The product of two subgroups is another important place where Lagrange's theorem and coset counting appear. If $H$ and $K$ are finite subgroups of a group $G$, the set $HK$ consists of all products $h\circ k$ with $h\in H$ and $k\in K$. Even when $HK$ is not itself a subgroup, its number of elements can still be counted exactly. The overlap between $H$ and $K$ controls the repeated representations of elements of $HK$. The result is the product formula $|HK|=|H||K|/|H\cap K|$. :::definition[Product of Two Subgroups] Let $(G,\circ)$ be a group and let $H$ and $K$ be subgroups of $G$. The $product$ of $H$ and $K$ is the subset $$\begin{equation*} HK=\{h\circ k:h\in H,\;k\in K\}. \end{equation*}$$ ::: :::theorem 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 Given that $(G,\circ)$ is a group and $H,K$ are finite subgroups of $G$. To prove that $$\begin{equation*} |HK|=\frac{|H||K|}{|H\cap K|}. \end{equation*}$$ Consider the map $\varphi:H\times K\to HK$ defined by $$\begin{equation*} \varphi(h,k)=h\circ k. \end{equation*}$$ By the definition of $HK$, the map $\varphi$ is onto. We determine the number of preimages of a fixed element of $HK$. Let $x\in HK$. Then there exist $h\in H$ and $k\in K$ such that $$\begin{equation*} x=h\circ k. \end{equation*}$$ Let $(h',k')\in H\times K$ such that $$\begin{equation*} h'\circ k'=x. \end{equation*}$$ Then $$\begin{equation*} h'\circ k'=h\circ k. \end{equation*}$$ Therefore $$\begin{align*} h'\circ k'=h\circ k &\implies h^{-1}\circ h'=k\circ (k')^{-1}. \end{align*}$$ Since $h^{-1}\circ h'\in H$ and $k\circ (k')^{-1}\in K$, we get $$\begin{equation*} h^{-1}\circ h'\in H\cap K. \end{equation*}$$ Let $$\begin{equation*} t=h^{-1}\circ h'. \end{equation*}$$ Then $t\in H\cap K$ and $$\begin{equation*} h'=h\circ t. \end{equation*}$$ Also, from $h'\circ k'=h\circ k$, we get $$\begin{align*} h\circ t\circ k'=h\circ k &\implies t\circ k'=k\\ &\implies k'=t^{-1}\circ k. \end{align*}$$ Thus each preimage of $x$ has the form $$\begin{equation*} (h\circ t,\;t^{-1}\circ k) \end{equation*}$$ where $t\in H\cap K$. Conversely, for every $t\in H\cap K$, the pair $$\begin{equation*} (h\circ t,\;t^{-1}\circ k) \end{equation*}$$ belongs to $H\times K$, and $$\begin{align*} \varphi(h\circ t,\;t^{-1}\circ k) &=(h\circ t)\circ(t^{-1}\circ k)\\ &=h\circ (t\circ t^{-1})\circ k\\ &=h\circ e\circ k\\ &=h\circ k\\ &=x. \end{align*}$$ Therefore every element of $HK$ has exactly $|H\cap K|$ preimages under $\varphi$. Since $$\begin{equation*} |H\times K|=|H||K|, \end{equation*}$$ we get $$\begin{equation*} |H||K|=|HK||H\cap K|. \end{equation*}$$ Therefore $$\begin{equation*} |HK|=\frac{|H||K|}{|H\cap K|}. \end{equation*}$$ Hence, $$\begin{equation*} |HK|=\frac{|H||K|}{|H\cap K|}. \end{equation*}$$ $\square$ ::: We observe the map from $H\times K$ onto the product set $HK$. Choose the sizes of $H$, $K$, and $H\cap K$, and notice how the grid of ordered pairs is grouped into equal-size fibers. As $|H\cap K|$ increases, more pairs represent the same product, so the number of distinct elements in $HK$ decreases. Try setting $|H|=4$, $|K|=3$, and $|H\cap K|=1$ to match the example in $\mathbb{Z}_{12}$. :::scientific-preview[Subgroup Product Representation Explorer] ::: The denominator $|H\cap K|$ is not an arbitrary correction factor. It is the exact number of repeated representations of each product element, so dividing by it converts ordered pairs into distinct elements of $HK$. :::corollary Let $(G,\circ)$ be a group and let $H,K$ be finite subgroups of $G$. If $H\cap K=\{e\}$, then $$\begin{equation*} |HK|=|H||K|. