An Analogue of Chinese Remainder Theorem for Groups I am trying to prove the following analogue of Chinese remainder theorem for groups:
Let $G$ be group and let $H_1, \dots, H_n$ be its normal subgroups such that their indices $[G : H_1], \dots, [G : H_n]$ are pairwise coprime. Then we have $$G/(H_1 \cap \cdots \cap H_n) \cong G/H_1 \times \cdots \times G/H_n.$$
I think that a good strategy would be to try to prove that the mapping $\phi$ defined by $$\phi(g(H_1 \cap \cdots \cap H_n)) = (gH_1, \dots, gH_n)$$ is an isomorphism, but I am not sure how to do this.
 A: The strategy you have is not the most rigorous one (why is $\phi$ well defined ?  why is $\phi$ a group morphism ?). I would suggest to begin with :
$$\psi:G\rightarrow G/H_1\times \dots\times G/H_n $$ 
$$g\mapsto (gH_1,\dots, gH_n) $$
This is a group morphism, that will factor by $H_1\cap\dots \cap H_n$ and this will prove at the same time (and rigorously) that $\phi$ is well defined, a group morphism and one-to-one.
Clearly $Ker(\psi)=H_1\cap\dots\cap H_n$. Indeed, $\psi(g)$ is trivial iff for all $1\leq i\leq n$ we have $gH_i=H_i$ iff for all $1\leq i\leq n$ $g\in H_i$. 
Hence $\psi$ factors through $H_1\cap\dots\cap H_n$ by :
$$\phi:G/H_1\cap\dots \cap H_n\rightarrow G/H_1\times \dots\times G/H_n $$ 
$$g H_1\cap \dots \cap H_n\mapsto (gH_1,\dots, gH_n) $$
Remark here that I did not use the fact that the indices are coprime to each other. Of course, we will use it to show that $\phi$ is onto. Denote $d_i:=[G:H_i]$. Denote $H:=H_1\cap\dots \cap H_n$.
Remark that :
$$[G:H]=[G:H_i][H_i:H]=d_i[H_i:H]$$
Hence $d_i$ divides $[G:H]$ for all $i$, since they are pairwise coprime, their product also divides $[G:H]$. Denoting $d:=d_1\dots d_n$ we get that :
$$d\text{ divides } |G/H| $$
But $G/H$ is isomorphic to $Im(\phi)$ included in $G/H_1\times\dots G/H_n$ which is of cardinal $d_1\times\dots\times d_n=d$ so $|Im(\phi)|=d$ and $\phi$ is onto. 
