The Product HK of Two Subgroups H and K of a Group G
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# The Product HK of Two Subgroups H and K of a Group G

 Definition: Let $G$ be a group and let $H$ and $K$ be two subgroups of $G$. The Product of $H$ and $K$ is $HK = \{ hk : h \in H, k \in K \}$.

Note that in the definition above, we are simply "multiplying" the subgroups $H$ and $K$ together to form a set $HK$. In general, $HK$ might not be a subgroup of $G$. We now outline criterion for when $HK$ is a subgroup of $G$.

 Proposition 1: Let $G$ be a group and let $H$ and $K$ be two subgroups of $G$. Then $HK$ is a subgroup of $G$ if and only if $HK = KH$.
• Proof: $\Rightarrow$ Suppose that $HK$ is a subgroup of $G$.
• Let $x \in HK$. Then $x = h \cdot k$ for some $h \in H$ and for some $k \in K$. Since $HK$ is a group and $x \in HK$ we have that $x^{-1} \in HK$. But $x^{-1} = (h \cdot k)^{-1} = k^{-1} \cdot h^{-1} \in KH$. So for all $x^{-1} \in HK$ we have that $x^{-1} \in KH$. Since $x \to x^{-1}$ is a bijection, we see that for all $x \in HK$ we have that $x \in KH$. So $HK \subseteq KH$.
• Now since $H$ and $K$ are subgroups of $G$ we have that $e \in H$ and $e \in K$. So for all $h \in H$ we have that $h = he \in KH$ and for all $k \in K$ we have that $k = eh \in HK$. Since $HK$ is a subgroup of $G$ and since $k, h \in KH$ we have that $kh \in HK$. Since this holds true for all $k \in K$ and for all $h \in H$ we have that $KH \subseteq HK$.
• Thus $HK = KH$.
• $\Leftarrow$ Suppose that $HK = KH$. We aim to show that $HK$ is a subgroup of $G$.
• First we show that $HK$ is closed under the operation $\cdot$. Let $x, y \in HK$. Then $x = h \cdot k$ and $y = h' \cdot k'$ for some $h, h' \in H$ and for some $k, k' \in K$. So:
(1)
\begin{align} \quad x \cdot y = h \cdot k \cdot h' \cdot k' \end{align}
• Since $HK = KH$, for $k \cdot h' \in KH$ there exists a $\tilde{h} \cdot \tilde{k} \in HK$ such that $\tilde{h} \cdot \tilde{k} = k \cdot h'$. Thus:
(2)
\begin{align} \quad x \cdot y = \underbrace{h \cdot \tilde{h}}_{\in H} \cdot \underbrace{\tilde{k} \cdot k'}_{\in K} \in HK \end{align}
• So $HK$ is closed under $\cdot$.
• Secondly, since $H$ and $K$ are subgroups of $G$ we have that $e \in H$ and $e \in K$. So $e = e \cdot e \in HK$.
• Lastly, if $x \in HK$ then $x = h \cdot k$ for some $h \in H$ and for some $k \in K$. So $x^{-1} = (h \cdot k)^{-1} = k^{-1} \cdot h^{-1} \in KH = HK$.
• So $HK$ is a subgroup of $G$. $\blacksquare$
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