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Let $G$ be a finite group and $N_1,\ldots,N_n$ are normal subgroups of $G$ such that $G = N_1,\ldots,N_n$ and $o(G) = o(N_1)\cdots o(N_n)$. Then $G$ is the internal direct product of $N_1,\ldots,N_n$.

I have tried:

We need to show that $N_i \cap N_1\cdots N_{i-1} N_{i+1}\cdots N_n = \{e\}$ for all $1 \leq i \leq n$

if there is some i such that $N_i \cap N_1\cdots N_{i-1} N_{i+1}\cdots N_n \neq \{e\}$ , then

$o(G) = o(N_1N_2\cdots N_n) = \frac{o(N_i) o( N_1\cdots N_{i-1} N_{i+1}\cdots N_n )}{o(N_i \cap N_1\cdots N_{i-1} N_{i+1}\cdots N_n )}$ = $o(N_1)\cdots o(N_n)$

What more needs to be done?

Thank you.

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    $\begingroup$ Do some induction. $\endgroup$ – Pedro Tamaroff Mar 20 '14 at 5:12
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May be you can first of all go for small $n$

$$|N_1||N_2||N_3|=|G|=|N_1N_2N_3|=\frac{|N_1||N_2N_3|}{|N_1\cap N_2N_3|}$$

$$\Rightarrow |N_2||N_3|= \frac{|N_2N_3|}{|N_1\cap N_2N_3|}= \frac{|N_2||N_3|}{|N_1\cap N_2N_3||N_2\cap N_3|}$$

$$\Rightarrow |N_1\cap N_2N_3||N_2\cap N_3|=1$$

What does this tell you?

Can you generalize this?

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