---
title: Group Axioms and First Examples
module: Groups and Symmetry
moduleNumber: 2
lessonNumber: 1
order: 201
summary: >
  A group is a set with one associative operation that has an identity and
  inverses. We state the axioms, prove that the identity, inverses, and
  cancellation behave as expected, define the order of a group and of an
  element, and catalogue the running examples: the integers, the additive group
  of residues mod n, and the multiplicative group of units mod n.
topics: [Groups and Symmetry]
sources:
  - book: Dummit & Foote
    ref: "Ch. 1 — Introduction to Groups; §1.1 Basic Axioms and Examples"
  - book: Judson
    ref: "Ch. 3 — Groups; §3.2 Definitions and Examples"
draft: false
---

Arithmetic on the integers, the symmetries of a triangle, and the invertible
matrices over a field have elements of entirely different kinds, yet they share a
common shape: one operation that combines two elements into a third, an element
that changes nothing, and for every element something that undoes it. Group
theory studies that shape in isolation. Reducing the axioms to those few facts
lets one theorem about groups apply at once to number systems, geometric
symmetry, and permutations.

## Binary operations

A group's operation is a rule for combining two elements of a set into one.

> **Definition (Binary operation).** A **binary operation** on a set $G$ is a
> function $\ast : G \times G \to G$. We write $a \ast b$ for $\ast(a, b)$. The
> operation is **associative** if $(a \ast b) \ast c = a \ast (b \ast c)$ for all
> $a, b, c \in G$, and **commutative** if $a \ast b = b \ast a$ for all
> $a, b \in G$; two particular elements $a, b$ **commute** when $a \ast b = b \ast a$.

Two conditions are hidden in the phrase "function $G \times G \to G$." The
operation must be defined for _every_ ordered pair, and its output must land back
_inside_ $G$. The second condition is **closure**; many candidate operations fail
it.[^df-binop]

- Ordinary $+$ and $\times$ are commutative binary operations on
  $\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}$.
- Subtraction is a binary operation on $\mathbb{Z}$ (noncommutative), but _not_
  on $\mathbb{Z}^+$: for $a < b$ with $a, b \in \mathbb{Z}^+$, $a - b \notin \mathbb{Z}^+$,
  so $-$ fails to map $\mathbb{Z}^+ \times \mathbb{Z}^+$ into $\mathbb{Z}^+$.
- The cross product on $\mathbb{R}^3$ is a binary operation that is neither
  associative nor commutative.

When $H \subseteq G$ and $a \ast b \in H$ for all $a, b \in H$, the subset $H$ is
**closed** under $\ast$. Associativity and commutativity are inherited for free:
if $\ast$ is associative on $G$ and restricts to a binary operation on $H$, it is
automatically associative on $H$.

## The group axioms

> **Definition (Group).** A **group** is an ordered pair $(G, \ast)$ of a set $G$
> and a binary operation $\ast$ on $G$ satisfying:
>
> - **(G1) Associativity.** $(a \ast b) \ast c = a \ast (b \ast c)$ for all $a, b, c \in G$.
> - **(G2) Identity.** There is an element $e \in G$ with $a \ast e = e \ast a = a$ for all $a \in G$.
> - **(G3) Inverses.** For each $a \in G$ there is $a^{-1} \in G$ with $a \ast a^{-1} = a^{-1} \ast a = e$.
>
> The group is **abelian** (or **commutative**) if in addition $a \ast b = b \ast a$
> for all $a, b \in G$. It is **finite** if $G$ is a finite set.