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H,K$ are finite subgroups of $G$, and $H\cap K=\{e\}$. To prove that $$\begin{equation*} |HK|=|H||K|. \end{equation*}$$ Since $H\cap K=\{e\}$, we get $$\begin{equation*} |H\cap K|=1. \end{equation*}$$ By the product formula for finite subgroups, $$\begin{align*} |HK| &=\frac{|H||K|}{|H\cap K|}\\ &=\frac{|H||K|}{1}\\ &=|H||K|. \end{align*}$$ Hence, $$\begin{equation*} |HK|=|H||K|. \end{equation*}$$ $\square$ ::: The formula counts the set $HK$. It does not automatically say that $HK$ is a subgroup. In many courses, students mistakenly treat every product of subgroups as a subgroup. That is not valid without extra hypotheses. The counting formula remains true because it counts representations $h\circ k$, not because $HK$ must be closed under the group operation. :::example[Product Formula in Integers Modulo Twelve] Let $G=\mathbb{Z}_{12}$ under addition modulo $12$. Let $$\begin{equation*} H=\{0,3,6,9\} \end{equation*}$$ and $$\begin{equation*} K=\{0,4,8\}. \end{equation*}$$ Then $$\begin{equation*} H\cap K=\{0\}. \end{equation*}$$ Therefore $$\begin{align*} |H+K| &=\frac{|H||K|}{|H\cap K|}\\ &=\frac{4\cdot 3}{1}\\ &=12. \end{align*}$$ Indeed, $$\begin{equation*} H+K=\mathbb{Z}_{12}. \end{equation*}$$ ::: :::solved-problem[Counting a Product of Subgroups] Let $(G,\circ)$ be a group and let $H,K$ be finite subgroups of $G$. Suppose $$\begin{equation*} |H|=12,\quad |K|=10,\quad |H\cap K|=2. \end{equation*}$$ Find $|HK|$. ::: :::solution[Solution] Let $(G,\circ)$ be a group and let $H,K$ be finite subgroups of $G$. By the product formula for finite subgroups, $$\begin{align*} |HK| &=\frac{|H||K|}{|H\cap K|}\\ &=\frac{12\cdot 10}{2}\\ &=60. \end{align*}$$ Therefore $$\begin{equation*} |HK|=60. \end{equation*}$$ ::: We compute $|HK|$ from the three numbers that control the product formula. Enter the orders of $H$, $K$, and their intersection, and check whether the division produces an integer. The output reminds us that the formula counts the subset $HK$, not necessarily a subgroup. This distinction keeps the calculation connected to the proof rather than to an unproved closure assumption. :::calculator[Product Formula Calculator] ::: :::exercise[Exercises] 1. Let $H,K$ be finite subgroups of a group $G$ such that $|H|=8$, $|K|=6$, and $|H\cap K|=2$. Find $|HK|$. 2. Let $H,K$ be finite subgroups of a group $G$ such that $H\cap K=\{e\}$, $|H|=5$, and $|K|=7$. Find $|HK|$. 3. Let $H,K$ be finite subgroups of a group $G$. Prove that $|H\cap K|$ divides $|H||K|$. ::: :::answer[Answers] 1. By the product formula, $$\begin{equation*} |HK|=\frac{8\cdot 6}{2}=24. \end{equation*}$$ 2. Since $H\cap K=\{e\}$, we get $$\begin{equation*} |HK|=|H||K|=5\cdot 7=35. \end{equation*}$$ 3. Since $H\cap K$ is a finite subgroup, $|H\cap K|$ is a positive integer. The product formula gives $$\begin{equation*} |H||K|=|HK||H\cap K|. \end{equation*}$$ Therefore $|H\cap K|$ divides $|H||K|$. ::: :::faq Q: Is $HK$ always a subgroup? A: No. The product $HK$ is always a subset, but it need not be a subgroup without extra conditions. Q: Why does $|H\cap K|$ appear in the denominator? A: Elements in $H\cap K$ account for repeated representations of the same element of $HK$. Q: Does the formula require $G$ to be finite? A: It requires $H$ and $K$ to be finite. The whole group $G$ need not be finite for this counting formula. ::: :::call-to-action subtitle: Continue with relatively prime subgroup orders and their consequences. button: Next: Relatively Prime Subgroup Orders | published/abstract-algebra/subgroups-and-normal-subgroups/applications-of-lagrange-theorem/relatively-prime-subgroup-orders.md :::