## Question

(a) Write down why the table below is a Latin square.

\[\begin{gathered}

\begin{array}{*{20}{c}}

{}&d&e&b&a&c

\end{array} \\

\begin{array}{*{20}{c}}

d \\

e \\

b \\

a \\

c

\end{array}\left[ {\begin{array}{*{20}{c}}

c&d&e&b&a \\

d&e&b&a&c \\

a&b&d&c&e \\

b&a&c&e&d \\

e&c&a&d&b

\end{array}} \right] \\

\end{gathered} \]

(b) Use Lagrange’s theorem to show that the table is not a group table.

**Answer/Explanation**

## Markscheme

(a) Each row and column contains all the elements of the set. *A1A1*

*[2 marks]*

* *

(b) There are 5 elements therefore any subgroup must be of an order that is a factor of 5 *R2*

But there is a subgroup \(\begin{gathered}

\begin{array}{*{20}{c}}

{}&e&a

\end{array} \\

\begin{array}{*{20}{c}}

e \\

a

\end{array}\left( {\begin{array}{*{20}{c}}

e&a \\

a&e

\end{array}} \right) \\

\end{gathered} \) of order 2 so the table is not a group table *R2*

**Note: **Award ** R0R2 **for “

*a*is an element of order 2 which does not divide the order of the group”.

*[4 marks]*

*Total [6 marks]*

## Examiners report

Part (a) presented no problem but finding the order two subgroups (Lagrange’s theorem was often quoted correctly) was beyond some candidates. Possibly presenting the set in non-alphabetical order was the problem.

## Question

Associativity and commutativity are two of the five conditions for a set *S *with the binary operation \( * \) to be an Abelian group; state the other three conditions.

The Cayley table for the binary operation \( \odot \) defined on the set *T *= {*p*, *q*, *r*, *s*, *t*} is given below.

(i) Show that exactly three of the conditions for {*T *, \( \odot \)} to be an Abelian group are satisfied, but that neither associativity nor commutativity are satisfied.

(ii) Find the proper subsets of *T *that are groups of order 2, and comment on your result in the context of Lagrange’s theorem.

(iii) Find the solutions of the equation \((p \odot x) \odot x = x \odot p\)* *.

**Answer/Explanation**

## Markscheme

closure, identity, inverse *A2*** **

**Note: **Award ** A1 **for two correct properties,

**otherwise.**

*A0**[2 marks]*

(i) closure: there are no extra elements in the table *R1*

identity: *s *is a (left and right) identity *R1*

inverses: all elements are self-inverse *R1*

commutative: no, because the table is not symmetrical about the leading diagonal, or by counterexample *R1*

associativity: for example, \((pq)t = rt = p\) *M1A1*

not associative because \(p(qt) = pr = t \ne p\) *R1*** **

**Note: **Award ** M1A1 **for 1 complete example whether or not it shows non-associativity.

(ii) \(\{ s,\,p\} ,{\text{ }}\{ s,\,q\} ,{\text{ }}\{ s,\,r\} ,{\text{ }}\{ s,\,t\} \) *A2*** **

**Note: **Award ** A1 **for 2 or 3 correct sets.

as 2 does not divide 5, Lagrange’s theorem would have been contradicted if *T *had been a group *R1*

* *

(iii) any attempt at trying values *(M1)*

the solutions are *q*, *r*, *s *and *t **A1A1A1A1*** **

**Note: **Deduct ** A1 **if

*p*is included.

*[15 marks]*

## Examiners report

This was on the whole a well answered question and it was rare for a candidate not to obtain full marks on part (a). In part (b) the vast majority of candidates were able to show that the set satisfied the properties of a group apart from associativity which they were also familiar with. Virtually all candidates knew the difference between commutativity and associativity and were able to distinguish between the two. Candidates were familiar with Lagrange’s Theorem and many were able to see how it did not apply in the case of this problem. Many candidates found a solution method to part (iii) of the problem and obtained full marks.

This was on the whole a well answered question and it was rare for a candidate not to obtain full marks on part (a). In part (b) the vast majority of candidates were able to show that the set satisfied the properties of a group apart from associativity which they were also familiar with. Virtually all candidates knew the difference between commutativity and associativity and were able to distinguish between the two. Candidates were familiar with Lagrange’s Theorem and many were able to see how it did not apply in the case of this problem. Many candidates found a solution method to part (iii) of the problem and obtained full marks.

