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 $G = (\mathbb{Z}/p\mathbb{Z})^\times$).