The subgroup criteria show how to verify a subgroup in general. For finite subsets, the verification becomes even shorter. The focus keyword for this lecture is finite subgroup. In a finite closed subset of a group, repeated powers of an element cannot remain distinct forever. This repetition forces the identity and inverse elements to appear. In this class note, we shall prove the finite subgroup theorem and apply it to examples where closure alone determines the subgroup property. :::theorem[Finite Subgroup Criterion] Let $(G,\circ)$ be a group and let $H$ be a non-empty finite subset of $G$. Then $H$ is a subgroup of $G$ if and only if $$\begin{equation*} a\circ b\in H\quad \forall\,a,b\in H. \end{equation*}$$ ::: :::proof[Proof of Finite Subgroup Criterion] Given that $(G,\circ)$ is a group and $H$ is a non-empty finite subset of $G$. To prove that $H$ is a subgroup of $G$ if and only if $a\circ b\in H$ for all $a,b\in H$. [1] Let $H$ be a subgroup of $G$. Then $H$ is closed under $\circ$. Therefore $$\begin{equation*} a\circ b\in H\quad \forall\,a,b\in H. \end{equation*}$$ [2] Let $$\begin{equation*} a\circ b\in H\quad \forall\,a,b\in H. \end{equation*}$$ To prove that $H$ is a subgroup of $G$. Let $a\in H$. Since $H$ is closed under $\circ$, we get $$\begin{equation*} a,a^2,a^3,\ldots \in H. \end{equation*}$$ Since $H$ is finite, there exist positive integers $r$ and $s$ such that $r<s$ and $$\begin{equation*} a^r=a^s. \end{equation*}$$ Multiplying by $a^{-r}$ in $G$, we get $$\begin{align*} a^r=a^s &\implies e=a^{s-r}. \end{align*}$$ Since $s-r\in\mathbb{N}$ and $a^{s-r}\in H$, we get $e\in H$. Now $$\begin{align*} a^{s-r}=e &\implies a^{-1}=a^{s-r-1}. \end{align*}$$ If $s-r=1$, then $a^{-1}=e\in H$. If $s-r>1$, then $a^{s-r-1}\in H$ by closure. Therefore $a^{-1}\in H$. Thus $a^{-1}\in H$ for all $a\in H$. By the subgroup criterion, $H$ is a subgroup of $G$. Hence, $H$ is a subgroup of $G$ if and only if $H$ is closed under $\circ$. $\square$ ::: The finite condition cannot be removed. For example, $\mathbb{N}$ is closed under addition, but it is not a subgroup of $(\mathbb{Z},+)$ because additive inverses are missing. In a finite set, closure under the operation traps all repeated powers inside the same finite set, so repetition is unavoidable. That repetition produces the identity and then the inverse. This is the main mathematical reason behind the finite subgroup criterion. :::example[Subgroup of Integers Modulo Six] Let $H=\{0,3\}$ be a subset of $\mathbb{Z}_6$ under addition modulo $6$. Since $H$ is finite and non-empty, it is enough to check closure. We compute $$\begin{equation*} 0+0\equiv 0,\quad 0+3\equiv 3,\quad 3+0\equiv 3,\quad 3+3\equiv 0\pmod 6. \end{equation*}$$ Every sum belongs to $H$. Hence, $H$ is a subgroup of $\mathbb{Z}_6$. ::: :::example[Failure of Closure in Integers Modulo Six] Let $K=\{0,2\}$ be a subset of $\mathbb{Z}_6$ under addition modulo $6$. Then $K$ is finite and non-empty. But $$\begin{equation*} 2+2\equiv 4\pmod 6 \end{equation*}$$ and $4\notin K$. Therefore $K$ is not closed under addition modulo $6$. Hence, $K$ is not a subgroup of $\mathbb{Z}_6$. ::: We observe the finite subgroup criterion in $\mathbb{Z}_n$. Choose a finite subset, and the preview checks every possible sum of two elements. In a finite group, if this closure table stays inside the subset, then identity and inverses are forced automatically. This is why the finite theorem needs only closure, while the general subgroup criterion also checks inverses. :::scientific-preview[Finite Closure Table Explorer] ::: The word finite does essential work. For infinite subsets such as $\mathbb{N}\subseteq\mathbb{Z}$, closure under addition does not force inverses. :::solved-problem[Finite