Let abin hk where ah1k1 and bh2k2 for h1h2in h and

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Let $a,b\in HK$ where $a=h_{1}k_{1 }$ and $b=h_{2}k_{2}$ for $h_{1},h_{2}\in H$ and $k_{1},k_{2}\in K$.\\ Consider $k_{1}h_{2}\in KH$, such that $k_{1}h_{2}=h{'}k{'}$ as $KH=HK$. \begin{align*} &ab=h_{1}k_{1}h_{2}k_{2}\\ &=h_{1}h{'}k{'}k_{2}\\ &\in HK &&\text{as $h_{1}h{'}\in H$ and $k{'}k_{2}\in K$} \end{align*} Therefore $HK$ is closed under binary operation.\\ \\ As $H,K \subseteq G$, by the uniqueness of inverses, $H$ and $K$ contain unique identities. Let the identity of $H=1$ and the identity of $K=1$. \footnote{University of Queensland, 2013, 'Chapter 5: Groups', University of Queensland, accessed 15th April 2018, }\\ As $(1)(1)=1 \in HK$, then $HK$ is non-empty.\\ \\ Let $a\in HK$ where $a=h_{1}k_{1 }$ for $h_{1}\in H$ and $k_{1}\in K$.\\ Consider $a^{-1}=(h_{1}k_{1})^{-1}$. Then, \begin{align*} &a^{-1}=(h_{1}k_{1 })^{-1}\\ &=k_{1}^{-1}h_{1}^{-1}\\ &\in KH &&\text{as $k^{-1}\in K$ and $h^{-1}\in H$} \end{align*} Since, $k_{1}^{-1}h_{1}^{-1} \in KH=HK$, then $\exists h_{2}\in H$ and $\exists k_{2}\in K$ such that, \begin{align*} &k_{1}^{-1}h_{1}^{-1}=h_{2}k_{2}\\ &\in HK &&\text{by definition}\\ &\Rightarrow a^{-1}=k_{1}^{-1}h_{1}^{-1}\in HK
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\end{align*} Hence $HK$ contains inverses.\\ Therefore $HK$ is a subgroup as it satisfies all conditions.\\ \\ Hence $HK \subseteq G$ if and only if $HK=KH$. \end{proof} \end{document}
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