After intersections and unions, the next natural construction is the product of two subgroups. The focus keyword for this lecture is product of subgroups. If $H$ and $K$ are subgroups of a group $G$, then $HK$ is the set of all products $h\circ k$ with $h\in H$ and $k\in K$. Unlike intersections, the product $HK$ is not always a subgroup. In this lesson, we prove the exact condition $HK=KH$ and derive the counting formula for finite subgroups. :::definition[Product of Subgroups] Let $(G,\circ)$ be a group and let $H$ and $K$ be two subgroups of $G$. The product of subgroups $H$ and $K$ is the subset $$\begin{equation*} HK=\{h\circ k:h\in H,\;k\in K\}. \end{equation*}$$ Similarly, $$\begin{equation*} KH=\{k\circ h:k\in K,\;h\in H\}. \end{equation*}$$ ::: In an abelian group, $h\circ k=k\circ h$, so $HK=KH$ automatically. In a non-abelian group, the order of factors matters, and $HK$ may differ from $KH$. This is why the subgroup property of $HK$ requires a special condition. Students often assume that a product of subgroups must be a subgroup because the notation resembles multiplication. The theorem below shows exactly when this assumption is valid. :::theorem[Product of Two Subgroups] Let $(G,\circ)$ be a group and let $H$ and $K$ be two subgroups of $G$. Then $HK$ is a subgroup of $G$ if and only if $$\begin{equation*} HK=KH. \end{equation*}$$ ::: :::proof[Proof of Product of Two Subgroups] Given that $(G,\circ)$ is a group and $H$ and $K$ are two subgroups of $G$. Let $$\begin{equation*} HK=\{h\circ k:h\in H,\;k\in K\} \end{equation*}$$ and $$\begin{equation*} KH=\{k\circ h:k\in K,\;h\in H\}. \end{equation*}$$ To prove that $HK$ is a subgroup of $G$ if and only if $HK=KH$. [1] Let $HK$ be a subgroup of $G$. To prove that $HK=KH$. Let $x\in HK$. Then there exist $h\in H$ and $k\in K$ such that $$\begin{equation*} x=h\circ k. \end{equation*}$$ Since $HK$ is a subgroup of $G$, we get $x^{-1}\in HK$. Thus there exist $h_1\in H$ and $k_1\in K$ such that $$\begin{equation*} x^{-1}=h_1\circ k_1. \end{equation*}$$ Taking inverse in $G$, we get $$\begin{align*} x &=(h_1\circ k_1)^{-1}\\ &=k_1^{-1}\circ h_1^{-1}. \end{align*}$$ Since $k_1^{-1}\in K$ and $h_1^{-1}\in H$, we get $x\in KH$. Therefore $HK\subseteq KH$. Let $y\in KH$. Then there exist $k\in K$ and $h\in H$ such that $$\begin{equation*} y=k\circ h. \end{equation*}$$ Now $$\begin{align*} y^{-1} &=h^{-1}\circ k^{-1}\in HK. \end{align*}$$ Since $HK$ is a subgroup of $G$, we get $y=(y^{-1})^{-1}\in HK$. Therefore $KH\subseteq HK$. Hence, $HK=KH$. [2] Let $HK=KH$. To prove that $HK$ is a subgroup of $G$. Since $e\in H$ and $e\in K$, we get $$\begin{equation*} e=e\circ e\in HK. \end{equation*}$$ Thus $HK$ is non-empty. Let $x,y\in HK$. Then there exist $h_1,h_2\in H$ and $k_1,k_2\in K$ such that $$\begin{equation*} x=h_1\circ k_1,\quad y=h_2\circ k_2. \end{equation*}$$ Now $$\begin{align*} x\circ y^{-1} &=(h_1\circ k_1)\circ(h_2\circ k_2)^{-1}\\ &=h_1\circ k_1\circ k_2^{-1}\circ h_2^{-1}. \end{align*}$$ Since $k_1\circ k_2^{-1}\in K$ and $h_2^{-1}\in H$, we get $$\begin{equation*} (k_1\circ k_2^{-1})\circ h_2^{-1}\in KH. \end{equation*}$$ Using $HK=KH$, we get $$\begin{equation*} (k_1\circ k_2^{-1})\circ h_2^{-1}\in HK. \end{equation*}$$ Thus there exist $h_3\in H$ and $k_3\in K$ such that $$\begin{equation*} (k_1\circ k_2^{-1})\circ h_2^{-1}=h_3\circ k_3. \end{equation*}$$ Therefore $$\begin{align*} x\circ y^{-1} &=h_1\circ h_3\circ k_3. \end{align*}$$ Since $h_1\circ h_3\in H$ and $k_3\in K$, we get $$\begin{equation*} x\circ y^{-1}\in HK. \end{equation*}$$ By the one-step subgroup criterion, $HK$ is a subgroup of $G$. Hence, $HK$ is a subgroup of $G$ if and only if $HK=KH$. $\square$ ::: :::corollary[Product in an Abelian Group] Let $(G,\circ)$ be an abelian group and let $H$ and $K$ be two subgroups of $G$. Then $HK$ is a subgroup of $G$. ::: :::proof[Proof of Product in an Abelian Group] Given that $(G,\circ)$ is an abelian group and $H$ and $K$ are two subgroups of $G$. To prove that $HK$ is a subgroup of $G$. Let $x\in HK$. Then there exist $h\in H$ and $k\in K$ such that $$\begin{equation*} x=h\circ k. \end{equation*}$$ Since $G$ is abelian, $$\begin{align*} x &=h\circ k\\ &=k\circ h. \end{align*}$$ Therefore $x\in KH$, and hence $HK\subseteq KH$. Similarly, $KH\subseteq HK$. Thus $HK=KH$. By the product theorem, $HK$ is a subgroup of $G$. Hence, $HK\leq G$. $\square$ ::: :::theorem[Cardinality of a Product] Let $(G,\circ)$ be a group and let $H$ and $K$ be finite subgroups of $G$. Then $$\begin{equation*} |HK|=\frac{|H||K|}{|H\cap K|}. \end{equation*}$$ ::: :::proof[Proof of Cardinality of a Product] Given that $(G,\circ)$ is a group and $H$ and $K$ are finite subgroups of $G$. To prove that $$\begin{equation*} |HK|=\frac{|H||K|}{|H\cap K|}. \end{equation*}$$ Define $f:H\times K\to HK$ by $$\begin{equation*} f(h,k)=h\circ k. \end{equation*}$$ Then $f$ is onto by definition of $HK$. Let $h\in H$ and $k\in K$. Suppose $$\begin{equation*} h_1\circ k_1=h\circ k. \end{equation*}$$ Then $$\begin{align*} h_1\circ k_1=h\circ k &\implies h^{-1}\circ h_1=k\circ k_1^{-1}. \end{align*}$$ Since $h^{-1}\circ h_1\in H$ and $k\circ k_1^{-1}\in K$, we get $$\begin{equation*} h^{-1}\circ h_1\in H\cap K. \end{equation*}$$ Let $$\begin{equation*} t=h^{-1}\circ h_1. \end{equation*}$$ Then $t\in H\cap K$ and $$\begin{equation*} h_1=h\circ t,\quad k_1=t^{-1}\circ k. \end{equation*}$$ Conversely, for each $t\in H\cap K$, $$\begin{align*} (h\circ t)\circ(t^{-1}\circ k) &=h\circ k. \end{align*}$$ Therefore each element of $HK$ has exactly $|H\cap K|$ inverse images under $f$. Thus $$\begin{align*} |H||K| &=|HK||H\cap K|\\ &\implies |HK|=\frac{|H||K|}{|H\cap K|}. \end{align*}$$ Hence, $$\begin{equation*} |HK|=\frac{|H||K|}{|H\cap K|}. \end{equation*}$$ $\square$ ::: We observe $HK$ and $KH$ inside the symmetric group $S_3$. Choose two small subgroups, and the preview lists the two product sets. If the sets are equal, the theorem says $HK$ is a subgroup. If the sets differ, the product set is not a subgroup, even though both $H$ and $K$ are subgroups. :::scientific-preview[Subgroup Product Explorer] ::: The product set is not just a union and not just a pairwise list. In non-abelian groups, reversing the order can change the product set and therefore change whether the product is a