Closed Subset] Let $G$ be a group and let $H=\{e,a,a^2\}$, where $a^3=e$. Prove that $H$ is a subgroup of $G$. ::: :::solution[Solution of Finite Closed Subset] Let $(G,\circ)$ be a group and let $a\in G$ such that $a^3=e$. Let $$\begin{equation*} H=\{e,a,a^2\}. \end{equation*}$$ To prove that $H$ is a subgroup of $G$. Since $e\in H$, the set $H$ is non-empty. Also $H$ is finite. It remains to check closure. The products are governed by the rule $$\begin{equation*} a^i\circ a^j=a^{i+j}. \end{equation*}$$ Since $a^3=e$, every power of $a$ reduces to one of $e,a,a^2$. Therefore $$\begin{equation*} a^i\circ a^j\in H\quad \forall\,a^i,a^j\in H. \end{equation*}$$ Thus $H$ is closed under $\circ$. By the finite subgroup criterion, $H$ is a subgroup of $G$. Hence, $H\leq G$. $\square$ ::: :::solved-problem[Squares 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 $H=\{x^2:x\in G\}$ is a subgroup of $G$. ::: :::solution[Solution of Squares 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 $$\begin{equation*} H=\{x^2:x\in G\}. \end{equation*}$$ To prove that $H$ is a subgroup of $G$. Since $e=e^2$, we get $e\in H$. Therefore $H$ is non-empty. Let $x^2,y^2\in H$, where $x,y\in G$. Using the given condition with $a=y$ and $b=x$, we get $$\begin{align*} (y\circ x)^3=y^3\circ x^3 &\implies y\circ x\circ y\circ x\circ y\circ x=y\circ y\circ y\circ x\circ x\circ x\\ &\implies x\circ y\circ x\circ y=y^2\circ x^2 \quad [\because \text{ cancellation laws}]. \end{align*}$$ Interchanging $x$ and $y$, we get $$\begin{equation*} y\circ x\circ y\circ x=x^2\circ y^2. \end{equation*}$$ Therefore $$\begin{align*} x^2\circ y^2 &=y\circ x\circ y\circ x\\ &=(y\circ x)^2\in H. \end{align*}$$ Thus $H$ is closed under $\circ$. Let $x^2\in H$. Then $$\begin{align*} (x^2)^{-1} &=x^{-2}\\ &=(x^{-1})^2\in H. \end{align*}$$ Therefore $(x^2)^{-1}\in H$ for all $x^2\in H$. By the subgroup criterion, $H$ is a subgroup of $G$. Hence, $H\leq G$. $\square$ ::: This calculator tests the finite subgroup criterion for subsets of $\mathbb{Z}_n$. Enter a modulus and a subset. The output says whether closure under addition modulo $n$ holds and gives the subgroup conclusion when it does. It is a closure calculator, not a general infinite-subset test. :::calculator[Finite Subgroup Closure Calculator] ::: :::exercise[Exercises on Finite Subgroups] 1. Prove that $\{0,2,4\}$ is a subgroup of $\mathbb{Z}_6$ under addition modulo $6$. 2. Show that $\{0,1,3\}$ is not a subgroup of $\mathbb{Z}_4$ under addition modulo $4$. 3. Let $a\in G$ and $a^4=e$. Prove that $\{e,a,a^2,a^3\}$ is a subgroup of $G$. ::: :::answer[Answers on Finite Subgroups] 1. The set is finite and non-empty. All sums modulo $6$ remain in $\{0,2,4\}$, so it is a subgroup. 2. Since $1+1\equiv 2\pmod 4$ and $2\notin\{0,1,3\}$, closure fails. 3. Products of elements have the form $a^i a^j=a^{i+j}$, and since $a^4=e$, every product reduces to one of $e,a,a^2,a^3$. ::: :::faq[Frequently Asked Questions] Q: Why does closure alone work for finite subsets? A: Because repeated powers must repeat in a finite set, forcing the identity and inverses to appear. Q: Does closure alone work for infinite subsets? A: No. The set $\mathbb{N}$ is closed under addition but is not a subgroup of $\mathbb{Z}$. Q: Must the finite subset be non-empty? A: Yes. A group must contain an identity element. ::: :::call-to-action[Continue to Intersections of Subgroups] subtitle: Learn why the common part of subgroups is again a subgroup. button: Study Intersections of Subgroups | published/abstract-algebra/subgroups-and-normal-subgroups/subgroups/intersections-of-subgroups.md button-ghost: Review Subgroup Criteria | published/abstract-algebra/subgroups-and-normal-subgroups/subgroups/subgroup-criteria.md :::