## Question

(a) Find the six roots of the equation \({z^6} – 1 = 0\) , giving your answers in the form \(r\,{\text{cis}}\,\theta {\text{, }}r \in {\mathbb{R}^ + }{\text{, }}0 \leqslant \theta < 2\pi \) .

(b) (i) Show that these six roots form a group *G *under multiplication of complex numbers.

(ii) Show that *G *is cyclic and find all the generators.

(iii) Give an example of another group that is isomorphic to *G*, stating clearly the corresponding elements in the two groups.

**Answer/Explanation**

## Markscheme

(a) \({z^6} = 1 = {\text{cis}}\,2n\pi \) *(M1)*

The six roots are

\({\text{cis}}\,0(1),{\text{ cis}}\frac{\pi }{3},{\text{ cis}}\frac{{2\pi }}{3},{\text{ cis}}\,\pi ( – 1),{\text{ cis}}\frac{{4\pi }}{3},{\text{ cis}}\frac{{5\pi }}{3}\) *A3*

**Note: **Award ** A2 **for 4 or 5 correct roots,

**for 2 or 3 correct roots.**

*A1** *

*[4 marks]*

* *

(b) (i) Closure: Consider any two roots \({\text{cis}}\frac{{m\pi }}{3},{\text{ cis}}\frac{{n\pi }}{3}\). *M1*

\({\text{cis}}\frac{{m\pi }}{3} \times {\text{cis}}\frac{{n\pi }}{3} = {\text{cis}}\,(m + n){\text{(mod6)}}\frac{\pi }{3} \in G\) *A1*

**Note: **Award ** M1A1 **for a correct Cayley table showing closure.

Identity: The identity is 1. *A1*

Inverse: The inverse of \({\text{cis}}\frac{{m\pi }}{3}{\text{ is cis}}\frac{{(6 – m)\pi }}{3} \in G\) . *A2*

Associative: This follows from the associativity of multiplication. *R1*

The 4 group axioms are satisfied. *R1*

* *

(ii) Successive powers of \({\text{cis}}\frac{\pi }{3}\left( {{\text{or cis}}\frac{{5\pi }}{3}} \right)\)

generate the group which is therefore cyclic. *R2*

The (only) other generator is \({\text{cis}}\frac{{5\pi }}{3}\left( {{\text{or cis}}\frac{\pi }{3}} \right)\) . *A1*

**Note: **Award ** A0 **for any additional answers.

(iii) The group of the integers 0, 1, 2, 3, 4, 5 under addition modulo 6. *R2*

The correspondence is

\(m \to {\text{cis}}\frac{{m\pi }}{3}\) *R1*

**Note: **Accept any other cyclic group of order 6.

* *

*[13 marks]*

*Total [17 marks]*

## Examiners report

This question was reasonably well answered by many candidates, although in (b)(iii), some candidates were unable to give another group isomorphic to *G*.

## Question

(a) Draw the Cayley table for the set of integers *G* = {0, 1, 2, 3, 4, 5} under addition modulo 6, \({ + _6}\).

(b) Show that \(\{ G,{\text{ }}{ + _6}\} \) is a group.

(c) Find the order of each element.

(d) Show that \(\{ G,{\text{ }}{ + _6}\} \) is cyclic and state its generators.

(e) Find a subgroup with three elements.

(f) Find the other proper subgroups of \(\{ G,{\text{ }}{ + _6}\} \).

**Answer/Explanation**

## Markscheme

(a) *A3*

**Note: **Award ** A2 **for 1 error,

**for 2 errors and**

*A1***for more than 2 errors.**

*A0*

*[3 marks]*

* *

(b) The table is closed *A1*

Identity element is 0 *A1*

Each element has a unique inverse (0 appears exactly once in each row and column) *A1*

Addition mod 6 is associative *A1*

Hence \(\{ G,{\text{ }}{ + _6}\} \) forms a group *AG*

*[4 marks]*

* *

(c) 0 has order 1 (0 = 0),

1 has order 6 (1 + 1 + 1 + 1 + 1 + 1 = 0),

2 has order 3 (2 + 2 + 2 = 0),

3 has order 2 (3 + 3 = 0),

4 has order 3 (4 + 4 + 4 = 0),

5 has order 6 (5 + 5 + 5 + 5 + 5 + 5 = 0). *A3*

**Note****: **Award ** A2 **for 1 error,

**for 2 errors and**

*A1***for more than 2 errors.**

*A0*

*[3 marks]*

* *

(d) Since 1 and 5 are of order 6 (the same as the order of the group) every element can be written as sums of either 1 or 5. Hence the group is cyclic. *R1*