Axiom (G2) forces $G$ to be nonempty, since it asserts $e$ exists. From here we
drop the operation symbol: for an abstract group we write $ab$ for $a \ast b$ and
denote the identity by $1$. When the operation is addition we write $0$ for the
identity and $-a$ for the inverse of $a$.

$$
% caption: The four conditions a set-with-operation must pass to be a group;
% associativity and closure are the two most often overlooked.
\begin{tikzpicture}[font=\small,
  chk/.style={draw, minimum width=52mm, minimum height=11mm, align=left, anchor=west}]
  \definecolor{acc}{HTML}{2A6F97}
  \node[chk] (c0) at (0,3.6)  {\textbf{Closure:} $ab$ lies in $G$};
  \node[chk] (c1) at (0,2.4)  {\textbf{Associativity:} $(ab)c = a(bc)$};
  \node[chk, draw=acc] (c2) at (0,1.2) {\textbf{Identity:} some $e$ leaves each $a$};
  \node[chk, draw=acc] (c3) at (0,0.0) {\textbf{Inverse:} every $a$ has an inverse};
  \foreach \y in {3.6,2.4,1.2,0.0}
    \draw[acc, very thick, line cap=round] (5.7,\y) -- (5.95,\y-0.25) -- (6.4,\y+0.35);
  \node[align=left, font=\footnotesize, text=black] at (2.75,-1.15)
    {closure is built into the operation;\\ the other three are the numbered axioms};
\end{tikzpicture}
$$

Everything that follows is a consequence of these three axioms. First, the words
"the identity" and "the inverse" are justified: both are unique.

> **Proposition (Uniqueness).** In a group $G$:
>
> - the identity $e$ is unique;
> - each $a \in G$ has a unique inverse $a^{-1}$, and $(a^{-1})^{-1} = a$;
> - $(ab)^{-1} = b^{-1} a^{-1}$;
> - the value of $a_1 a_2 \cdots a_n$ is independent of how it is bracketed
>   (the **generalized associative law**).

> **Proof of uniqueness of the identity.** Suppose $f$ and $g$ are both identities.
> Reading (G2) with $a = f, e = g$ gives $fg = f$; reading it with $a = g, e = f$
> gives $fg = g$. Hence $f = g$.[^df-prop1]

> **Proof of uniqueness of inverses.** Let $b, c$ both be inverses of $a$, with
> identity $e$. Then
>
> $$
> b = eb = (ca)b = c(ab) = ce = c,
> $$
>
> using $ca = e$, associativity, and $ab = e$ in turn. So $b = c$. The identity
> $(ab)^{-1} = b^{-1}a^{-1}$ follows by checking that $b^{-1}a^{-1}$ inverts $ab$:
> $(ab)(b^{-1}a^{-1}) = a(bb^{-1})a^{-1} = aa^{-1} = 1$, and uniqueness does the rest.
> The order reversal is not optional in a nonabelian group.

### Cancellation

> **Proposition (Cancellation).** For $a, b \in G$ the equations $ax = b$ and
> $ya = b$ have unique solutions $x = a^{-1}b$ and $y = ba^{-1}$. In particular:
>
> - if $au = av$ then $u = v$ (left cancellation), and
> - if $ub = vb$ then $u = v$ (right cancellation).

> **Proof.** Multiply $ax = b$ on the left by $a^{-1}$: $x = a^{-1}b$, and the
> solution is unique because $a^{-1}$ is. Setting $b = av$ in $au = av$ and
> cancelling gives $u = v$.[^df-prop2] Two shortcuts fall out. If $ab = e$ or
> $ba = e$ for a _single_ side, then $b = a^{-1}$ — one equation suffices. And if
> $ab = a$ for even one $a$, then $b = 1$: to check that an element is the identity,
> testing it against one element is enough.

## Order

Two different quantities carry the name "order," and they are related but not the
same.

> **Definition (Order of a group; order of an element).** The **order of a group**
> $G$, written $|G|$, is its cardinality as a set. The **order of an element**
> $x \in G$, written $|x|$, is the smallest positive integer $n$ with $x^n = 1$;
> if no positive power of $x$ equals $1$, then $x$ has **infinite order**.

The notation $x^n$ abbreviates the $n$-fold product $xx\cdots x$; by the
generalized associative law no bracketing is needed. Set $x^0 = 1$ and
$x^{-n} = (x^{-1})^n$. In an additive group $x^n$ becomes $nx$. The clash of
symbols with set cardinality is deliberate: the order of $x$ turns out to equal
the number of distinct powers of $x$, so the two meanings of "order" agree once
we reach [cyclic groups](/abstract-algebra/subgroups-and-quotients/cyclic-groups).

