_i$ is nonidentity. For any nonidentity coordinate, we have $a_m n ≡ 0 \mod p_m$. Clearly, it also fixes the identity coordinates. Therefore, $| \prod a_i | \mid n$, and in particular the order of every element is finite. This proves $R ≤ Tor(\prod_{i \in I} G_i)$.

Assume $\prod x_i \in Tor(\prod_{i \in I} G_i)$ such that $\prod x_i$ has infinitely many nonidentity coordinates. Observe its order $n$ and the $(n+1)^{th}$ nonidentity coordinate. We have $x_{m} n ≡ 0 \mod p_{m}$ with $x_{m} \not ≡ 0$, and we can see $0 < n < p_{n+1} ≤ p_{m}$. Therefore, $p_{m} \nmid x_{m}$ and $p_{m} \nmid n$, so that $p_{m} \nmid x_{m} n$, i.e. $x_{m} n \not ≡ 0 \mod p_{m}$, a contradiction. Therefore, $Tor(\prod_{i \in I} G_i) ≤ R$ and now $Tor(\prod_{i \in I} G_i) = R$. $\square$