Ytukyg

I am studying Internal Direct Products in Group Theory.

$mbox{Let}$ $G$ be a group and $H_{i}lhd G; forall i=1,2ldots
> n$

Suppose ${prod_{i=1}^{r} H_{i}} bigcap H_{r+1} = {e} ; forall
> r=1,2ldots (n-1)$ then prove that $H_{i}bigcap H_{j} = {e} ;
forall ineq j$

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 $ain H_{i}bigcap H_{j},; ; aneq e, ; ; f.s. ; ineq j$

$Rightarrow ain H_{i}$ and $a in H_{j}$

WLOG, assume $i < j $

Clearly, $a in prod_{i=1}^{j-1}H_{i}$ since we can write $a=e.eldots a.e.dots e$
$Rightarrow {prod_{i=1}^{j-1}H_{i} } bigcap H_{j} neq {e} ; ; mbox{contradiction}$

Converse is obviously true for $n=2$ but how can I show it is not true for $n>2$?

Counter example :
Group $Z_3times Z_3$ under +

$N_1=${$(0,0),(1,0),(2,0)$}

$N_2=${$(0,0)(0,1)(0,2)$}

$N_3=${$(0,0)(1,1)(2,2)$}

$N_1N_2cap N_3=${$(0,0),(1,1)(2,2)$}