The definition of subgroup is exact, but checking all group axioms every time is inefficient. Since a subset of a group inherits associativity from the larger group, the real work is to verify closure, identity, and inverses. The focus keyword for this lecture is subgroup criteria. We shall prove two standard tests: the two-condition subgroup criterion and the one-step subgroup criterion. These criteria are among the most frequently used tools in group theory. :::theorem[Subgroup Criterion] Let $(G,\circ)$ be a group and let $H$ be a non-empty subset of $G$. Then $H$ is a subgroup of $G$ if and only if the following conditions hold: (i) $a\circ b\in H$ for all $a,b\in H$. (ii) $a^{-1}\in H$ for all $a\in H$. ::: :::proof[Proof of Subgroup Criterion] Given that $(G,\circ)$ is a group and $H$ is a non-empty subset of $G$. To prove that $H$ is a subgroup of $G$ if and only if $a\circ b\in H$ for all $a,b\in H$ and $a^{-1}\in H$ for all $a\in H$. [1] Let $H$ be a subgroup of $G$. Then $(H,\circ)$ is a group. Therefore $H$ is closed under $\circ$, that is, $$\begin{equation*} a\circ b\in H\quad \forall\,a,b\in H. \end{equation*}$$ Also every element of $H$ has its inverse in $H$, that is, $$\begin{equation*} a^{-1}\in H\quad \forall\,a\in H. \end{equation*}$$ [2] Let $$\begin{equation*} a\circ b\in H\quad \forall\,a,b\in H \end{equation*}$$ and $$\begin{equation*} a^{-1}\in H\quad \forall\,a\in H. \end{equation*}$$ To prove that $H$ is a subgroup of $G$. Since $H$ is non-empty, there exists $h\in H$. By the inverse condition, $h^{-1}\in H$. By closure, $$\begin{align*} h\circ h^{-1}\in H &\implies e\in H. \end{align*}$$ Let $a,b,c\in H$. Since associativity holds in $G$, we get $$\begin{align*} (a\circ b)\circ c &=a\circ(b\circ c). \end{align*}$$ Also $e\in H$ and $$\begin{equation*} e\circ a=a\circ e=a\quad \forall\,a\in H. \end{equation*}$$ By the inverse condition, $a^{-1}\in H$ for every $a\in H$. Therefore $(H,\circ)$ is a group. Hence, $H$ is a subgroup of $G$. $\square$ ::: :::theorem[One-Step Subgroup Criterion] Let $(G,\circ)$ be a group and let $H$ be a non-empty subset of $G$. Then $H$ is a subgroup of $G$ if and only if $$\begin{equation*} a\circ b^{-1}\in H\quad \forall\,a,b\in H. \end{equation*}$$ ::: :::proof[Proof of One-Step Subgroup Criterion] Given that $(G,\circ)$ is a group and $H$ is a non-empty subset of $G$. To prove that $H$ is a subgroup of $G$ if and only if $a\circ b^{-1}\in H$ for all $a,b\in H$. [1] Let $H$ be a subgroup of $G$. Let $a,b\in H$. Since $H$ is a subgroup of $G$, $b^{-1}\in H$. Since $H$ is closed under $\circ$, we get $$\begin{equation*} a\circ b^{-1}\in H. \end{equation*}$$ Therefore $$\begin{equation*} a\circ b^{-1}\in H\quad \forall\,a,b\in H. \end{equation*}$$ [2] Let $$\begin{equation*} a\circ b^{-1}\in H\quad \forall\,a,b\in H. \end{equation*}$$ To prove that $H$ is a subgroup of $G$. Since $H$ is non-empty, there exists $h\in H$. Using the given condition with $a=h$ and $b=h$, we get $$\begin{align*} h\circ h^{-1}\in H &\implies e\in H. \end{align*}$$ Let $b\in H$. Using the given condition with $a=e$ and $b=b$, we get $$\begin{align*} e\circ b^{-1}\in H &\implies b^{-1}\in H. \end{align*}$$ Therefore $H$ is closed under inverses. Let $a,b\in H$. Since $b^{-1}\in H$, we may apply the given condition to $a$ and $b^{-1}$. Thus $$\begin{align*} a\circ (b^{-1})^{-1}\in H &\implies a\circ b\in H. \end{align*}$$ Therefore $H$ is closed under $\circ$. By the subgroup criterion, $H$ is a subgroup of $G$. Hence, $H$ is a subgroup of $G$ if and only if $a\circ b^{-1}\in H$ for all $a,b\in H$. $\square$ ::: The first