The product formula becomes especially simple when two subgroups have relatively prime orders. If $|H|$ and $|K|$ have no common divisor other than $1$, then the intersection $H\cap K$ must have only one element. Since every subgroup contains the identity element, this means $H\cap K=\{e\}$. This lecture shows how Lagrange's theorem controls intersections through divisibility and then uses that control to count products of subgroups. :::definition[Relatively Prime Subgroup Orders] Let $(G,\circ)$ be a group and let $H,K$ be finite subgroups of $G$. The subgroups $H$ and $K$ are said to have $relatively prime orders$ if $$\begin{equation*} \gcd(|H|,|K|)=1. \end{equation*}$$ ::: :::theorem Let $(G,\circ)$ be a group and let $H,K$ be finite subgroups of $G$. If $$\begin{equation*} \gcd(|H|,|K|)=1, \end{equation*}$$ then $$\begin{equation*} H\cap K=\{e\}. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H,K$ are finite subgroups of $G$, and $\gcd(|H|,|K|)=1$. To prove that $$\begin{equation*} H\cap K=\{e\}. \end{equation*}$$ Since $H\cap K$ is a subgroup of $H$, Lagrange's theorem gives $$\begin{equation*} |H\cap K|\mid |H|. \end{equation*}$$ Since $H\cap K$ is a subgroup of $K$, Lagrange's theorem gives $$\begin{equation*} |H\cap K|\mid |K|. \end{equation*}$$ Therefore $|H\cap K|$ is a common divisor of $|H|$ and $|K|$. Since $\gcd(|H|,|K|)=1$, we get $$\begin{equation*} |H\cap K|=1. \end{equation*}$$ Since $e\in H\cap K$, the subgroup $H\cap K$ has exactly one element. Therefore $$\begin{equation*} H\cap K=\{e\}. \end{equation*}$$ Hence, $$\begin{equation*} H\cap K=\{e\}. \end{equation*}$$ $\square$ ::: We observe why relatively prime subgroup orders force a trivial intersection. Choose the orders of $H$ and $K$, and compare their positive divisors side by side. The order of $H\cap K$ must appear in both divisor lists, because the intersection is a subgroup of both $H$ and $K$. When the only common divisor is $1$, the identity is the only possible element of the intersection. :::scientific-preview[Coprime Subgroup Intersection Explorer] ::: The proof is a divisibility argument. Once the only common divisor is $1$, Lagrange's theorem leaves no room for a larger intersection subgroup. :::corollary Let $(G,\circ)$ be a group and let $H,K$ be finite subgroups of $G$. If $\gcd(|H|,|K|)=1$, then $$\begin{equation*} |HK|=|H||K|. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $H,K$ are finite subgroups of $G$, and $\gcd(|H|,|K|)=1$. To prove that $$\begin{equation*} |HK|=|H||K|. \end{equation*}$$ By the previous theorem, $$\begin{equation*} H\cap K=\{e\}. \end{equation*}$$ Therefore $$\begin{equation*} |H\cap K|=1. \end{equation*}$$ 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, $$\begin{equation*} |HK|=|H||K|. \end{equation*}$$ $\square$ ::: :::corollary Let $(G,\circ)$ be a finite group and let $H,K$ be subgroups of $G$. If $\gcd(|H|,|K|)=1$, then $$\begin{equation*} |H||K|\leq |G|. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a finite group, $H,K$ are subgroups of $G$, and $\gcd(|H|,|K|)=1$. To prove that $$\begin{equation*} |H||K|\leq |G|. \end{equation*}$$ By the previous corollary, $$\begin{equation*} |HK|=|H||K|. \end{equation*}$$ Since $HK\subseteq G$, we get $$\begin{equation*} |HK|\leq |G|. \end{equation*}$$ Therefore $$\begin{equation*} |H||K|\leq |G|. \end{equation*}$$ Hence, $$\begin{equation*} |H||K|\leq |G|. \end{equation*}$$ $\square$ ::: :::example[Subgroups of Orders Four and Three] 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 $$\begin{equation*} |H|=4 \end{equation*}$$ and $$\begin{equation*} |K|=3. \end{equation*}$$ Since $\gcd(4,3)=1$, we get $$\begin{equation*} H\cap K=\{0\}. \end{equation*}$$ Also, $$\begin{equation*} |H+K|=|H||K|=12. \end{equation*}$$ ::: :::solved-problem[Trivial Intersection from Coprime Orders] Let $(G,\circ)$ be a group and let $H,K$ be finite subgroups of $G$. Suppose $|H|=15$ and $|K|=8$. Prove that $H\cap K=\{e\}$. ::: :::solution[Solution] Let $(G,\circ)$ be a group and let $H,K$ be finite subgroups of $G$. Since $$\begin{equation*} \gcd(15,8)=1, \end{equation*}$$ we get $$\begin{equation*} \gcd(|H|,|K|)=1. \end{equation*}$$ By the theorem on relatively prime subgroup orders, $$\begin{equation*} H\cap K=\{e\}. \end{equation*}$$ ::: We compute the greatest common divisor of $|H|$ and $|K|$ and use it to decide what the theorem guarantees. When the gcd is $1$, the calculator reports the trivial intersection and computes $|HK|=|H||K|$. If an ambient group order is entered, the same calculation checks whether the inequality $|H||K|\leq |G|$ is possible. Notice that when the gcd is not $1$, the theorem simply does not apply; it does not prove that the intersection is nontrivial. :::calculator[Coprime Product Calculator] ::: :::exercise[Exercises] 1. Let $H,K$ be finite subgroups of a group $G$ with $|H|=9$ and $|K|=10$. Find $H\cap K$. 2. Let $H,K$ be finite subgroups of a group $G$ with $|H|=7$ and $|K|=12$. Find $|HK|$. 3. Let $G$ be a finite group with subgroups $H,K$ such that $|H|=5$ and $|K|=9$. What inequality involving $|G|$ follows? ::: :::answer[Answers] 1. Since $\gcd(9,10)=1$, we get $$\begin{equation*} H\cap K=\{e\}. \end{equation*}$$ 2. Since $\gcd(7,12)=1$, we get $$\begin{equation*} |HK|=|H||K|=7\cdot 12=84. \end{equation*}$$ 3. Since $\gcd(5,9)=1$, we have $$\begin{equation*} |H||K|\leq |G|. \end{equation*}$$ Therefore $$\begin{equation*} 45\leq |G|. \end{equation*}$$ ::: :::faq Q: Why is the intersection trivial when subgroup orders are relatively prime? A: The order of the intersection divides both subgroup orders, so it must divide their greatest common divisor. Q: Does $H\cap K=\{e\}$ imply that $|H|$ and $|K|$ are relatively prime? A: No. Trivial intersection can occur even when the subgroup orders are not relatively prime. Q: Does $|HK|=|H||K|$ mean $HK$ is a subgroup? A: No. It counts the set $HK$; it does not automatically prove closure. ::: :::call-to-action subtitle: Continue with subgroup orders inside groups whose order is a power of a prime. button: Next: Subgroups of Prime Power Order Groups | published/abstract-algebra/subgroups-and-normal-subgroups/applications-of-lagrange-theorem/subgroups-of-prime-power-order-groups.md :::