$HK$ is a subgroup of $G$ if and only if $HK = KH$

Question

Let $H$ and $K$ be subgroups of $G$. Prove that $HK$ is a subgroup of $G$ if and only if $HK=KH$. In particular, the condition holds if $hk=kh$ for all $h$ in $H$ and $k$ in $K$.

Answer 1

Consider $f(x)=x^{-1}$, $f(HK)=KH$ and if $HK$ is a subgroup, $f(HK)=HK=KH$, on the other hand, $HK=KH$ implies that $h,h'\in H, $k,k'\in K$, $(hk)^{-1}h^{-1}\in HK$=k^{-1}$, $(hk)(h'k')=h(kh')k$, but $(kh')\in HK$ implies that $(hk)(h'k')\in HK$.

Answer 2

Theorem: If $H,K \leq G$ are subgroups, then $HK \leq G$ is a subgroup if and only if $HK=KH$.

Proof:

We prove the backwards direction first.

($\Leftarrow$) Suppose first that $HK=KH$. I claim $HK \leq G$ is a subgroup given $H,K \leq G$ are subgroups. Let $x,y \in HK$ then $x=h_1k_1,y=h_2k_2$ 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} && \text{definition of $x,y$}\\ &=h_1k_1k_2^{-1}h_2^{-1} && \text{definition of inverse}\\ &=h_1k_3h_3 && \text{putting $k_1k_2^{-1}=k_3 \in K$ and $h_2^{-1}=h_3 \in H$}\\ &=h_1h_4k_4 && \text{as $HK=KH$, $k_3h_3=h_4k_4$ some $h_4 \in H, k_4 \in K$}\\ &\in HK \end{align}

thus $HK \leq G$ is a subgroup.

($\Rightarrow$) Next suppose $HK \leq G$ is a subgroup by this assumption it is clear to see $KH \subset HK$. So let $x \in HK$, then as $HK \leq G$, $x$ is the inverse of an element of $HK$. I.e.,

\begin{align} x&=(hk)^{-1} && \text{as $x$ is the inverse of element in $HK$} \\ &=k^{-1}h^{-1} && \text{definition of inverse}\\ &\in KH \end{align} Thus $HK \subset KH$ forcing $HK = KH$. $\blacksquare$