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PROBLEM

Let $H$ and $K$ be subgroups of a group $G$. Prove that $HK = \{ab | a \in H, b \in K\}$ is a subgroup of $G$ if and only if $HK = KH$.

WHAT I KNOW

(1) If $H = \{e, \alpha, \alpha^2\}$ and $K = \{e, \beta, \beta^2\}$ are subgroups of $S_4$, one can show that $HK \neq KH$. We can compute $HK$ and $KH$ to verify that: In general, the product of two subgroups of a group $G$ is not a subgroup of $G$.

(2) If $H = \{h_1, h_2, \ldots, h_r\}$ and $K = \{b_1, b_2, \ldots, b_p\}$ are subgroups and one of $H$ or $K$ is normal in $G$, then (a) $HK = KH$, and (b) $HK$ is a subgroup of $G$.

(3) If $H$ and $K$ are normal subgroups of $G$, so also is $HK$.

QUESTION

How do I then solve the problem at hand, given that what I know seems to require that at least one of $H$ or $K$ is normal in $G$? That is, how does one prove the statement in the PROBLEM section in full generality (without requiring that one of $H$ or $K$ is normal in $G$)?

Archimedes Plutonium
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    You will not need any of the above knowledge. Just show that if $HK = KH$, then $HK$ is a subgroup (what do you have to check here?) and show that if $HK$ is a subgroup, then $HK = KH$ (hint: inversion). – Matthias Klupsch Sep 06 '17 at 12:05

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