I have a doubt in the following proof from Kurzweil and Stellmacher (transcribed below):
Why is every normal subgroup of $E_1$ also normal in $N$?
My attempt: For simplicity, assume $n=2$ and let $H\trianglelefteq E_1$. Let $x\in N$ and $h\in H$. We need to show $x^{-1}hx\in H$. Since $x\in N$, we have $e_i\in E_i$ such that $x=e_1e_2$. Thus, $x^{-1}hx= e_2^{-1} ke_2$ for $k=e_1^{-1}he_1\in H$.
How to show that $e_2^{-1} ke_2\in H$?
Transcribed text: Let $N$ be a minimal normal subgroup of $G$ and $E$ be a minimal normal subgroup of $N$, and assume that $M=\{E^g:g\in G\}$ is finite. Then $E$ is simple and $N=E_1\times\ldots \times E_n$ for some $E_1,\ldots,E_n\in M$.
Proof. The subgroup $\prod_{g\in G}E^g$ is normal in $G$ and thus equal to $N$. Hence, $N=E_1\times\ldots \times E_n$. Every normal subgroup of $E_1$ also is a normal subgroup of $N$. This shows that $E_1$ is simple, and then $E$ is simple as $E\cong E_1$.
Note that $N=E_1\times\cdots\times E_n$. In particular, if $x\in E_i$, $y\in E_j$, and $i\neq j$, then $xy=yx$. Thus, if $H\triangleleft E_1$, $h\in H$, and $n=x_1\cdots x_n\in N$, with $x_i\in E_i$, then $h$ commutes with $x_2,\ldots,x_n$, so $$\begin{align*} nhn^{-1} &= (x_1\cdots x_n)h(x_n^{-1}\cdots x_1^{-1})\\ &=x_1h(x_2\cdots x_n)(x_n^{-1}\cdots x_2^{-1})x_1^{-1}\\ &= x_1hx_1^{-1}, \end{align*}$$ but by assumption $x_1hx_1^{-1}\in H$, since $H\triangleleft E_1$. Thus, $H\triangleleft N$, as claimed.
In general, if $G=K_1\times\cdots \times K_n$ and $N\triangleleft K_i$, then $N\triangleleft G$, by the same argument.