FP2 1.2 - Cayley tables and cyclic groups | Edexcel A-Level Further Mathematics

Recognise groups in modular arithmetic, symmetry, matrices and permutations (FP2 1.2). You will construct and read Cayley tables, compose operations in the right order and test whether one element generates a whole group.

Read an operation from a Cayley table

A square can be turned, reflected or left alone. Can we record every way of combining these moves in the same format as adding clock readings? A Cayley table records the operation, regardless of what its elements represent.

You already know

A group needs closure, associativity, an identity and inverses in the set. Modular arithmetic preserves addition and multiplication. For transformations acting on column vectors, $XY$ means apply $Y$ first, then $X$ ; matrix multiplication is associative.

Start with a familiar small set: residues $0,1,2,3$ , added modulo $4$ . The sum $2+3=5$ is represented by $1$ , so the table stays inside the set.

Cayley table

A Cayley table lists the result of every ordered pair under an operation. In this lesson, the entry at row $a$ , column $b$ means $a\star b$ .

For a group, the identity's row and column reproduce the headings. The inverse of $a$ is found where its row meets the identity, with the reversed product checked in the corresponding column.

A table containing only members proves closure, but it does not automatically prove associativity. For modular addition, both bracketings reduce the ordinary sum $a+b+c$ , so associativity follows from integer addition.

Worked example 1: addition modulo four

Question. Construct the Cayley table for addition modulo $4$ on $\{0,1,2,3\}$ . Identify the identity and inverse pairs.

Plan. Add the row and column headings, reduce each result modulo $4$ , then locate zero outputs.

\begin{aligned} 2+3&\equiv1\pmod4&&\text{one reduced entry}\\ 3+3&\equiv2\pmod4&&\text{another reduced entry} \end{aligned}

$+_4$	$0$	$1$	$2$	$3$
$0$	$0$	$1$	$2$	$3$
$1$	$1$	$2$	$3$	$0$
$2$	$2$	$3$	$0$	$1$
$3$	$3$	$0$	$1$	$2$

Why this step? An inverse under addition must produce the additive identity $0$ , rather than the number $1$ .

Answer. The identity is $0$ . The inverses are $0\leftrightarrow0$ , $1\leftrightarrow3$ , $2\leftrightarrow2$ . Together with closure and associativity, these establish a group.

Check. Row and column $0$ reproduce the headings, and every input has a zero output in its row and column.

Change only the modulus from $4$ to $5$ , using its five residue representatives.

Worked example 2: addition modulo five

Question. Construct the addition table modulo $5$ and identify its identity and inverses.

Plan. Use the same row-plus-column rule with reductions modulo $5$ .

\begin{aligned} 3+4&\equiv2\pmod5&&\text{reduce the sum}\\ 4+4&\equiv3\pmod5&&\text{reduce another sum} \end{aligned}

$+_5$	$0$	$1$	$2$	$3$	$4$
$0$	$0$	$1$	$2$	$3$	$4$
$1$	$1$	$2$	$3$	$4$	$0$
$2$	$2$	$3$	$4$	$0$	$1$
$3$	$3$	$4$	$0$	$1$	$2$
$4$	$4$	$0$	$1$	$2$	$3$

Why this step? The modulus changes the representative of each sum and which pairs return to zero.

Answer. Identity $0$ ; inverse pairs $0\leftrightarrow0$ , $1\leftrightarrow4$ , $2\leftrightarrow3$ . This is a group under addition modulo $5$ .

Check. $1+4\equiv2+3\equiv0\pmod5$ , as the table shows.

The table-reading rules and the modular group conclusions are not in the formula booklet.

Watch out

An entry $7$ in the row $3$ , column $4$ of the modulo- $5$ table is unreduced. Write $2$ , since $7\equiv2\pmod5$ . The entry must be one of the stated group elements.

Use the same layout for one smaller modulus.

Choose the right set for modular multiplication

Multiplication needs a different identity and a carefully chosen set. Zero cannot have a multiplicative inverse modulo a modulus greater than one.

Write $U_n$ for the representatives between $1$ and $n-1$ that are coprime to $n$ . We will check two small multiplicative groups directly from their complete tables.

Multiplicative inverse modulo a modulus

For multiplication modulo $n>1$ , the identity is $1$ . An inverse of $a$ is a representative $b$ satisfying

ab\equiv1\pmod n

Use the operation and the allowed set from the question; “nonzero” alone does not guarantee invertibility for a composite modulus.