- An element has order $1$ exactly when it is the identity.
- In $(\mathbb{Z}, +)$ every nonzero element has infinite order.
- In $(\mathbb{R} \setminus \{0\}, \times)$ the element $-1$ has order $2$; every
  other nonidentity element has infinite order.
- In the additive group $\mathbb{Z}/9\mathbb{Z}$ the class $\bar 3$ has order $3$,
  since $\bar 3 + \bar 3 = \bar 6 \ne \bar 0$ but $\bar 3 + \bar 3 + \bar 3 = \bar 9 = \bar 0$.
- In the multiplicative group $(\mathbb{Z}/7\mathbb{Z})^\times$ the class $\bar 2$
  has order $3$ (its powers are $\bar 2, \bar 4, \bar 8 = \bar 1$), and $\bar 3$
  has order $6$.

$$
% caption: The powers of an element of order six close up into a cycle; each
% arrow multiplies by the generator, and the sixth step returns to the identity.
\begin{tikzpicture}[font=\small, >=stealth,
  v/.style={circle, draw, minimum size=9mm, inner sep=0pt}]
  \definecolor{acc}{HTML}{2A6F97}
  \node[v, draw=acc, text=acc] (p0) at (90:2.1) {$x^{0}$};
  \foreach \i/\ang in {1/30, 2/-30, 3/-90, 4/-150, 5/150}
    \node[v] (p\i) at (\ang:2.1) {$x^{\i}$};
  \foreach \i/\j in {0/1, 1/2, 2/3, 3/4, 4/5, 5/0}
    \draw[->, acc, thick] (p\i) to[bend left=14] (p\j);
  \node[align=center, font=\footnotesize] at (0,0)
    {multiply by $x$\\ along each arrow};
\end{tikzpicture}
$$

> **Worked example.** In $(\mathbb{Z}/7\mathbb{Z})^\times = \{\bar 1, \dots, \bar 6\}$
> the order of an element is the first power reaching $\bar 1$:
>
> $$
> |\bar 1| = 1,\quad |\bar 6| = 2,\quad |\bar 2| = |\bar 4| = 3,\quad |\bar 3| = |\bar 5| = 6.
> $$
>
> The powers of $\bar 3$ run $\bar 3, \bar 2, \bar 6, \bar 4, \bar 5, \bar 1$, reaching
> every unit before returning, so $\bar 3$ has order $6 = |(\mathbb{Z}/7\mathbb{Z})^\times|$.
> An element whose order equals the group order generates the whole group — the
> defining property of a [cyclic group](/abstract-algebra/subgroups-and-quotients/cyclic-groups).

## The running examples

Three groups built from the integers recur as test cases throughout.[^df-examples]

- **$(\mathbb{Z}, +)$ and its cousins.** $\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}$
  are groups under addition with $e = 0$ and inverse $-a$. Under multiplication,
  $\mathbb{Q} \setminus \{0\}, \mathbb{R} \setminus \{0\}, \mathbb{C} \setminus \{0\}$
  are groups with $e = 1$. But $\mathbb{Z} \setminus \{0\}$ is _not_ a group under
  $\times$: the element $2$ has no integer inverse.
- **$\mathbb{Z}/n\mathbb{Z}$ under addition.** The residue classes mod $n$ form an
  abelian group of order $n$ under addition of classes, with identity $\bar 0$ and
  inverse $-\bar a$. The construction of these classes is the content of
  [the integers and modular arithmetic](/abstract-algebra/foundations/integers-and-modular-arithmetic).
- **$(\mathbb{Z}/n\mathbb{Z})^\times$ under multiplication.** The classes $\bar a$
  that have a multiplicative inverse mod $n$ form an abelian group under
  multiplication, with identity $\bar 1$. The invertible classes are those $\bar a$ with
  $\gcd(a, n) = 1$; the group is the **units** mod $n$.

The superscript $\times$ always signals that the operation is multiplication, so
there is no confusion between $\mathbb{Z}/n\mathbb{Z}$ (all $n$ classes, added)
and $(\mathbb{Z}/n\mathbb{Z})^\times$ (the units, multiplied), even though the
latter is a subset of the former.

