After learning how to test a single subset for being a subgroup, we now study how subgroups interact with one another. The focus keyword for this lecture is intersection of subgroups. If two subgroups lie inside the same group, their intersection consists of the elements common to both. The important fact is that this common part is again a subgroup. This result is used frequently when constructing smaller subgroups from known ones. :::theorem[Intersection of Two Subgroups] Let $(G,\circ)$ be a group and let $H$ and $K$ be two subgroups of $G$. Then $H\cap K$ is a subgroup of $G$. ::: :::proof[Proof of Intersection of Two Subgroups] Given that $(G,\circ)$ is a group and $H$ and $K$ are two subgroups of $G$. To prove that $H\cap K$ is a subgroup of $G$. Since $H$ and $K$ are subgroups of $G$, the identity element $e$ of $G$ belongs to both $H$ and $K$. Therefore $$\begin{equation*} e\in H\cap K. \end{equation*}$$ Thus $H\cap K$ is non-empty. Let $a,b\in H\cap K$. Then $$\begin{equation*} a,b\in H \end{equation*}$$ and $$\begin{equation*} a,b\in K. \end{equation*}$$ Since $H$ is a subgroup of $G$, the one-step subgroup criterion gives $$\begin{equation*} a\circ b^{-1}\in H. \end{equation*}$$ Since $K$ is a subgroup of $G$, the one-step subgroup criterion gives $$\begin{equation*} a\circ b^{-1}\in K. \end{equation*}$$ Therefore $$\begin{equation*} a\circ b^{-1}\in H\cap K. \end{equation*}$$ By the one-step subgroup criterion, $H\cap K$ is a subgroup of $G$. Hence, $H\cap K\leq G$. $\square$ ::: :::corollary[Intersection of a Family of Subgroups] Let $(G,\circ)$ be a group and let $\{H_i:i\in I\}$ be a non-empty family of subgroups of $G$. Then $$\begin{equation*} \bigcap_{i\in I}H_i \end{equation*}$$ is a subgroup of $G$. ::: :::proof[Proof of Intersection of a Family of Subgroups] Given that $(G,\circ)$ is a group and $\{H_i:i\in I\}$ is a non-empty family of subgroups of $G$. To prove that $\bigcap_{i\in I}H_i$ is a subgroup of $G$. Since each $H_i$ is a subgroup of $G$, we get $$\begin{equation*} e\in H_i\quad \forall\,i\in I. \end{equation*}$$ Therefore $$\begin{equation*} e\in \bigcap_{i\in I}H_i. \end{equation*}$$ Thus $\bigcap_{i\in I}H_i$ is non-empty. Let $a,b\in \bigcap_{i\in I}H_i$. Then $$\begin{equation*} a,b\in H_i\quad \forall\,i\in I. \end{equation*}$$ Since each $H_i$ is a subgroup of $G$, we get $$\begin{equation*} a\circ b^{-1}\in H_i\quad \forall\,i\in I. \end{equation*}$$ Therefore $$\begin{equation*} a\circ b^{-1}\in \bigcap_{i\in I}H_i. \end{equation*}$$ By the one-step subgroup criterion, $\bigcap_{i\in I}H_i$ is a subgroup of $G$. Hence, $\bigcap_{i\in I}H_i\leq G$. $\square$ ::: An intersection keeps only the elements that satisfy all membership conditions at once. Since each subgroup contains the identity, the intersection is never empty. Since each subgroup is closed under $a\circ b^{-1}$, the intersection is also closed under $a\circ b^{-1}$. This is why intersections are safe operations for subgroups. Students should contrast this with unions, where combining elements from different subgroups may destroy closure. :::example[Intersection in Integers] In $(\mathbb{Z},+)$, consider $2\mathbb{Z}$ and $3\mathbb{Z}$. Their intersection is $$\begin{equation*} 2\mathbb{Z}\cap 3\mathbb{Z}=6\mathbb{Z}. \end{equation*}$$ Indeed, an integer divisible by both $2$ and $3$ is divisible by $6$, and every multiple of $6$ is divisible by both $2$ and $3$. Since $6\mathbb{Z}$ is a subgroup of $\mathbb{Z}$, this example agrees with the intersection