Intersections of subgroups are always subgroups, but unions behave very differently. The focus keyword for this lecture is union of subgroups. If $H$ and $K$ are subgroups of a group $G$, then $H\cup K$ contains elements from both subgroups. The difficulty is that multiplying one element from $H$ and one element from $K$ may produce an element outside the union. In this lesson, we prove the exact condition under which a union of two subgroups is again a subgroup. :::theorem[Union of Two Subgroups] Let $(G,\circ)$ be a group and let $H$ and $K$ be two subgroups of $G$. Then $H\cup K$ is a subgroup of $G$ if and only if $$\begin{equation*} H\subseteq K \end{equation*}$$ or $$\begin{equation*} K\subseteq H. \end{equation*}$$ ::: :::proof[Proof of Union of Two Subgroups] Given that $(G,\circ)$ is a group and $H$ and $K$ are two subgroups of $G$. To prove that $H\cup K$ is a subgroup of $G$ if and only if $H\subseteq K$ or $K\subseteq H$. [1] Let $H\cup K$ be a subgroup of $G$. To prove that $H\subseteq K$ or $K\subseteq H$. If possible let $H\nsubseteq K$ and $K\nsubseteq H$. Then there exist $h\in H$ and $k\in K$ such that $$\begin{equation*} h\notin K \end{equation*}$$ and $$\begin{equation*} k\notin H. \end{equation*}$$ Since $h,k\in H\cup K$ and $H\cup K$ is a subgroup of $G$, we get $$\begin{equation*} h\circ k\in H\cup K. \end{equation*}$$ Therefore $h\circ k\in H$ or $h\circ k\in K$. If $h\circ k\in H$, then $$\begin{align*} h\circ k\in H &\implies h^{-1}\circ(h\circ k)\in H\\ &\implies k\in H, \end{align*}$$ which contradicts $k\notin H$. If $h\circ k\in K$, then $$\begin{align*} h\circ k\in K &\implies (h\circ k)\circ k^{-1}\in K\\ &\implies h\in K, \end{align*}$$ which contradicts $h\notin K$. A contradiction. Hence, $H\subseteq K$ or $K\subseteq H$. [2] Let $H\subseteq K$ or $K\subseteq H$. If $H\subseteq K$, then $$\begin{equation*} H\cup K=K. \end{equation*}$$ Since $K$ is a subgroup of $G$, $H\cup K$ is a subgroup of $G$. If $K\subseteq H$, then $$\begin{equation*} H\cup K=H. \end{equation*}$$ Since $H$ is a subgroup of $G$, $H\cup K$ is a subgroup of $G$. Hence, $H\cup K$ is a subgroup of $G$ if and only if $H\subseteq K$ or $K\subseteq H$. $\square$ ::: The union contains all elements of $H$ and all elements of $K$, but it does not automatically contain products mixing the two subgroups. If neither subgroup contains the other, we can choose $h\in H\setminus K$ and $k\in K\setminus H$. Closure of the union would force $h\circ k$ to lie in $H$ or $K$, and either case gives a contradiction. This is the precise reason unions are not as stable as intersections. :::example[A Union That Is a Subgroup] Let $H=2\mathbb{Z}$ and $K=4\mathbb{Z}$ in $(\mathbb{Z},+)$. Since $4\mathbb{Z}\subseteq 2\mathbb{Z}$, we get $$\begin{equation*} H\cup K=2\mathbb{Z}. \end{equation*}$$ Therefore $H\cup K$ is a subgroup of $(\mathbb{Z},+)$. ::: :::example[A Union That Is Not a Subgroup] Let $H=2\mathbb{Z}$ and $K=3\mathbb{Z}$ in $(\mathbb{Z},+)$. Neither $H\subseteq K$ nor $K\subseteq H$. Also $2\in H$ and $3\in K$, but $$\begin{equation*} 2+3=5. \end{equation*}$$ Since $5\notin 2\mathbb{Z}$ and $5\notin 3\mathbb{Z}$, we get $5\notin H\cup K$. Therefore $H\cup K$ is not closed under addition. Hence, $H\cup K$ is not a subgroup of $(\mathbb{Z},+)$. ::: We observe the union condition with subgroups $m\mathbb{Z}$ and $n\mathbb{Z}$ of $\mathbb{Z}$. One subgroup contains the other exactly when one of the divisibility conditions is stronger than the other. If neither contains the other, mixed sums can leave the union. The preview