I am having difficulty following the proof of the following proposition proved by Bourbaki in his Algebra book:
Let $G$ be a group with operators in $\Omega$ and $(H_{i})_{i\in I}$ a finite family of stable subgroups of $G$. For $G$ to be the restricted sum of the family of subgroups $(H_{i})$, it is necessary and sufficient that each $H_{i}$ be normal and that $G$ be the product of the quotient groups $(G/H^{i})$, where $H^{i}$ is the subgroup generated by the $H_{j}$ for $j\neq i$.
Proof. The condition is obviously necessary. Conversely, suppose $G$ is the product of the $K_{i} = G/H^{i}$ and let let $G$ be identified with the product of the $K_{i}$. Then $H_{i}$ is identified with a subgroup of $K_{i}$, so that, for $j\neq i$, every element of $H_{i}$ is permutable with every element of $H_{j}$; on the other hand, $H^{i}$ is identified with the product of the $K_{j}$ for $j\neq i$, hence $H_{i}=K_{i}$ for all $i$ and $G$ is the direct product of the $H_{i}$.
(Algebra, Chapter 1, $\S \ 4$, no. $9$)
First of all, I cannot see why the condition is "obviously" necessary. For each $i\in I$, let
$$\iota_{i}:H_{i}\rightarrow \bigoplus H_{j}$$
be the injective homomorphism such that $\iota_{i}(x)(i)=x$ and $\iota_{i}(x)(j)=e$ for all $j\neq i$. Letting $incl_{i}$ denote the canonical injection of $H_{i}$ into $G$, we have $\Phi\circ\iota_{i}=incl_{i}$ by definition, where
$$\Phi: \bigoplus H_{i}\rightarrow G,\ (x_{i})\mapsto \Pi x_{i}$$
is the isomorphism whose existence we have assumed. Since $\iota_{i}(H_{i})\vartriangleleft \prod H_{i}$ and $\Phi$ is a homomorphism, we have $H_{i}\vartriangleleft G$ for every $i\in I$. Therefore, $(H_{i})_{i\in I}$ is a family of normal subgroups of G; whence, $\bigvee_{j\neq i} H_{j}=H^{i}\vartriangleleft G$ for each $i\in I$.
Ok, so far so good.
Now, I am not sure how to prove that
$$\Psi: G\rightarrow \prod G/H^{i},\ g\mapsto(p_{i}(g))_{i\in I}$$
is a bijection.
On the other hand, I also cannot follow Bourbaki's proof of the sufficiency of this condition: specifically, I am not sure how to construct the following identification
$$H^{i}\simeq \prod_{j\neq i} G/H^{j}.$$
I know that if $I$ is finite, we have $\bigvee \iota_{i}(H_{i})=\prod H_{i}$; hence, $$H^{i}=\bigvee_{j\neq i} \iota_{j}(H_{j})=\prod_{j\neq i} H_{j}\simeq\{x\in\prod H_{j}\ |\ x_{i}=e\},$$ and similarly we have $$H_{i}\simeq(\prod H_{j})\ / \{x\in\prod H_{j}\ |\ x_{i}=e\}.$$
But can these properties be used to prove the result?—I have no idea how to utilise the quotient groups $G/H^{i}$; it seems to me that this might be what's causing me so much trouble—but I am not sure.
I would appreciate any hints.
Some definitions from Bourbaki:
1) Let $G$ be a group with operators in $\Omega$ and $(H_{i})_{i\in I}$ a family of stable subgroups of $G$. $G$ is called the internal restricted sum (or restricted sum) of the family of the subgroups $(H_{i})$ if every element of $H_{i}$ is permutable with every element of $H_{j}$ for $j\neq i$ and the unique homomorphism of $\bigoplus H_{i}$ into G whose restriction to each of the $H_{i}$ is the canonical injection is an isomorphism.
2) Let $G$ be a group with operators in $\Omega$ and $(H_{i})_{i\in I}$ a family of normal stable subgroups of $G$. Let $p_{i}:G\rightarrow G/H_{i}$ be the canonical homomorphism. G is called the internal product (or product) of the family of quotient groups $(G/H_{i})$ if the homomorphism $g\mapsto(p_{i}(g))_{i\in I}$ is an isomorphism of G onto $\prod G/H_{i}$.