theorem. ::: We observe intersections of subgroups of $\mathbb{Z}$. Enter two positive integers $m$ and $n$ to compare $m\mathbb{Z}$ and $n\mathbb{Z}$. Their intersection is the set of all integers divisible by both $m$ and $n$, so it equals $\operatorname{lcm}(m,n)\mathbb{Z}$. The preview lists nearby multiples so the common pattern is visible. :::scientific-preview[Integer Intersection Explorer] ::: The listed elements are not chosen from either subgroup separately; they are exactly the elements satisfying both divisibility conditions at once. :::solved-problem[Intersection of Two Integer Subgroups] Prove that $4\mathbb{Z}\cap 6\mathbb{Z}=12\mathbb{Z}$ and hence it is a subgroup of $(\mathbb{Z},+)$. ::: :::solution[Solution of Intersection of Two Integer Subgroups] Let $m\in 4\mathbb{Z}\cap 6\mathbb{Z}$. Then $m$ is divisible by $4$ and $6$. Therefore $m$ is divisible by the least common multiple of $4$ and $6$, which is $12$. Thus $$\begin{equation*} m\in 12\mathbb{Z}. \end{equation*}$$ Hence $$\begin{equation*} 4\mathbb{Z}\cap 6\mathbb{Z}\subseteq 12\mathbb{Z}. \end{equation*}$$ Conversely, let $m\in 12\mathbb{Z}$. Then $m=12k$ for some $k\in\mathbb{Z}$. Now $$\begin{align*} m &=4(3k), \end{align*}$$ so $m\in 4\mathbb{Z}$. Also $$\begin{align*} m &=6(2k), \end{align*}$$ so $m\in 6\mathbb{Z}$. Therefore $m\in 4\mathbb{Z}\cap 6\mathbb{Z}$. Thus $$\begin{equation*} 12\mathbb{Z}\subseteq 4\mathbb{Z}\cap 6\mathbb{Z}. \end{equation*}$$ Therefore $$\begin{equation*} 4\mathbb{Z}\cap 6\mathbb{Z}=12\mathbb{Z}. \end{equation*}$$ Since $12\mathbb{Z}$ is a subgroup of $(\mathbb{Z},+)$, the intersection is a subgroup. $\square$ ::: This calculator returns the generator of $m\mathbb{Z}\cap n\mathbb{Z}$. It uses the least common multiple because the intersection must contain exactly the integers divisible by both $m$ and $n$. The result is automatically a subgroup of $(\mathbb{Z},+)$. :::calculator[Integer Subgroup Intersection Calculator] ::: :::exercise[Exercises on Intersections of Subgroups] 1. Find $2\mathbb{Z}\cap 5\mathbb{Z}$ in $(\mathbb{Z},+)$. 2. Prove that $m\mathbb{Z}\cap n\mathbb{Z}=\operatorname{lcm}(m,n)\mathbb{Z}$ for positive integers $m,n$. 3. Let $H,K,L\leq G$. Prove that $H\cap K\cap L\leq G$. ::: :::answer[Answers on Intersections of Subgroups] 1. $2\mathbb{Z}\cap 5\mathbb{Z}=10\mathbb{Z}$. 2. An integer belongs to both $m\mathbb{Z}$ and $n\mathbb{Z}$ if and only if it is divisible by both $m$ and $n$, equivalently by $\operatorname{lcm}(m,n)$. 3. Since the identity belongs to all three subgroups and $a\circ b^{-1}$ remains in each subgroup whenever $a,b$ lie in all three, the one-step criterion gives $H\cap K\cap L\leq G$. ::: :::faq[Frequently Asked Questions] Q: Why is the intersection of subgroups never empty? A: Every subgroup contains the identity element of the original group. Q: Does the intersection theorem work for more than two subgroups? A: Yes. Any non-empty family of subgroups has a subgroup as its intersection. Q: Is intersection the same as product? A: No. Intersection keeps common elements, while product forms elements of the form $h\circ k$. ::: :::call-to-action[Continue to Unions of Subgroups] subtitle: Intersections always work, but unions require a special containment condition. button: Study Unions of Subgroups | published/abstract-algebra/subgroups-and-normal-subgroups/subgroups/unions-of-subgroups.md button-ghost: Review Finite Subgroups | published/abstract-algebra/subgroups-and-normal-subgroups/subgroups/finite-subgroups.md :::