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Show that if $G$ is the internal direct product of $H_1,H_2,\dots ,H_n$ and $i\neq j$ with $1\leq i\leq n,1\leq j\leq n$, then $H_i\cap H_j=\{e\}$.

The definition that I follow is as follows:

Let $H_1,H_2,\dots,H_n$ be a finite collection of normal subgroups of $G$. We say that $G$ is the internal direct product of $H_1,H_2,\dots,H_n$ and write $G=H_1\times H_2\times \dots\times H_n,$ if

  1. $G=H_1H_2\dotsm H_n=\{h_1h_2\dotsm h_n\mid h_i\in H_i\}$
  2. $H_1H_2\dotsm H_i\cap H_{i+1}=\{e\}$, for $i=1,2,\dots,n-1$
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    $\begingroup$ The spacing comes out better if you use \mid instead of | and \ldots instead of periods. $\endgroup$ – joriki May 30 '16 at 9:35
  • $\begingroup$ In fact, $\cdots$ ($\text{\\cdots}$) is (strongly) recommended when indicating operations rather than listing elements. If you're lazy to remember, use $\text{\\dots}$, which smartly inserts $\ldots$ or $\cdots$ based on context. $\endgroup$ – M. Vinay May 30 '16 at 9:38
  • $\begingroup$ @M.Vinay In this case, I suggest \dotsm (dots for multiplication); just \dots is needed between \times tokens (this is a multiplication). $\endgroup$ – egreg May 30 '16 at 9:53
  • $\begingroup$ @egreg I don't think I understand what you're saying. $\endgroup$ – M. Vinay May 30 '16 at 9:56
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Hint: If $x \in H_i \cap H_j$ ($i < j$), then $x \in H_i \implies x \in H_1 H_2 \cdots H_{j-1}$.

Solution:

Let $x \in H_i \cap H_j$, for any $i \ne j$. Without loss of generality, let $i < j$.

Then, $x \in H_i \implies x \in H_1 H_2 \cdots H_{j-1}$, which further implies that $x \in H_1 H_2 \cdots H_{j-1} \cap H_j$.

But $H_1 H_2 \cdots H_{j-1} \cap H_j = \{ e \}$, and therefore, $x = e$. Thus, $H_i \cap H_j = \{ e \}$.

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  • $\begingroup$ Oh! since $x \in H_1 H_2 \cdots H_{j-1}$ is in contradiction with the second part of the definition....isn't it? @math.stackexchange.com/users/152030/m-vinay $\endgroup$ – Bijesh K.S May 30 '16 at 9:40
  • $\begingroup$ @BijeshK.S Yes, (if you assume $x \ne e$). Without the assumption, you're merely showing that every element of $H_i \cap H_j$ has to be $e$, and therefore, $H_i \cap H_j = \{e\}$, directly. $\endgroup$ – M. Vinay May 30 '16 at 9:43

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