criterion separates multiplication and inverses. The second criterion combines them into a single expression $a\circ b^{-1}$. In additive groups, the one-step condition becomes $a-b\in H$ for all $a,b\in H$. A common mistake is to test the condition only for a few convenient elements. The condition must hold for every pair of elements in the subset. We observe the one-step subgroup criterion in $\mathbb{Z}_n$ under addition. Enter a modulus and a finite subset, and the preview builds the table of all differences $a-b$. A subset passes the test exactly when every entry in this difference table lies back inside the subset. This turns the universal phrase for all $a,b\in H$ into a visible finite check. :::scientific-preview[One-Step Difference Table Explorer] ::: Each row-column entry is one required test of $a-b\in H$. A single outside entry is enough to disprove the subgroup property. :::solved-problem[Multiples of an Integer] Let $n\in\mathbb{Z}$. Prove that $n\mathbb{Z}$ is a subgroup of $(\mathbb{Z},+)$. ::: :::solution[Solution of Multiples of an Integer] Let $H=n\mathbb{Z}=\{nk:k\in\mathbb{Z}\}$. Given that $(\mathbb{Z},+)$ is a group. To prove that $H$ is a subgroup of $\mathbb{Z}$. Since $0=n\cdot 0$, we get $0\in H$. Therefore $H$ is non-empty. Let $a,b\in H$. Then there exist $r,s\in\mathbb{Z}$ such that $$\begin{equation*} a=nr,\quad b=ns. \end{equation*}$$ In additive notation, the one-step subgroup criterion requires $a-b\in H$. Now $$\begin{align*} a-b &=nr-ns\\ &=n(r-s). \end{align*}$$ Since $r-s\in\mathbb{Z}$, we get $a-b\in H$. Hence, $n\mathbb{Z}$ is a subgroup of $(\mathbb{Z},+)$. $\square$ ::: :::example[Natural Numbers Fail the Criterion] Let $H=\mathbb{N}$ be considered as a subset of $(\mathbb{Z},+)$. Take $a=1$ and $b=2$. Then $$\begin{align*} a-b &=1-2\\ &=-1. \end{align*}$$ Since $-1\notin\mathbb{N}$, the one-step subgroup criterion fails. Hence, $\mathbb{N}$ is not a subgroup of $(\mathbb{Z},+)$. ::: Use the calculator to test subsets of $\mathbb{Z}_n$ under addition modulo $n$. Enter a positive integer $n$ and a comma-separated subset such as $0,2,4$. Click the button to check the one-step condition $a-b\in H$. Try $n=6$ with $0,3$, and then compare it with $0,2$. The contrast shows why every pair must be tested. :::calculator[Subgroup Criterion Calculator] ::: :::exercise[Exercises on Subgroup Criteria] 1. Use the one-step subgroup criterion to prove that $2\mathbb{Z}\leq\mathbb{Z}$. 2. Decide whether $\{0,3\}$ is a subgroup of $\mathbb{Z}_6$ under addition modulo $6$. 3. Decide whether $\{0,2\}$ is a subgroup of $\mathbb{Z}_6$ under addition modulo $6$. ::: :::answer[Answers on Subgroup Criteria] 1. If $a=2r$ and $b=2s$, then $a-b=2(r-s)\in 2\mathbb{Z}$. 2. Yes. The differences modulo $6$ are $0$ and $3$, both of which lie in the set. 3. No. Since $0-2\equiv 4\pmod 6$ and $4\notin\{0,2\}$, the one-step criterion fails. ::: :::faq[Frequently Asked Questions] Q: Why is associativity not tested separately? A: Associativity is inherited from the group $G$. Q: What is the additive form of the one-step criterion? A: It is $a-b\in H$ for all $a,b\in H$. Q: Can the empty set be handled by the one-step criterion? A: No. The theorem requires $H$ to be non-empty. ::: :::call-to-action[Continue to Finite Subgroups] subtitle: In finite subsets, closure alone is enough to force the subgroup structure. button: Study Finite Subgroups | published/abstract-algebra/subgroups-and-normal-subgroups/subgroups/finite-subgroups.md button-ghost: Review Subgroups | published/abstract-algebra/subgroups-and-normal-subgroups/subgroups/subgroups.md :::