normal groups of a infinite product of groups I have a question regarding the quotient of a infinite product of groups.
Suppose $(G_{i})_{i \in I}$ are abelian groups with $|I|$ infinite and each $G_i$ has a normal subgroup $N_i$. Is it true in general that
$$\prod_{i \in I} G_i/ \prod_{i \in I} N_i \cong \prod_{i\in I} G_i/N_i$$
More specifically, is it true that 
$$\prod_{p_i \text{prime}} \mathbb{Z}_{p_i} / \prod_{p_i \text{prime}} p^{e_i}\mathbb{Z}_{p_{i}} \cong \prod_{p_i \text{prime},e_{i} \leq \infty} \mathbb{Z}/p_{i}^{e_i}\mathbb{Z} \times \prod_{p_i \text{prime},e_{i} = \infty}\mathbb{Z}_{p_i}$$
where $\mathbb{Z}_{p_i}$ stands for the $p_i$-adic integers and $p_i^\infty \mathbb{Z}_{p_i}=0$ and all $e_i$ belong to $\mathbb{N} \cup \{\infty\}$.
Any help would be appreciated.
 A: Here is a slightly more general statement. 
Let $(X_i)$ be a family of sets, and $X$ its product. For each $i$ let $E_i\subset X_i^2$ be an equivalence relation. Write $x\ \sim_i\ y$ for $(x,y)\in E_i$. Let $E$ be the product of the $E_i$. There is a canonical bijection between $X^2$ and the product of the $X_i^2$. Thus $E$ can be viewed as a subset of $X^2$. Write $x\sim y$ for $(x,y)\in E$. Let $x,y$ be in $X$. The followong is clear:
Lemma 1. We have $x\sim y\ \Leftrightarrow\ x_i\ \sim_i\ y_i\ \forall\ i$. 
In particular $\sim$ is an equivalence relation on $X$. 
Define $f_i:X\to X_i/E_i$ by mapping $x$ to the canonical image of $x_i$. Let 
$$ 
f:X\to\prod\ (X_i/E_i) 
$$ 
be the map attached to the family $(f_i)$. 
CLAIM. The map $f$ induces a bijection $g$ from $X/E$ to $\prod(X_i/E_i)$. 
Let $x,y$ be in $X$. 
Lemma 2. We have  $x\sim y\ \Leftrightarrow\ f(x)=f(y)$. 
Proof: This follows from Lemma 1. 
Conclusion: The map $f$ induces an injection $g$ from $X/E$ to $\prod(X_i/E_i)$. 
It only remains to prove that $g$ is surjective. To do this, let $a$ be in $\prod(X_i/E_i)$. For each $i$ choose a representative $x_i\in X_i$ of $a_i$, put $x:=(x_i)$, and check the equality $f(x)=a$. 
