I was wondering if the following statement is true:
Let $G$ be a group with normal subgroups $H_1,H_2,...H_n$. Suppose $H_iH_j=G$ for all $i\neq j$. Then $G/H_1\cap H_2...\cap H_n\cong G/H_1 \times...\times G/H_n.$
There is a similar question here, which is the case when $n=2$. There is also a similar question here, but the conditions involve the index of a subgroup. I want to get rid of such condition so that the conclusion is applicable to some other situations.
But when I tried to proceed proof by induction to get the conclusion with arbitrary $n$, things became not so approachable.
If you think this is false, please give a counter-example. If you think this is true, please share your ideas of the proof. Thank you!
This is false. For instance, let $G=(\mathbb{Z}/2\mathbb{Z})^2$ and let $H_1,H_2,H_3\subset G$ be the three subgroups of order $2$. Then $H_iH_j=G$ for all $i\neq j$ but $G/(H_1\cap H_2\cap H_3)\cong G$ has $4$ elements while $G/H_1\times G/H_2\times G/H_3\cong(\mathbb{Z}/2\mathbb{Z})^3$ has $8$ elements.