Binary operations are where algebra becomes precise. The phrase “an operation on a set” must mean a genuine function $S\times S\to S$, and that single statement already packages closure. This chapter isolates the properties of operations before the full group axioms appear.

§2.1 The definition

Definition 2.1 (Binary operation)

A binary operation on a set $S$ is a function

For $(a,b)\in S\times S$, the output is written $a\ast b$. The requirement that $\ast$ maps into $S$ means closure is built into the definition: $a\ast b\in S$ for all $a,b\in S$.

Remark 2.2 (Closure is not a separate axiom)

When we say ”$\ast$ is a binary operation on $S$,” we have already asserted closure. If a proposed rule sends some pair outside $S$ or is not defined on some pair, then it is not a binary operation on $S$. The discussion should stop there, before checking any further properties.

§2.2 Properties of binary operations

Definition 2.3 (Commutativity)

A binary operation $\ast$ on $S$ is commutative if

Definition 2.4 (Associativity)

A binary operation $\ast$ on $S$ is associative if

Figure: associativity compares the two ways of multiplying a triple.

The diagram says that whether we combine the first two entries first or the last two first, we must land at the same element of $S$.

Definition 2.5 (Identity element)

An element $e\in S$ is an identity for $\ast$ if

Definition 2.6 (Inverse)

Suppose $\ast$ has an identity $e$. An element $b\in S$ is an inverse of $a\in S$ if

§2.3 Examples and non-examples

Example 2.7 (Standard binary operations)

Example 2.8 (Subtraction is not associative)

On $\mathbb{Z}$, subtraction is a binary operation (closed: $a-b\in \mathbb{Z}$), but:

Since $1\ne 3$, subtraction is not associative. Also, $0$ is a right identity ($a-0=a$) but not a left identity ($0-a=-a\ne a$ for $a\ne 0$), so there is no two-sided identity.

Example 2.9 (Division is not a binary operation on $\mathbb{Z}$)

The rule $a/b$ is not defined for $b=0$, and even when defined, $1/2\notin \mathbb{Z}$. So $/:\mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}$ is not a function. The discussion stops here.

Example 2.10 (The left-projection operation)

On any nonempty set $S$, define $a\ast b=a$. This is a binary operation:

• Closure: $a\ast b=a\in S$. Yes.
• Associative: $(a\ast b)\ast c=a\ast c=a$, and $a\ast (b\ast c)=a\ast b=a$. Yes.
• Commutative: $a\ast b=a$ vs $b\ast a=b$; fails unless $a=b$. Not commutative (if $|S|\ge 2$).
• Identity: Need $e\ast a=a$ for all $a$. But $e\ast a=e$, so $e=a$ for all $a$, impossible if $|S|\ge 2$. No identity.

This example shows that associativity alone forces nothing about invertibility.

Example 2.11 (Max on $\{0,1,2\}$)

On $S=\{0,1,2\}$, define $a\ast b=\max \{a,b\}$. Associative and commutative, with $0$ as identity ($\max \{0,a\}=a$). But only $0$ has an inverse (itself), since $\max \{a,b\}=0$ forces $a=b=0$. Not a group.

§2.4 Operation tables for finite sets

Definition 2.12 (Operation table / Cayley table)

For a finite set $S=\{a_1,\dots ,a_n\}$ with binary operation $\ast$, the operation table (or Cayley table) is the $n\times n$ array whose $(i,j)$-entry is $a_i\ast a_j$.

Example 2.13 (Operation table for ${\mathbb{Z}}_4$ under addition)

Reading the table:

• Closure: automatic (every entry is in $\{0,1,2,3\}$).
• Commutativity: the table is symmetric across the main diagonal.
• Identity: the row for $0$ reproduces the header, and so does the column.
• Inverses: every element appears in every row (Latin square property).

Example 2.14 (A non-associative operation table)

Define $\ast$ on $\{a,b\}$ by: $a\ast a=a$, $a\ast b=b$, $b\ast a=b$, $b\ast b=a$.

