Let $H$ and $K$ be subgroups of a group $G$, and let $HK=\{hk: h \in H, k \in K\}$, $KH=\{kh: k \in K, h \in H\}$. How can we prove that $HK$ is a subgroup iff $HK=KH$?
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It is usually helpful to recall that $ L$ is a subgroup if and only if $ L^ 2=L $, $ L^{- 1}= L $ and $ L\neq \varnothing$.
If $HK$ is a subgroup, then $HK=(HK)^{-1}=K^{ -1}H^{-1}=KH$. Hence the condition is met. Conversely, $$ \begin{align}(HK)^{ -1} &= K^{ -1}H^{-1}\\&=KH =HK\text{, }\\(HK)(HK)=H(KH)K&=\\H( HK)K&= H^2K^ 2= HK\end{align} $$ and of course $ HK\neq \varnothing$. We made repeated use of the fact $ H $ and $ K $ are subgroups and $ H K = KH $.
Pedro
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Suppose that $HK$ is a subgroup. Then for every $x\in HK$ there exists $h\in H, k\in K$ such as $x=hk$. But $x^{-1}$ is in $HK$ too, and $x^{-1}=(hk)^{-1}=k^{-1}h^{-1}\in KH$, therefore $HK\subseteq KH$. Conversely, for $k\in K$ and $h\in H$ we have $k\in HK$ and $h\in HK$. Since $HK$ is a subgroup we get $kh\in HK$, so $KH\subseteq HK$.
Now suppose $HK=KH$.
1° $e\in H, e\in K$ so $e\in HK$.
2° For $x,y\in HK$ there exists $h_x,h_y\in H$ and $k_x,k_y\in K$ such that $x=h_xk_x$ and $y=h_yk_y$, so $xy=h_xk_xh_yk_y$, but $KH=HK$ so you can find $h\in H, k\in K$ such as $k_xh_y=hk$ and therefore $xy=h_xhkk_y\in HK$.
3° Let $x=hk$. Then $x^{-1}=k^{-1}h^{-1}\in KH$, but since $KH=HK$, $x^{-1}\in HK$.
So $HK$ is a subgroup.
user26857
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melodynelson
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2I do not think the argument for the first inclusion is correct. – Sawyer Sep 06 '19 at 00:56
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@SawyerR. Recall that in a group $G$ for all $x\in G$ there exists $y\in G$ such that $y^{-1}=x$. Because $HK$ is a subgroup, the argument for the first inclusion is correct. – Lord Vader Nov 03 '21 at 02:33
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1The argument for the first inclusion is not correct because what we have shown is that $x \in HK$ implies $x^{-1} \in KH$. But we cannot assume $KH$ is closed under inverses so that $x \in KH$ and $HK \subseteq KH$. To remedy this, note that since $k^{-1}h^{-1} \in HK$, there is $h' \in H, k' \in K$ such that $k^{-1}h^{-1} = h'k'$. So $x=hk=(k^{-1}h^{-1})^{-1} = (h'k')^{-1} = (k')^{-1}(h')^{-1} \in KH$. – alvnya Jun 27 '24 at 11:42
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"$\Leftarrow$" Clearly $e\in HK$ since $e\in H\cap K$. Let $hk, h'k'\in HK$. Then for $HK$ to be a group, we need $(hk)(h'k')\in HK$. This follows from $HK=KH$, since $$(hk)(h'k') = h(kh')k'=h(h''k'')k' = (hh'')(k''k')\in HK $$ (where $h''\in H$, $k''\in K$). You can similarly show that $HK=KH$ implies that $HK$ is closed with respect to inverses.
