8.03a–8.03d - Binary operations and group axioms | OCR A-Level Further Mathematics A

A group is a set with a way of combining its elements so that every action can be undone. You will test binary operations, construct Cayley tables, use the group axioms and explain the Latin square property. Numbers, functions, matrices and symmetries will turn out to share the same underlying rules.

Binary operations and their properties

Can $6$ act as a multiplicative identity? In ordinary arithmetic, certainly not: $6\times 3\ne 3$ . But on the set $\{3,6,9,12\}$ , with every product reduced modulo $15$ , multiplying by $6$ leaves every element unchanged. The identity depends on the set and the operation together. We will find the whole structure behind this surprise.

You already know

Congruent integers have the same remainder. When working modulo $n$ , use the representatives $0,1,\ldots,n-1$ . Matrix multiplication and composition of transformations both depend on the order in which you act.

Start with two inputs and specify one output. Ordinary addition takes two integers to an integer. Subtraction also does this, although swapping the inputs usually changes the answer. Subtraction on positive integers fails a different test: $2-5=-3$ is outside the set.

Binary operation

A binary operation $\circ$ on a set $S$ assigns exactly one element $a\circ b$ of $S$ to every ordered pair $(a,b)$ from $S$ . The rule is closed on $S$ when all its outputs belong to $S$ . For example, addition is a binary operation on $\mathbb Z$ . Division is not a binary operation on $\mathbb R$ , since inputs with second entry $0$ have no defined output.

Two properties answer different questions. Commutativity asks whether swapping two inputs changes the output. Associativity asks whether changing the brackets in a calculation with three inputs changes it. Associativity never swaps the inputs.

Commutativity and associativity

For every $a,b,c$ in the set:

\begin{aligned} a\circ b&=b\circ a &&\text{commutativity}\\ (a\circ b)\circ c&=a\circ(b\circ c) &&\text{associativity} \end{aligned}

An operation has the stated property only if the corresponding equality holds for all the inputs.
Not in the formula booklet: learn it.

To prove a property, calculate with arbitrary elements. To disprove it, one counterexample is enough.

Worked example 1: a shifted addition

Question. On $\mathbb Z$ , define $a\circ b=a+b+1$ . Show that this operation is closed, commutative and associative.

Plan. Use arbitrary integers, swap the two inputs, then expand both bracketings of three inputs.

\begin{aligned} a+b+1&\in\mathbb Z &&\text{integer sums are integers}\\ a\circ b&=a+b+1=b+a+1=b\circ a &&\text{ordinary addition commutes}\\ (a\circ b)\circ c&=(a+b+1)+c+1=a+b+c+2 &&\text{apply the rule twice}\\ a\circ(b\circ c)&=a+(b+c+1)+1=a+b+c+2 &&\text{expand the other brackets} \end{aligned}

Why this step? The symbol $\circ$ means the given rule each time it appears; it does not mean ordinary addition.

Answer. The operation has all three properties on $\mathbb Z$ .

Check. $(2\circ 3)\circ 4=6\circ 4=11$ , and $2\circ(3\circ 4)=2\circ 8=11$ .

Change just the coefficient of the second input from $1$ to $2$ . This keeps closure but can destroy the other properties.

Worked example 2: unequal roles for the inputs

Question. On $\mathbb Z$ , define $a*b=a+2b+1$ . Determine whether the operation is closed, commutative and associative.

Plan. Establish closure, then find small counterexamples to the two equalities.

\begin{aligned} a+2b+1&\in\mathbb Z &&\text{closure}\\ 0*1&=3,\qquad 1*0=2 &&\text{different reversed products}\\ (0*0)*0&=1*0=2 &&\text{left brackets}\\ 0*(0*0)&=0*1=3 &&\text{right brackets} \end{aligned}

Why these inputs? Zeros remove most of the arithmetic, so any difference comes directly from the rule.

Answer. The operation is closed, but is neither commutative nor associative.

Check. Algebra gives $(a*b)*c=a+2b+2c+2$ , whereas $a*(b*c)=a+2b+4c+3$ ; their difference is not identically zero.

A successful check on one triple does not prove associativity, just as one successful multiplication does not prove closure for an infinite set.

Watch out

Wrong: “ $(a\circ b)\circ c=a\circ(c\circ b)$ , so the operation is associative.” The inputs $b$ and $c$ have been swapped. Right: compare $(a\circ b)\circ c$ with $a\circ(b\circ c)$ , keeping their order fixed.

Now move the coefficient $2$ to the other input and check the consequences.

Constructing Cayley tables

For a finite set, a table can show every possible product, so closure can be checked exhaustively.

Cayley table

A Cayley table records a binary operation on a finite set. In this lesson the entry in row $a$ , column $b$ is $a\circ b$ : the row element comes first. A table contains the outputs, with one copy of the set labelling the rows and another labelling the columns.

The operation must be stated beside the set. The symbols $ab$ , $a\times_n b$ or $a\circ b$ may all mean multiplication modulo $n$ , when that convention is given. We will write $\times_n$ explicitly.

Worked example 3: multiplication modulo five

Question. Construct the Cayley table for $S=\{1,2,3,4\}$ under multiplication modulo $5$ . State whether the rule is closed and commutative.

Plan. Multiply the row and column labels, then replace each product by its least non-negative remainder modulo $5$ .

\begin{aligned} 2\times_5 3&=1 &&6\equiv 1\pmod 5\\ 3\times_5 3&=4 &&9\equiv 4\pmod 5\\ 4\times_5 4&=1 &&16\equiv 1\pmod 5 \end{aligned}

$\times_5$	$1$	$2$	$3$	$4$
$1$	$1$	$2$	$3$	$4$
$2$	$2$	$4$	$1$	$3$
$3$	$3$	$1$	$4$	$2$
$4$	$4$	$3$	$2$	$1$

Why reduce every product? The entries must be elements of the stated set, rather than ordinary products such as $6$ or $16$ .