> **Worked example.** Take $\{\bar 1, \bar 2, \bar 3, \bar 4\}$ under multiplication
> mod $5$. Products stay in the set because $5$ is prime, so no two nonzero classes
> multiply to $\bar 0$; multiplication is associative and commutative, inherited
> from $\mathbb{Z}$; $\bar 1$ is the identity; and every element has an inverse,
>
> $$
> \bar 1^{-1} = \bar 1,\qquad \bar 2 \cdot \bar 3 = \bar 1,\qquad \bar 4 \cdot \bar 4 = \overline{16} = \bar 1,
> $$
>
> so $(\mathbb{Z}/5\mathbb{Z})^\times$ is a group of order $4$. Dropping primality
> breaks this. Mod $6$, the class $\bar 2$ has no inverse, since $\gcd(2, 6) = 2$
> and no $\bar c$ solves $\bar 2\,\bar c = \bar 1$; and $\bar 2 \cdot \bar 3 = \bar 0$
> leaves the nonzero classes, so $\{\bar 1, \dots, \bar 5\}$ under multiplication mod
> $6$ fails both closure and inverses.

### Direct products

New groups assemble from old ones componentwise.

> **Definition (Direct product).** If $(A, \ast)$ and $(B, \diamond)$ are groups,
> their **direct product** $A \times B$ has underlying set the Cartesian product
> $\{(a, b) : a \in A, b \in B\}$ and operation
> $(a_1, b_1)(a_2, b_2) = (a_1 \ast a_2,\ b_1 \diamond b_2)$.

Each axiom holds in $A \times B$ because it holds in each coordinate separately.
The identity is $(1_A, 1_B)$ and $(a, b)^{-1} = (a^{-1}, b^{-1})$. Taking
$A = B = \mathbb{R}$ recovers the Euclidean plane $\mathbb{R} \times \mathbb{R}$
under vector addition. The order of $(a, b)$ is the least common multiple of
$|a|$ and $|b|$, a fact we take up again in the study of
[direct products](/abstract-algebra/products-and-group-structure/direct-products-and-finite-abelian-groups).

## Reading a group from its table

For a finite group the entire operation is a square array.

> **Definition (Multiplication table).** For a finite group
> $G = \{g_1, \dots, g_n\}$ with $g_1 = 1$, the **multiplication table** (or
> **Cayley table**) is the $n \times n$ array whose $(i, j)$ entry is $g_i g_j$.

Cancellation has a visible consequence in the table: no element repeats within a
row or within a column, so every row and every column is a permutation of $G$ —
a Latin square. The identity is the element whose row and column reproduce the
header, and $x^{-1}$ is found by locating $1$ in the row of $x$.

$$
% caption: The Cayley table of the additive group of residues mod four; the
% identity row and column reproduce the header, and shaded cells mark inverse
% pairs, where an element meets its partner at the identity.
\begin{tikzpicture}[font=\small, x=11mm, y=11mm]
  \definecolor{acc}{HTML}{2A6F97}
  % top-left corner holds the operation symbol
  \node at (0.5,-0.5) {$+$};
  % headers: top row and left column carry 0..3
  \foreach \v [count=\c from 1] in {0,1,2,3} {
    \node[acc] at (\c+0.5,-0.5) {$\v$};
    \node[acc] at (0.5,-\c-0.5) {$\v$};
  }
  % body: entry = (i+j) mod 4, shaded where it equals the identity
  \foreach \i in {0,1,2,3}
    \foreach \j in {0,1,2,3} {
      \pgfmathtruncatemacro{\s}{mod(\i+\j,4)}
      \ifnum\s=0 \fill[acc!14] (\j+1,-\i-1) rectangle (\j+2,-\i-2); \fi
      \node at (\j+1.5,-\i-1.5) {$\s$};
    }
  \draw (0,0) grid (5,-5);
  \draw[acc, thick] (0,0) rectangle (5,-1);
  \draw[acc, thick] (0,0) rectangle (1,-5);
\end{tikzpicture}
$$

The table is complete information about the group, but a poor way to _see_
structure: it grows as $|G|^2$ and reveals nothing at a glance. Generators,
subgroups, and symmetries expose the structure a table only encodes.