The generators are 1 and 5. *A1*

*[2 marks]*

* *

(e) A subgroup of order 3 is \(\left( {\{ 0,{\text{ }}2,{\text{ }}4\} ,{\text{ }}{ + _6}} \right)\) *A2*

**Note: **Award ** A1 **if only {0, 2, 4} is seen.

*[2 marks]*

* *

(f) Other proper subgroups are \(\left( {\{ 0\} { + _6}} \right),{\text{ }}\left( {\{ 0,{\text{ }}3\} { + _6}} \right)\) *A1A1*

**Note: **Award ** A1 **if only {0}, {0, 3} is seen.

**[2 marks]**

**Total [16 marks]**

## Examiners report

The table was well done as was showing its group properties. The order of the elements in (b) was done well except for the order of 0 which was often not given. Finding the generators did not seem difficult but correctly stating the subgroups was not often done. The notion of a ‘proper’ subgroup is not well known.

## Question

(a) Show that {1, −1, i, −i} forms a group of complex numbers *G* under multiplication.

(b) Consider \(S = \{ e,{\text{ }}a,{\text{ }}b,{\text{ }}a * b\} \) under an associative operation \( * \) where *e* is the identity element. If \(a * a = b * b = e\) and \(a * b = b * a\) , show that

(i) \(a * b * a = b\) ,

(ii) \(a * b * a * b = e\) .

(c) (i) Write down the Cayley table for \(H = \{ S{\text{ , }} * \} \).

(ii) Show that *H* is a group.

(iii) Show that *H* is an Abelian group.

(d) For the above groups, *G* and *H* , show that one is cyclic and write down why the other is not. Write down all the generators of the cyclic group.

(e) Give a reason why *G* and *H* are not isomorphic.

**Answer/Explanation**

## Markscheme

(a)

see the Cayley table, (since there are no new elements) the set is closed *A1*

1 is the identity element *A1*

1 and –1 are self inverses and i and -i form an inverse pair, hence every element has an inverse *A1*

multiplication is associative *A1*

hence {1, –1, i, –i} form a group *G* under the operation of multiplication *AG*

*[4 marks]*

(b) (i) *aba* = *aab*

= *eb* *A1*

= *b* *AG*

* *

(ii) *abab* = *aabb*

= *ee* *A1*

= *e* *AG*

*[2 marks]*

(c) (i)

* A2*

**Note:** Award ** A1** for 1 or 2 errors,

**for more than 2.**

*A0*

(ii) see the Cayley table, (since there are no new elements) the set is closed *A1*

*H* has an identity element *e* *A1*

all elements are self inverses, hence every element has an inverse *A1*

the operation is associative as stated in the question

hence {*e *, *a *, *b *, *ab*} forms a group *G* under the operation \( * \) *AG*

* *

(iii) since there is symmetry across the leading diagonal of the group table, the group is Abelian *A1*

*[6 marks]*

(d) consider the element i from the group *G* *(M1)*

\({{\text{i}}^2} = – 1\)

\({{\text{i}}^3} = – {\text{i}}\)

\({{\text{i}}^4} = 1\)

thus i is a generator for *G* and hence *G* is a cyclic group *A1*

–i is the other generator for *G* *A1*

for the group *H* there is no generator as all the elements are self inverses *R1*

*[4 marks]*

(e) since one group is cyclic and the other group is not, they are not isomorphic *R1*

*[1 mark]*

*Total [17 marks]*

## Examiners report

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. A number of candidates did not understand the term “Abelian”. Many candidates understood the conditions for a group to be cyclic. Many candidates did not realise that the answer to part (e) was actually found in part (d), hence the reason for this part only being worth 1 mark. Overall, a number of fully correct solutions to this question were seen.

## Question

Let *G* be a finite cyclic group.

(a) Prove that *G* is Abelian.

(b) Given that *a* is a generator of *G*, show that \({a^{ – 1}}\) is also a generator.

(c) Show that if the order of *G* is five, then all elements of *G*, apart from the identity, are generators of *G*.

**Answer/Explanation**

## Markscheme

(a) let *a* be a generator and consider the (general) elements \(b = {a^m},{\text{ }}c = {a^n}\) *M1*

then

\(bc = {a^m}{a^n}\) *A1*

\( = {a^n}{a^m}\) (using associativity) *R1*

\( = cb\) *A1*

therefore *G* is Abelian *AG*

*[4 marks]*

* *

(b) let *G* be of order *p* and let \(m \in \{ 1,…….,{\text{ }}p\} \), let *a* be a generator

consider \(a{a^{ – 1}} = e \Rightarrow {a^m}{({a^{ – 1}})^m} = e\) *M1R1*

this shows that \({({a^{ – 1}})^m}\) is the inverse of \({a^m}\) *R1*

as *m* increases from 1 to *p*, \({a^m}\) takes *p* different values and it generates *G* *R1*

it follows from the uniqueness of the inverse that \({({a^{ – 1}})^m}\) takes *p* different values and is a generator *R1*

