About -derived subgroup- of a group $G$

Question

This question is already solved, I am typing down the resolution from manual solutions. However there are some steps that I did not understand properly, could anyone help me?

P.S. My doubts are in ${\color{red}{\text{Color Red}}}$.

QUESTION: Let $G$ be a group and $\mathcal{H}$ be a family of normal subgroups $N$ of $G$ such that $G/N$ is abelian. Let $I=\displaystyle\bigcap_{n \in \mathcal{H}} N$ the intersection of subgroups of $G$ belonging to the family $\mathcal{H}$. Prove that $G/I$ is abelian.

$I$ is called by derived subgroup or commutative subgroup of $G$.

SOLUTION FROM MANUAL SOLUTIONS:

Let's consider the direct product $X=\displaystyle\Pi_{N \in \mathcal{H}} G/N$. We have the canonical homomorphism $\pi:G\rightarrow X$, given by $g\mapsto \displaystyle(gN)_{N\in \mathcal{H}}$, which the kernel is $I$, then by the isomorphism theorem we have $G/I$ is isomorph to a subgroup of $X$. However $X$ is abelian, and because this it is a direct product of abelian groups, it follows that $G/I$ is abelian too.

${\color{red}{\text{Why kernel$(\pi)=I$?}}}$

Answer 1

We have the following sequence of equivalent statements: $$\begin{align} g \in \ker \pi &\iff \forall N \in \mathcal H : gN = N\\ & \iff \forall N \in \mathcal H : g \in N \\ &\iff g \in \bigcap_{N \in \mathcal H}N. \end{align}$$

Answer 2

$gN=e\in G/N\iff gN=N\iff g\in N$.

But $g\in\operatorname{ker}\pi\iff gN=e\in G/N,\,\forall N\in\mathcal H$.

Thus $\operatorname{ker}\pi=\bigcap_{N\in\mathcal H}N$.