Associativity is inherited from ordinary multiplication and the congruence laws. The tables below check all remaining finite-set requirements.

Worked example 3: a multiplicative group modulo five

Question. Show that $U_5=\{1,2,3,4\}$ forms a group under multiplication modulo $5$ , using a Cayley table.

Plan. Multiply and reduce every pair, then find the identity and inverses.

\begin{aligned} 2(3)&\equiv1\pmod5&&\text{one inverse pair}\\ 4(4)&\equiv1\pmod5&&\text{a self-inverse element} \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 this step? Here inverses must give $1$ , because multiplication is the operation.

Answer. Closure is visible in the table, associativity is inherited, the identity is $1$ , and inverse pairs are $1\leftrightarrow1$ , $2\leftrightarrow3$ , $4\leftrightarrow4$ . Thus this is a group.

Check. Every row and column contains the multiplicative identity once.

Change only the modulus to $8$ , using its coprime representatives $U_8=\{1,3,5,7\}$ .

Worked example 4: a composite modulus can still give a group

Question. Show that $U_8$ forms a group under multiplication modulo $8$ .

Plan. Construct the table and inspect where each element returns to $1$ .

\begin{aligned} 3(5)&=15\equiv7\pmod8&&\text{reduce an ordinary product}\\ 3^2&=9\equiv1\pmod8&&\text{one inverse check} \end{aligned}

$\times_8$	$1$	$3$	$5$	$7$
$1$	$1$	$3$	$5$	$7$
$3$	$3$	$1$	$7$	$5$
$5$	$5$	$7$	$1$	$3$
$7$	$7$	$5$	$3$	$1$

Why this step? A composite modulus does not forbid a group; the choice of allowed elements determines whether closure and inverses work.

Answer. The table proves closure and gives identity $1$ . Every element is its own inverse; associativity is inherited. Hence $U_8$ is a group.

Check. $5^2=25\equiv1\pmod8$ and $7^2=49\equiv1\pmod8$ .

These group tables and the multiplicative-group tests are not in the formula booklet.

Watch out

“The nonzero residues always form a multiplicative group” is false. Modulo $6$ , $2(3)\equiv0$ , so the set $\{1,2,3,4,5\}$ is not closed. The coprime representatives here are $U_6=\{1,5\}$ .

Change the modulus again, and use the table's inverse test.

Compose the symmetries of a square

A symmetry maps a figure onto itself. Compose two symmetries and the figure still maps onto itself; undoing a symmetry is another symmetry. Composition is associative and doing nothing is the identity, so the symmetries of a figure form a group.

For a square centred at the origin, let $r$ be an anticlockwise quarter-turn and $f$ reflection in the $x$ -axis. Write $e$ for doing nothing.

The square's eight symmetries

With composition $XY$ meaning apply $Y$ first, then $X$ , the symmetry group is

\{e,r,r^2,r^3,f,rf,r^2f,r^3f\}

There are four rotations and four reflections, with

r^4=e,\qquad f^2=e

Here exponents mean repeated composition.

Not in the formula booklet: learn it.

The transformations are distinct, as their coordinate actions show.

Element	Image of $(x,y)$	Description
$e$	$(x,y)$	identity
$r$	$(-y,x)$	anticlockwise quarter-turn
$r^2$	$(-x,-y)$	half-turn
$r^3$	$(y,-x)$	clockwise quarter-turn
$f$	$(x,-y)$	reflection in the $x$ -axis
$rf$	$(y,x)$	reflection in $y=x$
$r^2f$	$(-x,y)$	reflection in the $y$ -axis
$r^3f$	$(-y,-x)$	reflection in $y=-x$

Recall the Core matrices

\mathbf R=\begin{pmatrix}0&-1\\1&0\end{pmatrix} \qquad \mathbf F=\begin{pmatrix}1&0\\0&-1\end{pmatrix}

The general planar rotation and reflection matrices are supplied in the formula booklet; these are their quarter-turn and horizontal-mirror special cases. The square-group composition rules are not supplied.

Worked example 5: reflection first

Question. Determine $rf$ , its matrix and its action on $P=(2,1)$ .

Plan. Apply $f$ first, then $r$ , and multiply matrices in that same functional order.

\begin{aligned} (2,1)&\xmapsto{f}(2,-1)&&\text{rightmost action first}\\ &\xmapsto{r}(1,2)&&\text{anticlockwise quarter-turn}\\ \mathbf R\mathbf F &=\begin{pmatrix}0&1\\1&0\end{pmatrix} &&\text{row-by-column product} \end{aligned}

Why this step? The rightmost matrix reaches a point vector first, so $rf$ is represented by $\mathbf R\mathbf F$ .