## Symmetry as a group

The examples so far are arithmetic. The other source of groups is geometry: the
**symmetries** of a figure, meaning the rigid motions that carry the figure onto
itself, form a group under composition.

For the equilateral triangle the symmetries are three rotations (by $0$, $120$,
and $240$ degrees) and three reflections across the medians. Composition of two
symmetries is another symmetry, the identity motion leaves everything fixed, and
every motion reverses, so the six form a group.

$$
% caption: The six symmetries of an equilateral triangle: rotation about the
% centre through multiples of a third turn, and reflection across each of the
% three axes through a vertex and the opposite midpoint.
\begin{tikzpicture}[font=\small, >=stealth]
  \definecolor{acc}{HTML}{2A6F97}
  % triangle vertices
  \coordinate (A) at (90:1.9);
  \coordinate (B) at (210:1.9);
  \coordinate (C) at (330:1.9);
  \draw[thick] (A) -- (B) -- (C) -- cycle;
  \node[above] at (A) {$1$};
  \node[below left] at (B) {$2$};
  \node[below right] at (C) {$3$};
  % reflection axes (median through each vertex)
  \draw[acc, dashed] (A) -- ($(B)!0.5!(C)$);
  \draw[acc, dashed] (B) -- ($(A)!0.5!(C)$);
  \draw[acc, dashed] (C) -- ($(A)!0.5!(B)$);
  % rotation arc
  \draw[->, thick, black] (0:0.85) arc (0:250:0.85);
  \node[black, font=\footnotesize] at (300:0.75) {rotate};
  \node[acc, font=\footnotesize, anchor=west] at (2.2,0.4) {three mirror axes};
  \node[black, font=\footnotesize, anchor=west] at (2.2,-0.4) {three rotations};
\end{tikzpicture}
$$

This group of six is the smallest nonabelian group, and it reappears under three
names: as the [dihedral group](/abstract-algebra/groups-and-symmetry/dihedral-and-symmetric-groups)
$D_6$ of the triangle, as the symmetric group $S_3$ of permutations of three
labels, and as a matrix group. That three descriptions coincide is a first illustration of
the abstract viewpoint, and it is the content of
[isomorphism](/abstract-algebra/groups-and-symmetry/homomorphisms-and-group-actions):
groups that look different are the same group when their tables match after
relabeling.

## Role of each axiom

Each axiom plays a distinct role.

| Axiom | Guarantees | Failure elsewhere |
| --- | --- | --- |
| Associativity | products of many elements are unambiguous | cross product on $\mathbb{R}^3$ |
| Identity | a fixed reference point $1$; nonemptiness | — |
| Inverses | cancellation; equations $ax = b$ solvable | $\mathbb{Z} \setminus \{0\}$ under $\times$ |
| Commutativity (optional) | order of factors irrelevant | $D_6$, $S_3$, matrix groups |

Dropping any one axiom leaves a weaker structure (a semigroup, a monoid) with a
much coarser theory.

[^df-binop]: **Dummit & Foote**, _Abstract Algebra_, §1.1 — binary operations, associativity, commutativity, and closure of a subset under an operation.
[^df-prop1]: **Dummit & Foote**, _Abstract Algebra_, §1.1, Proposition 1 — uniqueness of the identity and inverses, $(a^{-1})^{-1} = a$, $(ab)^{-1} = b^{-1}a^{-1}$, and the generalized associative law.
[^df-prop2]: **Dummit & Foote**, _Abstract Algebra_, §1.1, Proposition 2 — solvability of $ax = b$ and $ya = b$ and the left and right cancellation laws.
[^df-examples]: **Dummit & Foote**, _Abstract Algebra_, §1.1 — the examples $\mathbb{Z}$, $\mathbb{Z}/n\mathbb{Z}$, $(\mathbb{Z}/n\mathbb{Z})^\times$, direct products, order of an element, and the multiplication table; and **Judson**, §3.2, for the same catalogue with additional worked tables.