Question: Let $G$ be a group, $H \unlhd G$, $K \unlhd G$ and $H \subseteq K$. Show that $H \unlhd K$.
What I thought: I need to show that $kH=Hk$. (1) $kH \subseteq Hk$. Let $kh \in kH$, where $h\in H$. Note that $kh=khk^{-1}k$. So now, I need to show that $khk^{-1} \in H$. Now I know I need to use my hypotheses, but I got nowhere with those. So, I know:
- $gH=Hg, \forall g\in G$ or $H=gHg^{-1} \Rightarrow ghg^{-1} \in H, h\in H$;
- $gK=Kg, \forall g\in G$ or $K=gKg^{-1}$;
- $H \subseteq K \Rightarrow H=gHg^{-1} \subseteq K=gKg^{-1}$ and $H \subseteq K \Rightarrow ghg^{-1} \in K, h\in H$.
But I'm missing where these help me to figure it out $khk^{-1} \in H$. I need desperately a hint.
(2) $Hk\subseteq kH$ will probably be something as (1), right?
Best Answer
Since $H\trianglelefteq G$ we have $ghg^{-1}\in H$ for all $g\in G$. In particular $khk^{-1}\in H$ for all $k\in K$, because $K\leq G$.