# 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!

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What are the $H_i$ in your example? – azimut Mar 6 '13 at 8:59

$Q_{8}$ 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 $V$.

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Thanks, so it was easy...sorry i thought i would find it difficult. – Basil R Mar 6 '13 at 9:03
@BasilR, you're welcome! I think the exercise was given so that you understand why the intersection condition for a direct product is $H_i \cap H_1 \cdots H_{i-1} H_{i+1} \cdots H_{n} = \{ e \}$ for all $i$. Hence the three factors in this example. – Andreas Caranti Mar 6 '13 at 9:06
Oh, I see that. – Basil R Mar 6 '13 at 9:15

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 $|N_1\times N_2\times N_3|=16$.

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Thank you for the hint. – Basil R Mar 6 '13 at 9:16
$$\;+1\quad \ddot\smile$$ – amWhy Mar 6 '13 at 15:50
I slept...beautiful slumber! How are you this day? – amWhy Mar 6 '13 at 15:57
@amWhy: It was good. Al the time at home. Thanks God – Babak S. Mar 6 '13 at 16:00

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.

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+ Nice point of views here. – Babak S. Mar 6 '13 at 14:38