subgroup. :::solved-problem[Sixth Powers Under a Cubic Condition] Let $(G,\circ)$ be a group such that $$\begin{equation*} (a\circ b)^3=a^3\circ b^3\quad \forall\,a,b\in G. \end{equation*}$$ Prove that $H=\{x^6:x\in G\}$ is a subgroup of $G$. ::: :::solution[Solution of Sixth Powers Under a Cubic Condition] Let $(G,\circ)$ be a group such that $$\begin{equation*} (a\circ b)^3=a^3\circ b^3\quad \forall\,a,b\in G. \end{equation*}$$ Let $$\begin{equation*} H=\{x^6:x\in G\}. \end{equation*}$$ To prove that $H$ is a subgroup of $G$. Since $e=e^6$, we get $e\in H$. Therefore $H$ is non-empty. From the earlier result, the set $$\begin{equation*} K=\{x^2:x\in G\} \end{equation*}$$ is a subgroup of $G$. Let $x^6,y^6\in H$, where $x,y\in G$. Then $x^2,y^2\in K$. Using the given condition with $a=x^2$ and $b=y^2$, we get $$\begin{align*} (x^2\circ y^2)^3 &=(x^2)^3\circ (y^2)^3\\ &=x^6\circ y^6. \end{align*}$$ Since $K$ is a subgroup of $G$ and $x^2,y^2\in K$, we get $$\begin{equation*} x^2\circ y^2\in K. \end{equation*}$$ Therefore there exists $z\in G$ such that $$\begin{equation*} x^2\circ y^2=z^2. \end{equation*}$$ Thus $$\begin{align*} x^6\circ y^6 &=(x^2\circ y^2)^3\\ &=(z^2)^3\\ &=z^6\in H. \end{align*}$$ Therefore $H$ is closed under $\circ$. Let $x^6\in H$. Then $$\begin{align*} (x^6)^{-1} &=x^{-6}\\ &=(x^{-1})^6\in H. \end{align*}$$ Therefore $(x^6)^{-1}\in H$ for every $x^6\in H$. By the subgroup criterion, $H$ is a subgroup of $G$. Hence, $H\leq G$. $\square$ ::: For finite subgroups, the size of $HK$ depends on the overlap $H\cap K$. Enter the orders of $H$, $K$, and their intersection. The calculator checks divisibility and then applies $|HK|=|H||K|/|H\cap K|$. This formula counts the product set even when $HK$ is not itself a subgroup. :::calculator[Product Cardinality Calculator] ::: :::exercise[Exercises on Products of Subgroups] 1. Let $G$ be abelian and let $H,K\leq G$. Prove directly that $HK\leq G$. 2. Let $H,K$ be finite subgroups of $G$ with $|H|=6$, $|K|=10$, and $|H\cap K|=2$. Find $|HK|$. 3. If $H\cap K=\{e\}$ and $H,K$ are finite subgroups, prove that $|HK|=|H||K|$. ::: :::answer[Answers on Products of Subgroups] 1. Since $G$ is abelian, $HK=KH$. Therefore $HK$ is a subgroup by the product theorem. 2. $|HK|=\frac{6\cdot 10}{2}=30$. 3. Since $|H\cap K|=1$, the cardinality formula gives $|HK|=\frac{|H||K|}{1}=|H||K|$. ::: :::faq[Frequently Asked Questions] Q: Is $HK$ always a subgroup? A: No. It is a subgroup exactly when $HK=KH$. Q: Is $HK=KH$ automatic in abelian groups? A: Yes. Since all elements commute, the two products are equal. Q: Does the cardinality formula require $HK$ to be a subgroup? A: No. The formula counts the product set $HK$ when $H$ and $K$ are finite subgroups. ::: :::call-to-action[Proceed to Special Subgroups] subtitle: Continue with important subgroup constructions such as centre, centraliser, and normaliser. button: Study Special Subgroups | published/abstract-algebra/subgroups-and-normal-subgroups/special-subgroups/centre-of-a-group.md button-ghost: Review Unions of Subgroups | published/abstract-algebra/subgroups-and-normal-subgroups/subgroups/unions-of-subgroups.md :::