A subgroup is, at its core, a group hiding inside another group — a subset that inherits the ambient operation and closes under it without escaping to anything new. The concept is simple to state but surprisingly rich: the subgroup structure of a group encodes much of the group's internal architecture. Before we can understand a group through its quotients, its homomorphisms, or its symmetries, we need to understand which of its pieces are themselves groups. The integers $\mathbb{Z}$ contain the even integers $2\mathbb{Z}$, and those even integers are closed under addition, contain $0$, and admit additive inverses — they are a subgroup. The rotations of a regular polygon sit inside the full symmetry group of the polygon as a distinguished subgroup. The kernel of any homomorphism is a subgroup of the domain. These examples share a common pattern, and isolating that pattern precisely is what the definition of subgroup accomplishes. [example: Even Integers Inside the Integers] Consider the group $(\mathbb{Z}, +)$ of integers under addition, with identity $e = 0$. The subset $2\mathbb{Z} = \{\ldots, -4, -2, 0, 2, 4, \ldots\}$ of even integers satisfies: - Closure: if $m, n \in 2\mathbb{Z}$, then $m = 2j$ and $n = 2k$ for some $j, k \in \mathbb{Z}$, so \begin{align*} m + n = 2j + 2k = 2(j + k) \in 2\mathbb{Z}. \end{align*} - Identity: $0 = 2 \cdot 0 \in 2\mathbb{Z}$. - Inverses: if $m = 2k \in 2\mathbb{Z}$, then $-m = -2k = 2(-k) \in 2\mathbb{Z}$. So $2\mathbb{Z} \le \mathbb{Z}$. By the same argument, $n\mathbb{Z} = \{nk : k \in \mathbb{Z}\} \le \mathbb{Z}$ for any $n \in \mathbb{N}$. These are in fact all subgroups of $\mathbb{Z}$ — a fact we will recover as a consequence of the structure theorem for cyclic groups. The more instructive lesson, however, comes from what *fails*. Consider the odd integers $2\mathbb{Z} + 1 = \{\ldots, -3, -1, 1, 3, \ldots\}$. This set looks like a natural companion to $2\mathbb{Z}$ — it covers the other half of $\mathbb{Z}$ — but it is not a subgroup. The sum of two odd integers is even, not odd: \begin{align*} (2j + 1) + (2k + 1) = 2(j + k + 1) \in 2\mathbb{Z}. \end{align*} Closure fails at the very first step: adding two elements of $2\mathbb{Z} + 1$ lands outside $2\mathbb{Z} + 1$. The set is also missing an identity, since $0 \notin 2\mathbb{Z} + 1$. The closure axiom is not a formality — it is a genuine constraint that most subsets of a group fail. [/example] ## Definition The motivating observation is this: given a group $(G, \cdot)$ and a subset $H \subset G$, when does $H$ carry a group structure using the same operation? The answer requires exactly that $H$ not escape when you apply the operation or take inverses, and that it contain the identity. [definition: Subgroup] Let $(G, \cdot)$ be a group with identity element $e$. A subset $H \subset G$ is a **subgroup** of $G$, written $H \le G$, if the following three conditions hold: 1. **Identity**: $e \in H$. 2. **Closure**: for all $h_1, h_2 \in H$, $h_1 \cdot h_2 \in H$. 3. **Inverses**: for all $h \in H$, $h^{-1} \in H$. A subgroup $H$ of $G$ is **proper** if $H \neq G$, written $H < G$. [/definition] When these conditions hold, $H$ becomes a group in its own right under the restriction of the operation: associativity is inherited from $G$ for free, and conditions (1)–(3) supply the remaining group axioms. Both $\{e\}$ and $G$ itself are always subgroups of $G$, called the **trivial subgroup** and the **improper subgroup**, respectively. [remark: Notation] We write $H \le G$ for "H is a subgroup of