Let G be a group and H,K is a normal subgroup of G
Prove that <HUK>=HK
$\displaystyle x\in\left<H\cup K\right>\ \Rightarrow\ x=y^n$ for some $\displaystyle y\in H\cup K,\ n\in\mathbb{Z}$
$\displaystyle y\in H\ \Rightarrow\ x=y^n\cdot e\in HK$
$\displaystyle y\in K\ \Rightarrow\ x=e\cdot y^n\in HK$
$\displaystyle \therefore\ \left<H\cup K\right>\subseteq HK$
For the reverse inclusion, use the fact that $\displaystyle H$ and $\displaystyle K$ are normal in $\displaystyle G$.
$\displaystyle hk\in HK\ \Rightarrow\ hk=(hkh^{-1})h\in H$ and $\displaystyle hk=k(k^{-1}hk)\in K$
$\displaystyle \therefore\ HK\subseteq \left<H\cup K\right>$