compares the containment relation and gives a typical mixed-sum warning when the union is not a subgroup. :::scientific-preview[Integer Subgroup Union Explorer] ::: A union can pass all tests inside $H$ and inside $K$ separately, yet fail when one element comes from each subgroup. This is the key difference from intersections. :::solved-problem[Union of Integer Subgroups] Determine whether $6\mathbb{Z}\cup 3\mathbb{Z}$ is a subgroup of $(\mathbb{Z},+)$. ::: :::solution[Solution of Union of Integer Subgroups] Let $H=6\mathbb{Z}$ and $K=3\mathbb{Z}$. Given that $H$ and $K$ are subgroups of $(\mathbb{Z},+)$. To determine whether $H\cup K$ is a subgroup of $\mathbb{Z}$. Since every multiple of $6$ is a multiple of $3$, we get $$\begin{equation*} 6\mathbb{Z}\subseteq 3\mathbb{Z}. \end{equation*}$$ Therefore $$\begin{equation*} 6\mathbb{Z}\cup 3\mathbb{Z}=3\mathbb{Z}. \end{equation*}$$ Since $3\mathbb{Z}$ is a subgroup of $(\mathbb{Z},+)$, we get that $6\mathbb{Z}\cup 3\mathbb{Z}$ is a subgroup of $(\mathbb{Z},+)$. $\square$ ::: :::solved-problem[Union Fails Without Containment] Determine whether $4\mathbb{Z}\cup 6\mathbb{Z}$ is a subgroup of $(\mathbb{Z},+)$. ::: :::solution[Solution of Union Fails Without Containment] Let $H=4\mathbb{Z}$ and $K=6\mathbb{Z}$. Given that $H$ and $K$ are subgroups of $(\mathbb{Z},+)$. To determine whether $H\cup K$ is a subgroup of $\mathbb{Z}$. We have $4\in H$ and $6\in K$. But $$\begin{align*} 4+6 &=10. \end{align*}$$ Since $10$ is not divisible by $4$ and $10$ is not divisible by $6$, we get $$\begin{equation*} 10\notin H\cup K. \end{equation*}$$ Therefore $H\cup K$ is not closed under addition. Hence, $4\mathbb{Z}\cup 6\mathbb{Z}$ is not a subgroup of $(\mathbb{Z},+)$. $\square$ ::: This calculator decides whether $m\mathbb{Z}\cup n\mathbb{Z}$ is a subgroup of $\mathbb{Z}$. The union is a subgroup exactly when $m\mathbb{Z}\subseteq n\mathbb{Z}$ or $n\mathbb{Z}\subseteq m\mathbb{Z}$. For integer subgroups, this is checked by divisibility of the generators. :::calculator[Union of Integer Subgroups Calculator] ::: :::exercise[Exercises on Unions of Subgroups] 1. Decide whether $2\mathbb{Z}\cup 6\mathbb{Z}$ is a subgroup of $(\mathbb{Z},+)$. 2. Decide whether $2\mathbb{Z}\cup 5\mathbb{Z}$ is a subgroup of $(\mathbb{Z},+)$. 3. Let $H,K\leq G$. Prove that if $H\subseteq K$, then $H\cup K\leq G$. ::: :::answer[Answers on Unions of Subgroups] 1. Yes, because $6\mathbb{Z}\subseteq 2\mathbb{Z}$, so the union is $2\mathbb{Z}$. 2. No, because neither subgroup contains the other. Also $2+5=7$ is in neither $2\mathbb{Z}$ nor $5\mathbb{Z}$. 3. If $H\subseteq K$, then $H\cup K=K$, and $K$ is a subgroup of $G$. ::: :::faq[Frequently Asked Questions] Q: Is the union of two subgroups always a subgroup? A: No. It is a subgroup exactly when one subgroup is contained in the other. Q: Why does intersection always work but union does not? A: Intersections preserve all subgroup conditions, while unions may fail closure for mixed products. Q: Can the union of three subgroups be a subgroup? A: It can be, but the two-subgroup containment theorem does not directly describe all three-subgroup cases. ::: :::call-to-action[Continue to Products of Subgroups] subtitle: The next lesson studies sets of products of elements from two subgroups. button: Study Products of Subgroups | published/abstract-algebra/subgroups-and-normal-subgroups/subgroups/products-of-subgroups.md button-ghost: Review Intersections of Subgroups | published/abstract-algebra/subgroups-and-normal-subgroups/subgroups/intersections-of-subgroups.md :::