I'm trying to prove If $H$, $K$ and $N$ are subgroups of a group $G$ such that $H\lt N$, then $HK\cap N=H(K\cap N).$ I'm trying sets inclusion to prove it, am I in the right way? I need help.

Thanks a lot.

  • $\begingroup$ Is this homework ? $\endgroup$ – Kasper Apr 26 '13 at 11:28
  • $\begingroup$ @Kasper Absolutely not. $\endgroup$ – user42912 Apr 26 '13 at 11:29
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    $\begingroup$ Hahaha, sounds very convincing ;) $\endgroup$ – Kasper Apr 26 '13 at 11:30
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    $\begingroup$ @Kasper It's not a homework at all, I have to do as many question I can to the exam in May, that's all ;) $\endgroup$ – user42912 Apr 26 '13 at 11:33
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    $\begingroup$ Google "Dedekind's Modular Law". $\endgroup$ – Martin Brandenburg Apr 26 '13 at 12:11

Let $H,K,N < G$ and $H < N$ then $HK \cap N = H(K \cap N)$.

An element of $H(K \cap N)$ is of the form $h k = h n$ which by $H < N$ means $h k = n$ - and that is exactly the form of elements of $HK \cap N$ so the two sets are equal.

Note that $H < N$ is essential, if we drop that hypothesis it does not hold that $HK \cap HN = H(K \cap N)$.

  • $\begingroup$ Why $H(K \cap N)$ is of the form $h k = h n$? Thank you for your answer. $\endgroup$ – user42912 Apr 26 '13 at 11:39
  • $\begingroup$ @user42912, to specify some element $y \in H(K \cap N)$ you can give $h$,$k$,$n$ where $y = hk = hn$. because y is the product of some $h$ and some element $x$ which lies in $K$ and $N$. $\endgroup$ – shobon Apr 26 '13 at 11:43
  • $\begingroup$ yes, of course, thank you for the clarification. $\endgroup$ – user42912 Apr 26 '13 at 11:44
  • $\begingroup$ we don't need the converse of the inclusion to say that these sets are equal? $\endgroup$ – user42912 Apr 26 '13 at 11:50
  • $\begingroup$ I proved both way in one step. $\endgroup$ – shobon Apr 26 '13 at 11:57

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