Answer. Every output belongs to $S$ , so the rule is closed. The table is symmetric about its main diagonal, so it is commutative.

Check. The $(2,3)$ and $(3,2)$ entries both equal $1$ , agreeing with ordinary multiplication being commutative before reduction.

For the second table, replace the residue system by the multiples of $3$ modulo $15$ . The calculation is still “multiply, then reduce”, but the identity will look different.

Worked example 4: an unexpected identity

Question. Construct the table for $T=\{3,6,9,12\}$ under multiplication modulo $15$ . Find an element that leaves every element unchanged on either side.

Plan. Reduce all sixteen products modulo $15$ , then look for a row and a column that reproduce their headings.

\begin{aligned} 3\times_{15}3&=9 &&\text{already a representative}\\ 3\times_{15}12&=6 &&36\equiv 6\pmod{15}\\ 6\times_{15}9&=9 &&54\equiv 9\pmod{15} \end{aligned}

$\times_{15}$	$3$	$6$	$9$	$12$
$3$	$9$	$3$	$12$	$6$
$6$	$3$	$6$	$9$	$12$
$9$	$12$	$9$	$6$	$3$
$12$	$6$	$12$	$3$	$9$

Why inspect both a row and a column? An identity must work on both sides, whether or not the operation is commutative.

Answer. The required element is $6$ : $6\times_{15}t=t\times_{15}6=t$ for every $t\in T$ .

Check. $6\times_{15}12=12$ , because $72\equiv12\pmod{15}$ . Notice that ordinary $1$ is not in $T$ .

The identity is defined by its effect. It need not be a number called $1$ , or even a number at all.

Watch out

Wrong: “The top-left entry is the identity.” The order of the headings is arbitrary. Right: find the element whose entire row and entire column reproduce the opposite headings.

A different modulus gives another short table to build from the rule itself.

The group axioms

The familiar operations are examples of a wider pattern: a rule can have its own identity and its own way of undoing an input.

Group and abelian group

A group $(G,\circ)$ is a non-empty set $G$ with an operation satisfying:

Closure: $a\circ b\in G$ for every $a,b\in G$ .
Associativity: $(a\circ b)\circ c=a\circ(b\circ c)$ for every $a,b,c\in G$ .
Identity: there is an element $e\in G$ such that $e\circ a=a\circ e=a$ for every $a\in G$ .
Inverses: each $a\in G$ has an element $a^{-1}\in G$ such that $a\circ a^{-1}=a^{-1}\circ a=e$ .

An abelian group is a group whose operation is also commutative. Commutativity is not a group axiom.

These requirements do separate jobs. Closure keeps the calculation inside the set; associativity removes ambiguity in a product of several elements; the identity does nothing; an inverse undoes an action. In additive notation the identity is often $0$ and the inverse of $a$ is $-a$ . In multiplicative notation the inverse symbol $a^{-1}$ means the group inverse, which need not be the ordinary reciprocal of a displayed label.

The axioms and group terminology are not in the formula booklet: learn them. When a question asks you to show that a structure is a group, justify all four axioms; a valid counterexample to any one axiom proves that a structure is not a group.

Worked example 5: finding an identity and inverses algebraically

Question. On $\mathbb R$ , define $a\circ b=a+b+2$ . Show that this gives an abelian group.

Plan. Check closure and associativity, solve for the identity and inverse, then check commutativity.

\begin{aligned} a+b+2&\in\mathbb R &&\text{closure}\\ (a\circ b)\circ c&=a+b+c+4=a\circ(b\circ c) &&\text{expand both bracketings}\\ a\circ e=a&\ \Rightarrow\ a+e+2=a\ \Rightarrow\ e=-2 &&\text{solve the identity condition}\\ a\circ x=e&\ \Rightarrow\ a+x+2=-2\ \Rightarrow\ x=-a-4 &&\text{solve the inverse condition} \end{aligned}

The value $-2$ belongs to $\mathbb R$ and works on both sides. For every real $a$ , its inverse $-a-4$ is real. Also $a\circ b=a+b+2=b+a+2=b\circ a$ .

Why solve two different equations? An identity leaves $a$ unchanged, whereas an inverse combines with $a$ to give that identity, which here is $-2$ .

Answer. All four group axioms hold, and the group is abelian. Its identity is $-2$ , and the group inverse of $a$ is $-a-4$ .

Check. The inverse of $3$ is $-7$ , since $3\circ(-7)=3-7+2=-2=e$ . This inverse is not the ordinary reciprocal $\frac13$ .

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$\times_5$	$1$	$2$	$3$	$4$
$1$	$1$	$2$	$3$	$4$
$2$	$2$	$4$	$1$	$3$
$3$	$3$	$1$	$4$	$2$
$4$	$4$	$3$	$2$	$1$

$\times_5$	$1$	$2$	$3$	$4$
$1$	$1$	$2$	$3$	$4$
$2$	$2$	$4$	$1$	$3$
$3$	$3$	$1$	$4$	$2$
$4$	$4$	$3$	$2$	$1$

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8.03a–8.03d - Binary operations and group axioms

Binary operations and their properties

Constructing Cayley tables

The group axioms

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$\times_5$	$1$	$2$	$3$	$4$
$1$	$1$	$2$	$3$	$4$
$2$	$2$	$4$	$1$	$3$
$3$	$3$	$1$	$4$	$2$
$4$	$4$	$3$	$2$	$1$