Example of a group $G$ such that $G = N_1\cdots N_n$ and $N_i \cap N_j = \{e\}$ for all $i \neq j$ but $G$ is not the internal direct product of them.

Question

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 $i \neq j$

but

(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....

Answer 1

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

Answer 2

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.