G does not have to be isomorphic to direct product of trivially intersecting normal subgroups of G

Question

From Dan Saracino, abstract algebra (Exercise 14.8)

Let $G_1,...,G_n$ be subgroups of G such that:

i) $G_1,...,G_n$ are all normal;

ii) $G = G_1.G_2...G_n$, that is, every element of G can be written as $g_l.g_2...g_n$, with $ g_i \in G_i$;

iii) for $1\leq i\leq n$, $G_i\cap G_1 G_2 . . G_{i-1}=\{e\}$.

Show that $G\cong G_1 \times G_2 \times \dots \times G_n$

I have proved the above, but I am not able to find a counterexample for the below statement.

Show, by an example, that if we replace (iii) by the weaker condition $G_i\cap G_j = \{e\}$ for $i \neq j$, then G does not have to be isomorphic to $\cong G_1 \times G_2 \times \dots \times G_n$

Answer 1

The group $C_2\times C_2$ contains three proper nontrivial subgroups, any two of which intersect trivially. But the direct product of all of them has order $8$.