Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $h_i=h_1\ldots h_{i-1}=1$

share|improve this question
    
That looks OK - but what is your question? –  Derek Holt Feb 12 '13 at 11:52
    
I did not understand why the conditions 1,2,3 are equivalent to 1,2,3'. But then I've find the above solution. –  Galoisfan Feb 12 '13 at 11:55
add comment

1 Answer

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 $3'$.

share|improve this answer
    
+1$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ –  amWhy Feb 12 '13 at 16:45
    
@amWhy: Thanks :-) –  B. S. Feb 12 '13 at 19:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.