# $HK\cap N=H(K\cap N)$

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.

• Is this homework ? – Kasper Apr 26 '13 at 11:28
• @Kasper Absolutely not. – user42912 Apr 26 '13 at 11:29
• Hahaha, sounds very convincing ;) – Kasper Apr 26 '13 at 11:30
• @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 ;) – user42912 Apr 26 '13 at 11:33
• Google "Dedekind's Modular Law". – 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 $HK \cap HN = H(K \cap N)$.

• Why $H(K \cap N)$ is of the form $h k = h n$? Thank you for your answer. – user42912 Apr 26 '13 at 11:39
• @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 $N$. – shobon Apr 26 '13 at 11:43
• yes, of course, thank you for the clarification. – user42912 Apr 26 '13 at 11:44
• we don't need the converse of the inclusion to say that these sets are equal? – user42912 Apr 26 '13 at 11:50
• I proved both way in one step. – shobon Apr 26 '13 at 11:57