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!

Honestly the last point in simple completing @Andreas's answer made me a hint as following: Take $G=\mathbb Z_2\times\mathbb Z_4$. there are 3 normal subgroups of $G$ as: $$N_1=\{(0,0),(1,0)\},~N_2=\{(0,0),(1,2)\},~N_3=\{(0,0),(0,2),(1,1),(1,3)\}$$ Some calculations will lead you that, $G\cong N_1N_2N_3$ but $G\ncong N_1\times N_2\times N_3$. Since $|G|=8$ but $|N_1\times N_2\times N_3|=16$.