G," following the notation standard that $\subset$ denotes set containment while $\le$ denotes the subgroup relation. The distinction matters: $\{2, 4\} \subset \mathbb{Z}$ but $\{2, 4\} \not\le \mathbb{Z}$ (it contains no identity element and is not closed under addition). A proper subgroup is $H < G$, analogous to strict set containment. [/remark] In practice, checking all three conditions separately is sometimes redundant. The following criterion consolidates them into a single statement, which is especially convenient when the group is finite. [quotetheorem:932] The one-step criterion is efficient but worth unpacking. Setting $a = b$ gives $b \cdot b^{-1} = e \in H$, so the identity is present. Setting $a = e$ gives $e \cdot b^{-1} = b^{-1} \in H$, so inverses are present. Then closure follows: for $a, b \in H$, we have $b^{-1} \in H$, and applying the criterion with $a$ and $(b^{-1})^{-1} = b$ gives $a \cdot b \in H$. So the single condition does indeed imply all three. The converse direction is clear from the definition. [example: Subgroup Criterion Applied] We verify that $H = \{e, r, r^2\} \le D_6$, where $D_6$ is the dihedral group of symmetries of an equilateral triangle, with $r$ denoting rotation by $120°$ and $e$ the identity. The elements satisfy $r^3 = e$. The set $H$ consists of the three rotations. Take any $a, b \in H$; then $a = r^i$ and $b = r^j$ for $i, j \in \{0, 1, 2\}$. Since $b^{-1} = r^{-j} = r^{3-j}$ (using $r^3 = e$), we have \begin{align*} a \cdot b^{-1} = r^i \cdot r^{3-j} = r^{i + 3 - j}. \end{align*} Since $i, j \in \{0, 1, 2\}$, the exponent $i + 3 - j$ lies in $\{1, 2, 3, 4, 5\}$, and reducing modulo $3$ gives a value in $\{0, 1, 2\}$. So $a \cdot b^{-1} \in H$. By the subgroup criterion, $H \le D_6$. Note that the reflections in $D_6$ — call them $s, sr, sr^2$ — do not form a subgroup: the product of two distinct reflections is a nontrivial rotation, which lies outside the set of reflections. [/example] [remark: Finite Subgroup Criterion] When $G$ is a finite group, a nonempty subset $H \subset G$ closed under the group operation is automatically a subgroup: finiteness forces every element to have finite order, and $h^{-1} = h^{n-1}$ where $n = \operatorname{ord}(h)$ (so $h^n = e$). Thus inverses are free. This simplification disappears for infinite groups: the positive integers $\mathbb{N} \subset \mathbb{Z}$ are closed under addition but form no subgroup, since $1 \in \mathbb{N}$ has no inverse in $\mathbb{N}$. [/remark] ## Cosets and the Theorem of Lagrange One of the most useful tools for understanding subgroups is the coset decomposition. Given a subgroup $H \le G$, the group $G$ partitions into disjoint translates of $H$, much as the integers partition into residue classes modulo $n$. To make this precise, we need to pick an element $g \in G$ and look at everything you can reach from $g$ by multiplying on the right by elements of $H$. The reason to fix a representative $g$ and form the set $gH$ — rather than working with the partition directly — is that $g$ selects which "sheet" of the partition we are on: two elements $g$ and $g'$ lie in the same piece precisely when $g^{-1}g' \in H$, and naming that piece $gH$ gives us a concrete set to manipulate. Before stating this precisely, we need the definition. [definition: Left Coset] Let $H \le G$ and $g \in G$. The **left coset** of $H$ with representative $g$ is the set \begin{align*} gH = \{g \cdot h : h \in H\}. \end{align*} The collection of all left cosets of $H$ in $G$ is denoted $G/H$ (as a set of cosets, not a group — that requires more structure). The **index** of $H$ in $G$, written $[G : H]$, is the number of distinct left cosets of $H$ in $G$. [/definition] <!