## Question

The following Cayley table for the binary operation multiplication modulo 9, denoted by \( * \), is defined on the set \(S = \{ 1,{\text{ }}2,{\text{ }}4,{\text{ }}5,{\text{ }}7,{\text{ }}8\} \).

Copy and complete the table.

Show that \(\{ S,{\text{ }} * \} \) is an Abelian group.

Determine the orders of all the elements of \(\{ S,{\text{ }} * \} \).

(i) Find the two proper subgroups of \(\{ S,{\text{ }} * \} \).

(ii) Find the coset of each of these subgroups with respect to the element 5.

Solve the equation \(2 * x * 4 * x * 4 = 2\).

**Answer/Explanation**

## Markscheme

*A3*

**Note: **Award ** A3 **for correct table,

**for one or two errors,**

*A2***for three or four errors and**

*A1***otherwise.**

*A0**[3 marks]*

the table contains only elements of \(S\), showing closure *R1*

the identity is 1 *A1*

every element has an inverse since 1 appears in every row and column, or a complete list of elements and their correct inverses *A1*

multiplication of numbers is associative *A1*

the four axioms are satisfied therefore \(\{ S,{\text{ }} * \} \) is a group

the group is Abelian because the table is symmetric (about the leading diagonal) *A1*

*[5 marks]*

*A3*

**Note: **Award ** A3 **for all correct values,

**for 5 correct,**

*A2***for 4 correct and**

*A1***otherwise.**

*A0**[3 marks]*

(i) the subgroups are \(\{ 1,{\text{ }}8\} \); \(\{ 1,{\text{ }}4,{\text{ }}7\} \) *A1A1*

(ii) the cosets are \(\{ 4,{\text{ }}5\} \); \(\{ 2,{\text{ }}5,{\text{ }}8\} \) *A1A1*

*[4 marks]*

**METHOD 1**

use of algebraic manipulations *M1*

and at least one result from the table, used correctly *A1*

\(x = 2\) *A1*

\(x = 7\) *A1*

**METHOD 2**

testing at least one value in the equation *M1*

obtain \(x = 2\) *A1*

obtain \(x = 7\) *A1*

explicit rejection of all other values *A1*

*[4 marks]*

## Examiners report

The majority of candidates were able to complete the Cayley table correctly.

Generally well done. However, it is not good enough for a candidate to say something along the lines of ‘the operation is closed or that inverses exist by looking at the Cayley table’. A few candidates thought they only had to prove commutativity.

Often well done. A few candidates stated extra, and therefore incorrect subgroups.

[N/A]

The majority found only one solution, usually the obvious \(x = 2\), but sometimes only the less obvious \(x = 7\).

## Question

The binary operation \( * \) is defined by

\(a * b = a + b – 3\) for \(a,{\text{ }}b \in \mathbb{Z}\).

The binary operation \( \circ \) is defined by

\(a \circ b = a + b + 3\) for \(a,{\text{ }}b \in \mathbb{Z}\).

Consider the group \(\{ \mathbb{Z},{\text{ }} \circ {\text{\} }}\) and the bijection \(f:\mathbb{Z} \to \mathbb{Z}\) given by \(f(a) = a – 6\).

Show that \(\{ \mathbb{Z},{\text{ }} * \} \) is an Abelian group.

Show that there is no element of order 2.

Find a proper subgroup of \(\{ \mathbb{Z},{\text{ }} * \} \).

Show that the groups \(\{ \mathbb{Z},{\text{ }} * \} \) and \(\{ \mathbb{Z},{\text{ }} \circ \} \) are isomorphic.

**Answer/Explanation**

## Markscheme

closure: \(\{ \mathbb{Z},{\text{ }} * \} \) is closed because \(a + b – 3 \in \mathbb{Z}\) *R1*

identity: \(a * e = a + e – 3 = a\) *(M1)*

\(e = 3\) *A1*

inverse: \(a * {a^{ – 1}} = a + {a^{ – 1}} – 3 = 3\) *(M1)*

\({a^{ – 1}} = 6 – a\) *A1*

associative: \(a * (b * c) = a * (b + c – 3) = a + b + c – 6\) *A1*

\(\left( {a{\text{ }}*{\text{ }}b} \right){\text{ }}*{\text{ }}c{\text{ }} = \left( {a{\text{ }} + {\text{ }}b{\text{ }} – {\text{ }}3} \right)*{\text{ }}c{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c{\text{ }} – {\text{ }}6\) *A1*

