Prove $N\subseteq H$.

Question

Problem: Let $H,K,N$ be subgroups of a group $G$ such that $H\leq K$, $H\cap N=K\cap N$ and $HN=KN$. Show that $H=K$.

Here is my attempt: $H=K\Leftrightarrow$ $H\subseteq K$ and $K\subseteq H$.

Clearly, $H\subseteq K$ as $H\leq K$. To show $K\subseteq H$;

$HN=KN\Rightarrow k=hn$ where $k\in K, h\in H$ and $n\in N$, so $K\subseteq HN$. $$\color{blue}{If\ N\subseteq H,\ then\ K\subseteq HN \Rightarrow K\subseteq H.}$$ Hence the proof completes!

Even though it is clear that $N\subseteq H$ (intuitively) I don't know how to prove it rigorously. Or is my argument invalid at all(Is there another way to approach the problem)?

It isn't necessary in the problem that $N\subseteq H$. In fact $N=G$ is possible. — Peter Woolfitt, Dec 11 '15 at 14:32
Apply Dedekind's Modular Law: $K= KN \cap K = HN \cap K = H(N \cap K)=H(N \cap H)=H$. — Nicky Hekster, Dec 11 '15 at 22:07

score 2 · Accepted Answer · answered Dec 11 '15 at 14:36

You don't know anywhere that $N\subseteq H$ so you cannot make that leap, however you can do this $$k=hn\implies h^{-1}k=n$$ we have that intersection of subgroups is a subgroup itself and from $$H\cap N=K\cap N$$ we get then that $$h^{-1}k=n\in K\cap N=H\cap N \subseteq H$$ ergo all $k\in H$ and the rest follows

Prove $N\subseteq H$.

1 Answers1