-- NOTATION PROPOSAL: The notation $G/H$ for the set of left cosets is used here with the caveat that it does not denote a group unless $H$ is normal. This warning should be explicit, since the same notation $G/H$ means the quotient group when $H \trianglelefteq G$. Consider adding a note to the notation standards clarifying the dual usage. --> The crucial observation is that cosets partition $G$: every element of $G$ belongs to exactly one left coset of $H$, since two cosets either coincide or are disjoint (this follows from the equivalence relation $g \sim g'$ iff $g^{-1}g' \in H$). Each coset $gH$ has exactly $|H|$ elements (since $h \mapsto gh$ is a bijection $H \to gH$). Counting gives Lagrange's theorem, the first great structural result about subgroups. [quotetheorem:841] The proof is exactly the counting argument sketched above: $G$ is a disjoint union of $[G:H]$ left cosets, each of size $|H|$. The equality $|G| = [G:H] \cdot |H|$ is then immediate. Lagrange's theorem is a powerful constraint. It immediately implies that the order of any element $g \in G$ — which equals $|\langle g \rangle|$, the order of the cyclic subgroup generated by $g$ — must divide $|G|$. In particular, $g^{|G|} = e$ for every $g \in G$, which is the group-theoretic version of Fermat's little theorem (take $G = (\mathbb{Z}/p\mathbb{Z})^\times$). [example: Subgroups of a Group of Prime Order] Let $G$ be a group with $|G| = p$ where $p$ is prime. By Lagrange's theorem, any subgroup $H \le G$ has order dividing $p$. Since $p$ is prime, the only divisors of $p$ are $1$ and $p$, so $|H| = 1$ or $|H| = p$. This means $H = \{e\}$ or $H = G$. Therefore $G$ has no proper nontrivial subgroups. Moreover, for any $g \neq e$ in $G$, the cyclic subgroup $\langle g \rangle$ is a subgroup of order greater than $1$, hence $\langle g \rangle = G$. So $G$ is cyclic, generated by any non-identity element. Every group of prime order is cyclic, isomorphic to $\mathbb{Z}/p\mathbb{Z}$. For comparison, the group $\mathbb{Z}/6\mathbb{Z}$ has order $6 = 2 \cdot 3$, and Lagrange allows subgroups of orders $1, 2, 3, 6$. Indeed, $\langle \bar{3} \rangle = \{0, 3\}$ has order $2$, and $\langle \bar{2} \rangle = \{0, 2, 4\}$ has order $3$. [/example] [remark: The Converse of Lagrange's Theorem Fails] Lagrange's theorem tells us that if $H \le G$ then $|H|$ divides $|G|$, but the converse is false in general: if $n \mid |G|$, there need not exist a subgroup of order $n$. The standard counterexample is the alternating group $A_4$, which has order $12$. Lagrange allows a subgroup of order $6$, but no such subgroup exists. This gap motivates the Sylow theorems, which recover partial converses for prime power divisors. [/remark] ## Generated Subgroups and Cyclic Groups Given any subset $S \subset G$, one often wants the smallest subgroup of $G$ containing $S$. This is the subgroup *generated* by $S$, and its existence requires knowing that intersections of subgroups are subgroups. The intersection property is the key fact that makes generated subgroups well-defined, and it deserves to be stated as a theorem in its own right. [quotetheorem:767] The verification is direct. Each $H_i$ contains $e$, so $e \in \bigcap H_i$. If $a, b \in \bigcap H_i$, then $a, b \in H_i$ for every $i$, so $ab^{-1} \in H_i$ for every $i$ (since each $H_i \le G$), and hence $ab^{-1} \in \bigcap