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.


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 :


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.

  • $\begingroup$ $G$ needn't be finite, right? Does $H_1\cap\ldots\cap H_n=H_1\cdots H_n$ hold? $\endgroup$ – Leon Feb 1 '16 at 17:42
  • 1
    $\begingroup$ @Leon, I have never used the fact that $G$ is finite, haven't I? What is true though, is that $[G:H]$ is finite, in general it is smaller than the product of $d_i$'s so it is. As for $H_1\cap\dots H_n=H_1\dots H_n$ it is not true. Take $G$ to be $S_3$ (the symmetric group), $H_1$ to be $\langle (1,2,3)\rangle$ and $H_2=\langle (1,2)\rangle$ then $[G:H_1]=2$ and $[G:H_2]=3$ but $H_1\cap H_2$ is trivial whereas $H_1H_2=G$. $\endgroup$ – Clément Guérin Feb 2 '16 at 7:32

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