This looks like a group table (it is in fact ${\mathbb{Z}}_2$): under the identification $a\mapsto 0$, $b\mapsto 1$, the operation $\ast$ is addition mod $2$. One can also read the group axioms directly from the table: $a$ is the identity (its row and column reproduce the header), and every element is its own inverse ($a\ast a=a$, $b\ast b=a$).

But change one entry: let $b\ast a=a$ instead.

Now check: $(b{\ast }^{\prime }b){\ast }^{\prime }b=a{\ast }^{\prime }b=b$, but $b{\ast }^{\prime }(b{\ast }^{\prime }b)=b{\ast }^{\prime }a=a$. Not associative. A valid-looking table does not guarantee associativity.

Remark 2.15 (Associativity cannot be read from the table at a glance)

Closure, identity, inverses, and commutativity can all be checked directly from a Cayley table. Associativity cannot: it requires checking $n^3$ triples. For small tables one checks by hand; for larger structures one inherits associativity from a known associative ambient operation.

§2.5 Uniqueness of identity and inverse

Theorem 2.16 (Uniqueness of identity)

If a binary operation $\ast$ on $S$ has a two-sided identity, it is unique.

Proof

Suppose $e$ and $f$ are both two-sided identities. Then:

$$e=e\ast f=f,$$

where the first equality uses ”$f$ is a right identity” and the second uses ”$e$ is a left identity.” Hence $e=f$. $■$

Theorem 2.17 (Uniqueness of inverse in an associative structure)

If $\ast$ is associative with identity $e$, and $a\in S$ has a two-sided inverse, that inverse is unique.

Proof

Suppose $b$ and $c$ are both two-sided inverses of $a$:

$$a\ast b=b\ast a=e,a\ast c=c\ast a=e.$$

Then:

$$b=b\ast e=b\ast (a\ast c)=(b\ast a)\ast c=e\ast c=c.$$

The middle step uses associativity. Hence $b=c$. $■$

Remark 2.18 (Associativity is essential for uniqueness of inverses)

Without associativity, inverses need not be unique. The proof above uses associativity in exactly one place: the regrouping $b\ast (a\ast c)=(b\ast a)\ast c$. If this fails, the argument collapses.

§2.6 One-sided data can force two-sided data

Theorem 2.19 (Left identity + left inverses $⟹$ group)

Let $(S,\ast )$ be an associative binary structure with a left identity $e$ (so $e\ast x=x$ for all $x$) and left inverses (for each $a$, there exists $\ell$ with $\ell \ast a=e$). Then $e$ is a two-sided identity and every left inverse is a two-sided inverse.

Proof

Fix $a\in S$ with left inverse $\ell$ ($\ell \ast a=e$). Let $m$ be a left inverse of $\ell$ ($m\ast \ell =e$). Then:

$$a=e\ast a=(m\ast \ell )\ast a=m\ast (\ell \ast a)=m\ast e=m.$$

That argument is too fast, because at this stage we have not yet proved that $e$ is a right identity. So we restart with a computation that uses only the given hypotheses.

We have $\ell \ast a=e$ and $m\ast \ell =e$. Compute:

$$a\ast \ell =(e)\ast (a\ast \ell )=(m\ast \ell )\ast (a\ast \ell )=m\ast (\ell \ast a)\ast \ell =m\ast e\ast \ell =m\ast (e\ast \ell )=m\ast \ell =e.$$

So $\ell$ is also a right inverse of $a$. Now:

$$a\ast e=a\ast (\ell \ast a)=(a\ast \ell )\ast a=e\ast a=a.$$

So $e$ is also a right identity. $■$

This theorem shows how much associativity controls the algebra: purely one-sided hypotheses become two-sided.

Mastery Checklist

• State the definition of binary operation and explain why closure is part of the definition, not a separate axiom.
• Distinguish associativity from commutativity with examples where one holds but not the other.
• Give a non-example where the proposed rule fails to be a binary operation (closure fails).
• Prove uniqueness of the identity element.
• Prove uniqueness of inverses in the presence of associativity.
• Read a Cayley table and extract identity, inverses, and commutativity; explain why associativity cannot be read off.
• State and explain the “left identity + left inverses” theorem (Theorem 2.19).