H_i$. By the subgroup criterion, $\bigcap H_i \le G$. This theorem is what licenses the following definition: the intersection is taken over a nonempty collection of subgroups (at minimum, $G$ itself contains $S$), and the theorem guarantees the result is again a subgroup. [definition: Generated Subgroup] Let $G$ be a group and $S \subset G$ a nonempty subset. The **subgroup generated by $S$**, written $\langle S \rangle$, is the intersection of all subgroups of $G$ that contain $S$: \begin{align*} \langle S \rangle = \bigcap_{\substack{H \le G \\ S \subset H}} H. \end{align*} This is the smallest subgroup of $G$ containing $S$. [/definition] Concretely, $\langle S \rangle$ consists of all finite products of elements of $S$ and their inverses: $\langle S \rangle = \{s_1^{\varepsilon_1} \cdots s_k^{\varepsilon_k} : k \ge 0,\, s_i \in S,\, \varepsilon_i = \pm 1\}$, where the empty product is $e$. This is the set-theoretic description; the intersection definition is the conceptually cleaner one. When $S = \{g\}$ is a single element, the generated subgroup takes a particularly simple form. [definition: Cyclic Subgroup] Let $G$ be a group and $g \in G$. The **cyclic subgroup generated by $g$** is \begin{align*} \langle g \rangle = \{g^n : n \in \mathbb{Z}\}, \end{align*} where $g^0 = e$, $g^n = g \cdot g \cdots g$ ($n$ times) for $n \ge 1$, and $g^{-n} = (g^{-1})^n$ for $n \ge 1$. The **order** of $g$, written $\operatorname{ord}(g)$, is the smallest positive integer $n$ such that $g^n = e$, or $\infty$ if no such $n$ exists. [/definition] When $\operatorname{ord}(g) = n < \infty$, we have $\langle g \rangle = \{e, g, g^2, \ldots, g^{n-1}\}$, a group of order $n$ isomorphic to $\mathbb{Z}/n\mathbb{Z}$. When $\operatorname{ord}(g) = \infty$, the powers $g^n$ for $n \in \mathbb{Z}$ are all distinct and $\langle g \rangle \cong \mathbb{Z}$. [example: Cyclic Subgroups of the Symmetric Group] In the symmetric group $S_4$ (permutations of $\{1, 2, 3, 4\}$), consider the 4-cycle $\sigma = (1\; 2\; 3\; 4)$. Computing powers: \begin{align*} \sigma^1 &= (1\;2\;3\;4), \\ \sigma^2 &= (1\;3)(2\;4), \\ \sigma^3 &= (1\;4\;3\;2), \\ \sigma^4 &= \operatorname{id}. \end{align*} To verify: $\sigma^2$ sends $1 \mapsto 3 \mapsto 1$ and $2 \mapsto 4 \mapsto 2$, confirming it is the product of two 2-cycles. So $\operatorname{ord}(\sigma) = 4$ and $\langle \sigma \rangle = \{\operatorname{id}, (1\;2\;3\;4), (1\;3)(2\;4), (1\;4\;3\;2)\}$, a cyclic subgroup of order $4$ in $S_4$. Now consider $\tau = (1\;2)(3\;4)$. Then $\tau^2 = \operatorname{id}$ (applying $\tau$ twice returns every element to its start), so $\operatorname{ord}(\tau) = 2$ and $\langle \tau \rangle = \{\operatorname{id}, (1\;2)(3\;4)\}$. Lagrange's theorem is consistent: $4$ and $2$ both divide $|S_4| = 24$. [/example] ## Normal Subgroups and Quotient Groups Not every subgroup interacts with the ambient group in the same way. The left cosets $gH$ and right cosets $Hg$ of a subgroup $H$ can differ, and when they do, the set of cosets $G/H$ cannot be made into a group in any natural way. The subgroups for which left and right cosets coincide are exactly the ones that allow quotient group constructions, and they are important enough to deserve their own definition. The failure of left and right cosets to coincide has a concrete meaning: there exist $g \in G$ and $h \in H$ such that $ghg^{-1} \notin H$. The element $ghg^{-1}$ is the *conjugate* of $h$ by $g$. A subgroup that is closed under all conjugations — that is, $gHg^{-1} = H$ for every $g \in G$ — is normal. [citedefinition:Normal Subgroup] [illustration:cosets-normal-vs-non-normal] Every subgroup of an abelian group is normal, since $ghg^{-1} = h$ for all $g, h$ when the group is commutative. The center $Z(G) = \{g \in G : gx = xg \text{ for all } x \in G\}$ is always normal. The kernel of any group homomorphism is normal. [explanation: Why Normality Makes Cosets a Group] Given $N \trianglelefteq G$, the set of left cosets $G/N = \{gN : g \in G\}$ becomes a group under the operation \begin{align*} (g_1 N) \cdot (g_2 N) = (g_1 g_2) N. \end{align*} For this to be well-defined, we need the product to be independent of the choice of coset representatives. Suppose $g_1 N = g_1' N$ and $g_2 N = g_2' N$, so $g_1' = g_1 n_1$ and $g_2' = g_2 n_2$ for some $n_1, n_2 \in N$. Then: \begin{align*} g_1' g_2' = g_1 n_1 g_2 n_2 = g_1 g_2 (g_2^{-1} n_1 g_2) n_2. \end{align*} Since $N$ is normal, $g_2^{-1} n_1 g_2 \in N$. So $(g_2^{-1} n_1 g_2) n_2 \in N$, and thus $g_1' g_2' N = g_1 g_2 N$. The operation is well-defined precisely because $N$ is closed under conjugation. If $N$ is not normal, this argument breaks down. For instance, take $G = S_3$ and $H = \langle (1\;2) \rangle = \{e, (1\;2)\}$. The cosets are $H$, $(1\;3)H = \{(1\;3), (1\;2\;3)\}$, and $(2\;3)H = \{(2\;3), (1\;3\;2)\}$. Attempting to multiply cosets, say $((1\;3)H) \cdot ((1\;3)H)$: choosing representative $(1\;3)$ twice gives $(1\;3)(1\;3) = e \in H$; choosing representatives $(1\;3)$ and $(1\;2\;3)$ gives $(1\;3)(1\;2\;3) = (2\;3) \in (2\;3)H$. The product depends on the choice of representative, so the operation is not well-defined. [/explanation] [quotetheorem:790] The quotient group $G/N$ has order $[G:N]$, and when $G$ is finite, $|G/N| = |G|/|N|$ by Lagrange's theorem. The quotient "collapses" $N$ to a single point (the identity) and identifies elements that differ by an element of $N$. [example: Quotient of the Integers] The subgroup $n\mathbb{Z} \trianglelefteq \mathbb{Z}$ (normal automatically, since $\mathbb{Z}$ is abelian). The quotient group is \begin{align*} \mathbb{Z} / n\mathbb{Z} = \{\bar{0}, \bar{1}, \ldots, \overline{n-1}\}, \end{align*} where $\bar{k} = k + n\mathbb{Z} = \{k + mn : m \in \mathbb{Z}\}$ is the residue class of $k$ modulo $n$. The group operation is addition modulo $n$: $\bar{j} + \bar{k} = \overline{j+k}$. This is well-defined because $n\mathbb{Z}$ is normal (in fact, it is the only reason we can form this quotient using the coset product — but since $\mathbb{Z}$ is abelian, this is automatic). The quotient map $\pi: \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ sends $k \mapsto \bar{k}$. Its kernel is $n\mathbb{Z}$, consistent with the theorem. [/example] ## The Isomorphism Theorems The relationship between subgroups, normal subgroups, and homomorphisms is crystallized by the isomorphism theorems. These are the key structural results that explain how subgroups behave under maps between groups. The most fundamental is the first isomorphism theorem, which says that every homomorphism factors through a quotient by its kernel. [quotetheorem:842] The first isomorphism theorem tells us that understanding homomorphisms is equivalent to understanding normal subgroups and quotients. Every surjective homomorphism $G \to H$ realizes $H$ as a quotient of $G$; every normal subgroup $N$ arises as the kernel of the quotient map $G \to G/N$. [example: Symmetric Group Surjecting onto Cyclic Group of Order Two] Define $\varphi: S_n \to \{1, -1\}$ (multiplicative group) by $\varphi(\sigma) = \operatorname{sgn}(\sigma)$, the sign of the permutation. This is a homomorphism: $\operatorname{sgn}(\sigma\tau) = \operatorname{sgn}(\sigma)\operatorname{sgn}(\tau)$. The kernel is $\ker \varphi = \{\sigma \in S_n : \operatorname{sgn}(\sigma) = 1\} = A_n$, the alternating group of even permutations. By the first isomorphism theorem: \begin{align*} S_n / A_n \cong \{1, -1\} \cong \mathbb{Z}/2\mathbb{Z}. \end{align*} Since $|S_n / A_n| = 2$, we get $|A_n| = |S_n|/2 = n!