I am studying Internal Direct Products in Group Theory.
$\mbox{Let}$ $G$ be a group and $H_{i}\lhd G\; \forall i=1,2\ldots > n$
Suppose $\{\prod_{i=1}^{r} H_{i}\} \bigcap H_{r+1} = \{e\} \; \forall > r=1,2\ldots (n-1)$ then prove that $H_{i}\bigcap H_{j} = \{e\} \; \forall i\neq j$
But converse is true only if $n=2$
First of all, I am really not sure about the validity of these statements since I am reading from a not-so-standard material.
I tried to prove $(\Rightarrow)$ like this$:$
Let $a\in H_{i}\bigcap H_{j},\; \; a\neq e, \; \; f.s. \; i\neq j$
$\Rightarrow a\in H_{i}$ and $a \in H_{j}$
WLOG, assume $i < j $
Clearly, $a \in \prod_{i=1}^{j-1}H_{i}$ since we can write $a=e.e\ldots a.e.\dots e$ $\Rightarrow \{\prod_{i=1}^{j-1}H_{i} \} \bigcap H_{j} \neq \{e\} \; \; \mbox{contradiction}$
Converse is obviously true for $n=2$ but how can I show it is not true for $n>2$?
Counter example : Group $Z_3\times Z_3$ under +
$N_1=${$(0,0),(1,0),(2,0)$}
$N_2=${$(0,0)(0,1)(0,2)$}
$N_3=${$(0,0)(1,1)(2,2)$}
$N_1N_2\cap N_3=${$(0,0),(1,1)(2,2)$}