# 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.

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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 $f(x)=a$.

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Thanks for the general proof, it was precisely what I needed. –  KevinDL Oct 9 '11 at 10:07
Dear @Kevin, you're welcome! –  Pierre-Yves Gaillard Oct 9 '11 at 10:35