Counter example to "$(P_1...P_{r-1})\cap P_r = \{ e \}$ if $P_i \cap P_j = \{ e \}$ and $P_i$ is normal in $G$"

Question

Let $G$ be a group, and $P_1, ..., P_r$ its normal subgroups such that $i \neq j \Rightarrow P_i \cap P_j = \{ e \}$. I need to prove that then

$(P_1...P_{r-1})\cap P_r = \{ e \}$

Edit: the claim is wrong. So I edited the question(instead of showing a proof it now asks of finding a counterexample to the claim).

Answer 1

It is indeed wrong. Let $G=(\mathbb{Z}/2\mathbb{Z})^2$. Take $r=3$. Let $P_1$ be the subgroup generated by $(1,0)$, $P_2$ by $(0,1)$, and $P_3$ by $(1,1)$.