I was trying to find an example of a group $G$ and normal subgroups $N_1 , \ldots , N_n$ such that $G = N_1\cdots N_n$ and $N_i \cap N_j = \{e\}$ for all $i \neq j$ and yet $G$ is not the internal direct product of $N_1, \ldots , N_n$ .

It is easy to show that G is the internal direct product of normal subgroups $N_1 , \ldots , N_n$ iff

(i) $G = N_1\cdots N_n$

(ii) $N_i\cap (N_1 \cdots N_{i-1}\cdots N_{i+1} \cdots N_n) = \{e\}$ for $i = 1, 2, \ldots, n$.

So I guess the question boils down to produce an example of a group $G$ and its normal subgroups $N_1 , \ldots , N_n$ such that

(1) $G = N_1\cdots N_n$

(2) $N_i \cap N_j = \{e\}$ for all $i \neq j$


(3) $N_i \cap (N_1 \cdots N_{i-1} N_{i+1} \cdots N_n) \neq \{e\}$ for at least one $1 \le i \le n.$ Thanks in advance for help....


Try the Klein 4-group $\mathbb{Z}_2 \times \mathbb{Z}_2=\{e,a,b,c\},$ and $N_1=\{e,a\},$ $N_2=\{e,b\},$ $N_3=\{e,c\}.$

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    $\begingroup$ I think you should ask that as a different question. $\endgroup$ – positrón0802 Jul 23 '17 at 7:13
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    $\begingroup$ It is not usually a good idea to alter a correctly answered question. I could answer your edited question but then perhaps you would change it again and we would go on forever. In any case you should spend some time trying to answer your new question yourself. $\endgroup$ – Derek Holt Jul 23 '17 at 8:25

Let $H$ be a group with nontrivial center $Z$ and identity element $e$. Let $G=\{(a,b)\in H^2\mid a\equiv b\pmod{Z}\}$. Let $D = \{(h,h)\in H^2\mid h\in H\}$ be the diagonal subgroup of $H^2$.

The following are normal subgroups of $G$: $N_1=Z\times \{e\}$, $N_2 = D$, and $N_3=\{e\}\times Z$. For these choices, $N_i\cap N_j=\{(e,e)\}$ for $i\neq j$, and $N_1N_2N_3=G$. Note that $N_1\cap (N_2N_3)=N_1\neq \{(e,e)\}$.

Since $H\cong D\leq G\leq H^2$, the group $G$ is abelian iff $H$ is abelian, so one can build nonabelian examples by starting with a nonabelian group $H$ with nontrivial center.

  • $\begingroup$ what does it means $a\equiv b\mod Z$ outside the modular arithmetic? $\endgroup$ – Aaron Lenz Mar 28 at 21:32
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    $\begingroup$ @AaronLenz: If $E$ is an equivalence relation on $A$, and $a, b\in A$, then $a\equiv b\pmod{E}$ means that $a$ and $b$ are $E$-equivalent (i.e., lie in the same $E$-class). If $N\lhd G$, then $a\equiv b\pmod{N}$ means $a$ and $b$ lie in the same coset of $N$. $\endgroup$ – Keith Kearnes Mar 29 at 2:12

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