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!

Under additional assumptions it becomes true:
If $i \neq j$, then the elements of $H_i$ and $H_j$ commute with each other since $x \in H_i, y \in H_j$ implies $xyx^{-1} y^{-1} \in H_i \cap H_j = \{1\}$. It follows that the multiplication map $\omega : H_1 \times \dotsc \times H_n \to G$ is a homomorphism, and by assumption it is surjective.
If $G$ is finite and $|G|= \prod_i |H_i|$, then $\omega$ is an isomorphism. This can be easily checked in practice. For example, it happens to be the case when $G$ is nilpotent and $H_p$ is the $p$-Sylow subgroup of $G$ for all primes $p$ dividing $|G|$. This then shows that every nilpotent group is a finite direct product of $p$-groups.
In general, the map is an isomorphism if and only if $H_i \cap \prod_{j \neq i} H_j = \{1\}$ for all $1 \leq i \leq n$. This is the usual characterization of internal direct products. In particular, if $G$ is finite, it suffices to demand that the orders of the $H_i$ are coprime to each other.
