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I have recently been wondering about the group-theoretic Chinese remainder theorem. In particular, I was wondering whether this analogue can be generated to the case of more than two subgroups:

Showing that $G/(H\cap K)\cong (G/H)\times (G/K)$

An answer to following question demonstrates that it cannot, by giving an example where $G=(\mathbb Z/2\mathbb Z)^2$:

Validation for a conjecture about Chinese Remainder Theorem for groups

However, I was wondering whether anyone had a counterexample to this in an infinite group where we restrict the normal subgroups to having finite index. In particular, what I would like is an example of an infinite group $G$ and normal subgroups $L,M,N\triangleleft G$ all having finite index and satisfying $LM=LN=MN=G$ but $L(M\cap N)\neq G$ (or a proof that no such thing is possible). Can something be said of groups in which this cannot occur (e.g. $\mathbb Z$)?

Compare also the following question, in which it is shown that the Chinese remainder theorem still works if we take the indices to be coprime:

An Analogue of Chinese Remainder Theorem for Groups

jakehuryn
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  • Just take your counterexample with $G$ finite, and take the direct product of all groups/subgroups with a fixed infinite group. – YCor Oct 13 '19 at 19:05
  • You have an example with the Klein $4$-group, so just find an infinite group that projects onto the Klein $4$-group and lift the examples. – Arturo Magidin Oct 13 '19 at 19:06
  • One way to do this is what YCor suggest; e.g., $C_2\times C_2\times \mathbb{Z}$, and take your subgroups to be $C_2\times{e}\times \mathbb{Z}$, ${e}\times C_2\times \mathbb{Z}$, and $\langle (x,x)\rangle\times \mathbb{Z}_2$. Another is to take $\mathbb{Z}\times\mathbb{Z}$, and take the relevant overgroups of $2\mathbb{Z}\times 2\mathbb{Z}$. – Arturo Magidin Oct 13 '19 at 19:07
  • Thanks all three for your great responses! – jakehuryn Oct 14 '19 at 00:04

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