associative because \(a * (b * c) = (a * b) * c\) *R1*

\(b * a = b + a – 3 = a + b – 3 = a * b\) therefore commutative hence Abelian *R1*

hence \(\{ \mathbb{Z},{\text{ }} * \} \) is an Abelian group *AG*

*[9 marks]*

if \(a\) is of order 2 then \(a * a = 2a – 3 = 3\) therefore \(a = 3\) *A1*

which is a contradiction

since \(e = 3\) and has order 1 *R1*

**Note:** ** R1 **for recognising that the identity has order 1.

*[2 marks]*

for example \(S = \{ – 6,{\text{ }} – 3,{\text{ }}0,{\text{ }}3,{\text{ }}6 \ldots \} \) or \(S = \{ \ldots ,{\text{ }} – 1,{\text{ }}1,{\text{ }}3,{\text{ }}5,{\text{ }}7 \ldots \} \) *A1R1*

**Note:** ** R1 **for deducing, justifying or verifying that \(\left\{ {S, * } \right\}\) is indeed a proper subgroup.

*[2 marks]*

we need to show that \(f(a * b) = f(a) \circ f(b)\) *R1*

\(f(a * b) = f(a + b – 3) = a + b – 9\) *A1*

\(f(a) \circ f(b) = (a – 6) \circ (b – 6) = a + b – 9\) *A1*

hence isomorphic *AG*

**Note:** ** R1** for recognising that \(f\) preserves the operation; award

**for an attempt to show that \(f(a \circ b) = f(a) * f(b)\).**

*R1A0A0**[3 marks]*

## Examiners report

[N/A]

[N/A]

[N/A]

[N/A]

## Question

The binary operation multiplication modulo 10, denoted by ×_{10}, is defined on the set *T* = {2 , 4 , 6 , 8} and represented in the following Cayley table.

Show that {*T*, ×_{10}} is a group. (You may assume associativity.)

By making reference to the Cayley table, explain why* T* is Abelian.

Find the order of each element of {*T*, ×_{10}}.

Hence show that {*T*, ×_{10}} is cyclic and write down all its generators.

The binary operation multiplication modulo 10, denoted by ×_{10} , is defined on the set *V* = {1, 3 ,5 ,7 ,9}.

Show that {*V*, ×_{10}} is not a group.

**Answer/Explanation**

## Markscheme

closure: there are no new elements in the table **A1**

identity: 6 is the identity element **A1**

inverse: every element has an inverse because there is a 6 in every row and column (2^{−1} = 8, 4^{−1} = 4, 6^{−1} = 6, 8^{−1} = 2) **A1**

we are given that (modulo) multiplication is associative **R1**

so {*T*, ×_{10}} is a group **AG**

**[4 marks]**

the Cayley table is symmetric (about the main diagonal) **R1**

so* T* is Abelian ** AG**

**[1 mark]**

considering powers of elements **(M1)**

**A2**

**Note:** Award * A2 *for all correct and

*for one error.*

**A1****[3 marks]**

**EITHER**

{*T*, ×_{10}} is cyclic because there is an element of order 4 **R1**

**Note:** Accept “there are elements of order 4”.

**OR**

{*T*, ×_{10}} is cyclic because there is generator **R1**

**Note:** Accept “because there are generators”.

**THEN**

2 and 8 are generators **A1A1**

**[3 marks]**

**EITHER**

considering singular elements **(M1)**

5 has no inverse (5 ×_{10} a = 1, a∈*V* has no solution) **R1**

**OR**

considering Cayley table for {*V*, ×_{10}}

**M1**

the Cayley table is not a Latin square (or equivalent) **R1**

**OR**

considering cancellation law

*eg*, 5 ×_{10}_{ }9 = 5 ×_{10} 1 = 5 **M1**

if {*V*, ×_{10}} is a group the cancellation law gives 9 = 1 **R1**

**OR**

considering order of subgroups

*eg*, {1, 9} is a subgroup **M1**

it is not possible to have a subgroup of order 2 for a group of order 5 (Lagrange’s theorem) **R1**

**THEN**

so {*V*, ×_{10}} is not a group ** AG**

**[2 marks]**

## Examiners report

[N/A]

[N/A]

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[N/A]