/2$. The index $[S_n : A_n] = 2$ is also immediately clear: the two cosets are the even permutations $A_n$ and the odd permutations $S_n \setminus A_n$. Note that $A_n \trianglelefteq S_n$: it is the kernel of a homomorphism and hence normal. This gives the cleanest proof that $A_n$ is normal in $S_n$. [/example] The second and third isomorphism theorems give further control over how subgroups interact. [quotetheorem:843] [quotetheorem:844] The third isomorphism theorem is sometimes phrased as "cancellation of quotients": $N$ in the numerator and denominator cancel. It captures the intuition that if you have a chain of normal subgroups $N \trianglelefteq M \trianglelefteq G$, then the quotient by $M$ sees the same thing whether you first mod out by $N$ or not. ## The Correspondence Theorem and Subgroup Lattices One of the most powerful structural results is the correspondence theorem, which completely describes the subgroups of a quotient group $G/N$ in terms of the subgroups of $G$ that contain $N$. [quotetheorem:854] The correspondence theorem says that the quotient map does not create or destroy subgroup structure — it simply restricts our view to those subgroups that contain $N$. Understanding $G/N$ is exactly as hard as understanding which subgroups of $G$ lie above $N$ in the **subgroup lattice**, the partially ordered set of all subgroups of $G$ ordered by inclusion. [illustration:s3-subgroup-lattice] [example: Subgroups of the Cyclic Group of Order Twelve] Consider $G = \mathbb{Z}/12\mathbb{Z} = \langle \bar{1} \rangle$. The subgroups of $G$ correspond bijectively to the divisors of $12$, since $\mathbb{Z}/12\mathbb{Z}$ is cyclic: for each divisor $d$ of $12$, there is a unique subgroup of order $d$, generated by $\overline{12/d}$. | Divisor $d$ | Subgroup of order $d$ | Generator | |---|---|---| | $1$ | $\{\bar{0}\}$ | $\bar{0}$ | | $2$ | $\{\bar{0}, \bar{6}\}$ | $\bar{6}$ | | $3$ | $\{\bar{0}, \bar{4}, \bar{8}\}$ | $\bar{4}$ | | $4$ | $\{\bar{0}, \bar{3}, \bar{6}, \bar{9}\}$ | $\bar{3}$ | | $6$ | $\{\bar{0}, \bar{2}, \bar{4}, \bar{6}, \bar{8}, \bar{10}\}$ | $\bar{2}$ | | $12$ | $\mathbb{Z}/12\mathbb{Z}$ | $\bar{1}$ | All subgroups are normal (since $G$ is abelian). The subgroup $\langle \bar{4} \rangle$ of order $3$ is contained in the subgroup $\langle \bar{2} \rangle$ of order $6$, reflecting the divisibility $3 \mid 6$. The quotient $\langle \bar{2} \rangle / \langle \bar{4} \rangle$ has order $2$, consistent with $[6:3] = 2$. The correspondence theorem applies to $N = \langle \bar{4} \rangle \cong \mathbb{Z}/3\mathbb{Z}$: the subgroups of $G/N \cong \mathbb{Z}/4\mathbb{Z}$ are in bijection with subgroups of $G$ containing $\langle \bar{4} \rangle$, namely $\langle \bar{4} \rangle$, $\langle \bar{2} \rangle$, and $G$ itself — three subgroups corresponding to the three subgroups of $\mathbb{Z}/4\mathbb{Z}$. [/example] ## References - J. Rotman, *An Introduction to the Theory of Groups* (1995). - D. Dummit and R. Foote, *Abstract Algebra* (2004). - M. Artin, *Algebra* (2011). - I. Herstein, *Topics in Algebra* (1975).