*[5 marks]*

* *

(c) **EITHER**

by Lagrange, the order of any element divides the order of the group, *i.e.* 5 *R1*

the only numbers dividing 5 are 1 and 5 *R1*

the identity element is the only element of order 1 *R1*

all the other elements must be of order 5 *R1*

so they all generate *G* *AG*

**OR**

let *a* be a generator.

successive powers of *a* and therefore the elements of *G* are

\(a,{\text{ }}{a^2},{\text{ }}{a^3},{\text{ }}{a^4}{\text{ and }}{a^5} = e\) *A1*

successive powers of \({a^2}\) are \({a^2},{\text{ }}{a^4},{\text{ }}a,{\text{ }}{a^3},{\text{ }}{a^5} = e\) *A1*

successive powers of \({a^3}\) are \({a^3},{\text{ }}a,{\text{ }}{a^4},{\text{ }}{a^2},{\text{ }}{a^5} = e\) *A1*

successive powers of \({a^4}\) are \({a^4},{\text{ }}{a^3},{\text{ }}{a^2},{\text{ }}a,{\text{ }}{a^5} = e\) *A1*

this shows that \({a^2},{\text{ }}{a^3},{\text{ }}{a^4}\) are also generators in addition to *a* *AG*

*[4 marks]*

*Total [13 marks]*

## Examiners report

Solutions to (a) were often disappointing with some solutions even stating that a cyclic group is, by definition, commutative and therefore Abelian. Explanations in (b) were often poor and it was difficult in some cases to distinguish between correct and incorrect solutions. In (c), candidates who realised that Lagrange’s Theorem could be used were generally the most successful. Solutions again confirmed that, in general, candidates find theoretical questions on this topic difficult.

## Question

Set \(S = \{ {x_0},{\text{ }}{x_1},{\text{ }}{x_2},{\text{ }}{x_3},{\text{ }}{x_4},{\text{ }}{x_5}\} \) and a binary operation \( \circ \) on *S* is defined as \({x_i} \circ {x_j} = {x_k}\), where \(i + j \equiv k(\bmod 6)\).

(a) (i) Construct the Cayley table for \(\{ S,{\text{ }} \circ \} \) and hence show that it is a group.

(ii) Show that \(\{ S,{\text{ }} \circ \} \) is cyclic.

(b) Let \(\{ G,{\text{ }} * \} \) be an Abelian group of order 6. The element \(a \in {\text{G}}\) has order 2 and the element \(b \in {\text{G}}\) has order 3.

(i) Write down the six elements of \(\{ G,{\text{ }} * \} \).

(ii) Find the order of \({\text{a}} * b\) and hence show that \(\{ G,{\text{ }} * \} \) is isomorphic to \(\{ S,{\text{ }} \circ \} \).

**Answer/Explanation**

## Markscheme

(a) (i) Cayley table for \(\{ S,{\text{ }} \circ \} \)

\(\begin{array}{*{20}{c|cccccc}}

\circ &{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}} \\

\hline

{{x_0}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}} \\

{{x_1}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}} \\

{{x_2}}&{{x_2}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}} \\

{{x_3}}&{{x_3}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}} \\

{{x_4}}&{{x_4}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}} \\

{{x_5}}&{{x_5}}&{{x_0}}&{{x_1}}&{{x_2}}&{{x_3}}&{{x_4}}

\end{array}\) *A4*

**Note:** Award ** A4** for no errors,

**for one error,**

*A3***for two errors,**

*A2***for three errors and**

*A1***for four or more errors.**

*A0*

*S* is closed under \( \circ \) *A1*

\({x_0}\) is the identity *A1*

\({x_0}\) and \({x_3}\) are self-inverses, *A1*

\({x_2}\) and \({x_4}\) are mutual inverses and so are \({x_1}\) and \({x_5}\) *A1*

modular addition is associative *A1*

hence, \(\{ S,{\text{ }} \circ \} \) is a group *AG*

(ii) the order of \({x_1}\) (or \({x_5}\)) is 6, hence there exists a generator, and \(\{ S,{\text{ }} \circ \} \) is a cyclic group *A1R1*

*[11 marks]*

* *

(b) (i) *e*, *a*, *b*, *ab* *A1*

and \({b^2},{\text{ }}a{b^2}\) *A1A1*

**Note:** Accept \(ba\) and \({b^2}a\).

(ii) \({(ab)^2} = {b^2}\) *M1A1*

\({(ab)^3} = a\) *A1*

\({(ab)^4} = b\) *A1*

hence order is 6 *A1*

groups *G* and *S* have the same orders and both are cyclic *R1*

hence isomorphic *AG*

*[9 marks]*

*Total [20 marks]*

## Examiners report

a) Most candidates had the correct Cayley table and were able to show successfully that the group axioms were satisfied. Some candidates, however, simply stated that an inverse exists for each element without stating the elements and their inverses. Most candidates were able to find a generator and hence show that the group is cyclic.

b) This part was answered less successfully by many candidates. Some failed to find all the elements. Some stated that the order of *ab* is 6 without showing any working.

