Let $H_1,H_2,\ldots, H_n$ be some subrgoups of a (finite) group $G$, then if:

  1. $G=H_1H_2\ldots H_n$
  2. $[H_i,H_j]=\{1\}$ for $i\neq j$
  3. $H_i\cap H_1H_2\ldots H_{i-1}H_{i+1}\ldots H_n=\{1\}$ for every $i$

we can conclude that $G\cong H_1\times H_2\times\ldots\times H_n$. Now, why the condition $3.$ can be substituted by:

$3'$. $H_{i}\cap H_1H_2\ldots H_{i-1}=\{1\}$ (for every $i$) ?

Edited with my personal solution to the problem

Maybe I have the solution (I'll prove that $1,2,3'\Rightarrow 1,2,3$, the other implication is trivial):

If $h_i=h_1\ldots h_{i-1}h_{i+1}h_n$ with $h_j\in H_j$ then, by 2. $$h_n=h_1^{-1}h_2^{-1}\ldots h_{i-1}^{-1}h_ih_{i+1}^{-1}\ldots h_{n-1}^{-1}\in H_1H_2\ldots H_n$$ So by $3'$ $h_n=1$ and $h_i=h_1\ldots h_{i-1}h_{i+1}h_{n-1}$. We can do the same procedure to show that $h_{n-1}=1, h_{n-2}=1,\ldots, h_{i+1}=1$ and by $3'$ conclude that $h_i=h_1\ldots h_{i-1}=1$

  • $\begingroup$ That looks OK - but what is your question? $\endgroup$ – Derek Holt Feb 12 '13 at 11:52
  • $\begingroup$ I did not understand why the conditions 1,2,3 are equivalent to 1,2,3'. But then I've find the above solution. $\endgroup$ – Dubious Feb 12 '13 at 11:55

If $x\in H_i\cap H_1H_2...H_{i-1}$ then $\exists( h_i\in H_i, h_1\in H_1,...h_{i-1}\in H_{i-1});$ $$x=h_i~\text{and}~ x=h_1h_2...h_{i-1}$$ but $$x=h_1h_2...h_{i-1}=h_1h_2...h_{i-1}e_{i+1}...e_{n}\in H_1H_2...H_{i-1}\cdot H_{i+1}...H_n$$ where in $e_k=e_G$. So $x\in H_i\cap H_1H_2...H_{i-1}\cdot H_{i+1}...H_n=\{e\}$. This means that $3$ leads us to $3'$.


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