# normal

• August 3rd 2008, 05:07 AM
mathemanyak
normal
Let G be a group and H,K is a normal subgroup of G
Prove that <HUK>=HK
• August 3rd 2008, 08:09 AM
JaneBennet
$x\in\left\ \Rightarrow\ x=y^n$ for some $y\in H\cup K,\ n\in\mathbb{Z}$

$y\in H\ \Rightarrow\ x=y^n\cdot e\in HK$
$y\in K\ \Rightarrow\ x=e\cdot y^n\in HK$

$\therefore\ \left\subseteq HK$

For the reverse inclusion, use the fact that $H$ and $K$ are normal in $G$.

$hk\in HK\ \Rightarrow\ hk=(hkh^{-1})h\in H$ and $hk=k(k^{-1}hk)\in K$

$\therefore\ HK\subseteq \left$