Proving that $HK$ is a subgroup iff $HK = KH$. Is my proof correct?

Question

Here $HK = \{hk :h \in H, k \in K\}$, $KH= \{kh :h \in H, k \in K\}$ and $H$ and $K$ are subgroups of a group $G$.

$\implies$: Let $HK$ be a subgroup and $e$ be its neutral element. Let $x \in KH$. Then there exist $h \in H$ and $k \in K$ such that $x = kh$. Since $k = \underbrace{e}_{\displaystyle{\in H \text{ }}} \underbrace{k}_{\displaystyle{\in K}}$, $k \in HK$. Similarly, we have that $h \in HK.$ Since $HK$ is a group, it's closed, therefore $kh \in HK$. So $KH \subseteq HK$. Now, let $y \in HK$. In a group, every element is the inverse of another element in that group, so we can write $y = (\underbrace{hk}_{\displaystyle{\in HK}})^{-1} = k^{-1}h^{-1}$, so $y \in KH$ and we're done.

$\impliedby$: Suppose $HK = KH$. Take $a = h_1k_1 \in HK$ and $b = h_2k_2 \in HK$. Then $ab = h_1\underbrace{k_1 h_2}_{\displaystyle{=hk}}k_2 = h_1hkk_2 \in HK$. Since $H$ and $K$ are subgroups, then $e \in HK$. At last, for any $x = hk \in HK$, $x^{-1} = k^{-1}h^{-1} \in HK$, since $HK = KH$.

Answer 1

It is all correct.

For $\Longrightarrow$, I first thought that when you arrived at $HK\subseteq KH$ you could have concluded by symmetry, but the assumption is not symmetric in that case. This makes your proof even more beautiful.

Answer 2

For a cleaner proof:

First suppose $HK = KH$, let $x,y \in HK$ I claim $xy^{-1} \in HK$. We have that $x=h_1k_1, y=h_2k_2$ for some $h_1,h_2 \in H, k_1,k_2 \in K$. Then

\begin{align} xy^{-1}&=(h_1k_1)(h_2k_2)^{-1}\\ &=h_1k_1k_2^{-1}h_2^{-1}\\ &= h_1k_3h_3 & \text{putting $k_3:=k_1k_2^{-1},h_3:=h_2^{-1}$}\\ &= h_1h_4k_4 & \text{as $HK=KH$}\\ &\in HK \end{align}

Thus $HK \leq G$ is a subgroup.

Conversely suppose $HK \leq G$ is a subgroup. It is clear by closures that $KH \subset HK$. To see the reverse containment let $x \in HK$ then as $HK$ is a subgroup, $x$ is the inverse of some element of $HK$. That is,

\begin{align} x&= (hk)^{-1}\\ &= k^{-1}h^{-1}\\ &\in KH \end{align}

Thus $HK \subset KH$ and we have $HK=KH$.