Suppose $N_1,\ldots, N_m$ be sequence of normal subgroups of a group $G$. For each $1\leq i\leq m$, write $K_i=\langle\bigcup_{j\ne i} N_j\rangle$.
Suppose $\varphi:G\rightarrow G/K_1\times\ldots\times G/K_m, g\mapsto(gK_1,\ldots,gK_m)$, is an isomorphism.
The mapping $\phi:N_1\times\ldots\times N_m\rightarrow G, (g_1,\ldots,g_m)\mapsto g_1\ldots g_m$, is an isomorphism. Moreover, elements of $N_i$ and $N_j$ are permutable for every $i\ne j$.
I am having difficulty proving that $\phi$ is surjective. Is this actually true?
P.S: I don't understand the use of the groups $K_i$. Why not work directly with $N_i$? What is gained by using the $K_i$ in this specific situation?
Proof of surjectivity:
Let $g\in G$. Then $(gK_1,\ldots,gK_m)\in G/K_1\times\ldots\times G/K_m$. Then $$(gK_1,\ldots,gK_m)=(gK_1,\ldots,K_m)\cdot\ldots\cdot(K_1,\ldots,gK_m)$$
Thus there exist $a_i\in G$ such that $\varphi(a_i)=(\ldots,gK_i,\ldots)$ for each $i$ and $a_iK_j=K_j$ for $i\ne j$.
Where can I go from here?
I guess my idea is to show that $g=a_1...a_m$. However, I am not able to prove that $a_i\in N_i$.