Finding $G$ with normal $H_1,\ldots,H_n$ such that $G=H_1\cdots H_n$ and $H_i\cap H_j=\{e\}$ for $i\neq j$, but $G\not\cong H_1\times\cdots\times H_n$ How can I find a group $G$ with normal subgroups $H_1,\ldots,H_n$ such that $G=H_1H_2\cdots H_n$ and $H_i\cap H_j=\{e\}$ for all $i\neq j$, but $G\not\cong H_1\times H_2\times\cdots\times H_n$. 
I'm thinking about $Q_8$. Is that right? Thank you very much!

$Q_{8}$ is not ok, as in it two non-trivial normal subgroups intersect nontrivially.
Take instead the Klein $4$-group $V = \{e, a_1, a_2, a_3\}$, and $H_i = \langle a_i \rangle = \{ 1, a_i \}$ for $i = 1, 2, 3$.
You have indeed $G = H_1 H_2 H_3$ (actually, two factors already suffice) and $H_i \cap H_j = \{ e \}$ for $i \ne j$, but $H_1 \times H_2 \times H_3$ has order $8$, not $4$ like $V$.