Answer. $rf$ is reflection in $y=x$ , with the displayed matrix; it sends $P$ to $(1,2)$ .

Check. The matrix interchanges coordinates, agreeing with $(x,y)\mapsto(y,x)$ .

Change only the order of the two same transformations.

Worked example 6: rotation first

Question. Determine $fr$ , its matrix and its action on $P=(2,1)$ .

Plan. Apply $r$ first, then $f$ , so use $\mathbf F\mathbf R$ .

\begin{aligned} (2,1)&\xmapsto{r}(-1,2)&&\text{rightmost action first}\\ &\xmapsto{f}(-1,-2)&&\text{reflect in the }x\text{-axis}\\ \mathbf F\mathbf R &=\begin{pmatrix}0&-1\\-1&0\end{pmatrix} &&\text{multiply in reversed order} \end{aligned}

Why this step? Reversing the instructions reverses the product; associativity does not allow factors to swap.

Answer. $fr$ is reflection in $y=-x$ , and sends $P$ to $(-1,-2)$ .

Check. Its matrix sends a general $(x,y)$ to $(-y,-x)$ , matching the named reflection.

The two paths begin at the same point but finish at different images. This gives a concrete group that is not commutative.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

$+_4$	$0$	$1$	$2$	$3$
$0$	$0$	$1$	$2$	$3$
$1$	$1$	$2$	$3$	$0$
$2$	$2$	$3$	$0$	$1$
$3$	$3$	$0$	$1$	$2$

$+_5$	$0$	$1$	$2$	$3$	$4$
$0$	$0$	$1$	$2$	$3$	$4$
$1$	$1$	$2$	$3$	$4$	$0$
$2$	$2$	$3$	$4$	$0$	$1$
$3$	$3$	$4$	$0$	$1$	$2$
$4$	$4$	$0$	$1$	$2$	$3$

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

$\times_8$	$1$	$3$	$5$	$7$
$1$	$1$	$3$	$5$	$7$
$3$	$3$	$1$	$7$	$5$
$5$	$5$	$7$	$1$	$3$
$7$	$7$	$5$	$3$	$1$

$+_4$	$0$	$1$	$2$	$3$
$0$	$0$	$1$	$2$	$3$
$1$	$1$	$2$	$3$	$0$
$2$	$2$	$3$	$0$	$1$
$3$	$3$	$0$	$1$	$2$

$+_5$	$0$	$1$	$2$	$3$	$4$
$0$	$0$	$1$	$2$	$3$	$4$
$1$	$1$	$2$	$3$	$4$	$0$
$2$	$2$	$3$	$4$	$0$	$1$
$3$	$3$	$4$	$0$	$1$	$2$
$4$	$4$	$0$	$1$	$2$	$3$

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

$\times_8$	$1$	$3$	$5$	$7$
$1$	$1$	$3$	$5$	$7$
$3$	$3$	$1$	$7$	$5$
$5$	$5$	$7$	$1$	$3$
$7$	$7$	$5$	$3$	$1$

Last Accessed

Core Pure Mathematics

Further Pure Mathematics 1

Further Pure Mathematics 2

Further Statistics 1

Further Statistics 2

Further Mechanics 1

Further Mechanics 2

Decision Mathematics 1

Decision Mathematics 2

FP2 1.2 - Cayley tables and cyclic groups

Read an operation from a Cayley table

Choose the right set for modular multiplication

Compose the symmetries of a square

Continue this lesson.

Learn faster with Nova

$+_4$	$0$	$1$	$2$	$3$
$0$	$0$	$1$	$2$	$3$
$1$	$1$	$2$	$3$	$0$
$2$	$2$	$3$	$0$	$1$
$3$	$3$	$0$	$1$	$2$

$+_5$	$0$	$1$	$2$	$3$	$4$
$0$	$0$	$1$	$2$	$3$	$4$
$1$	$1$	$2$	$3$	$4$	$0$
$2$	$2$	$3$	$4$	$0$	$1$
$3$	$3$	$4$	$0$	$1$	$2$
$4$	$4$	$0$	$1$	$2$	$3$

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

$\times_8$	$1$	$3$	$5$	$7$
$1$	$1$	$3$	$5$	$7$
$3$	$3$	$1$	$7$	$5$
$5$	$5$	$7$	$1$	$3$
$7$	$7$	$5$	$3$	$1$