## Question

The group \(\{ G,{\rm{ }} * {\rm{\} }}\) has identity \({e_G}\) and the group \(\{ H,{\text{ }} \circ \} \) has identity \({e_H}\). A homomorphism \(f\) is such that \(f:G \to H\). It is given that \(f({e_G}) = {e_H}\).

Prove that for all \(a \in G,{\text{ }}f({a^{ – 1}}) = {\left( {f(a)} \right)^{ – 1}}\).

Let \(\{ H,{\text{ }} \circ \} \) be the cyclic group of order seven, and let \(p\) be a generator.

Let \(x \in G\) such that \(f(x) = {p^{\text{2}}}\).

Find \(f({x^{ – 1}})\).

Given that \(f(x * y) = p\), find \(f(y)\).

**Answer/Explanation**

## Markscheme

\(f({e_G}) = {e_H} \Rightarrow f(a * {a^{ – 1}}) = {e_H}\) *M1*

\(f\) is a homomorphism so \(f(a * {a^{ – 1}}) = f(a) \circ f({a^{ – 1}}) = {e_H}\) *M1A1*

by definition \(f(a) \circ {\left( {f(a)} \right)^{ – 1}} = {e_H}\) so \(f({a^{ – 1}}) = {\left( {f(a)} \right)^{ – 1}}\) (by the left-cancellation law) *R1*

*[4 marks]*

from (a) \(f({x^{ – 1}}) = {\left( {f(x)} \right)^{ – 1}}\)

hence \(f({x^{ – 1}}) = {({p^2})^{ – 1}} = {p^5}\) *M1A1*

*[2 marks]*

\(f(x * y) = f(x) \circ f(y)\;\;\;\)(homomorphism) *(M1)*

\({p^2} \circ f(y) = p\) *A1*

\(f(y) = {p^5} \circ p\) *(M1)*

\( = {p^6}\) *A1*

*[4 marks]*

*Total [10 marks]*

## Examiners report

Part (a) was well answered by those who understood what a homomorphism is. However many candidates simply did not have this knowledge and consequently could not get into the question.

Part (b) was well answered, even by those who could not do (a). However, there were many who having not understood what a homomorphism is, made no attempt on this easy question part. Understandably many lost a mark through not simplifying \({p^{ – 2}}\) to \({p^5}\).

Those who knew what a homomorphism is generally obtained good marks in part (c).

## 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

An Abelian group, \(\{ G,{\text{ }} * \} \), has 12 different elements which are of the form \({a^i} * {b^j}\) where \(i \in \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }}4\} \) and \(j \in \{ 1,{\text{ }}2,{\text{ }}3\} \). The elements \(a\) and \(b\) satisfy \({a^4} = e\) and \({b^3} = e\) where \(e\) is the identity.

Let \(\{ H,{\text{ }} * \} \) be the proper subgroup of \(\{ G,{\text{ }} * \} \) having the maximum possible order.

State the possible orders of an element of \(\{ G,{\text{ }} * \} \) and for each order give an example of an element of that order.

(i) State a generator for \(\{ H,{\text{ }} * \} \).

(ii) Write down the elements of \(\{ H,{\text{ }} * \} \).

(iii) Write down the elements of the coset of \(H\) containing \(a\).

**Answer/Explanation**

## Markscheme

orders are 1 2 3 4 6 12 *A2*

**Note: A1 **for four or five correct orders.

**Note: **For the rest of this question condone absence of xxx and accept equivalent expressions.

\(\begin{array}{*{20}{l}} {{\text{order:}}}&1&{{\text{element:}}}&2&{A1} \\ {}&2&{}&{{a^2}}&{A1} \\ {}&3&{}&{b{\text{ or }}{{\text{b}}^2}}&{A1} \\ {}&4&{}&{a{\text{ or }}{a^3}}&{A1} \\ {}&6&{}&{{a^2} * b{\text{ or }}{a^2} * {b^2}}&{A1} \\ {}&{12}&{}&{a * b{\text{ or }}a * {b^2}{\text{ or }}{a^3} * b{\text{ or }}{a^3} * {b^2}}&{A1} \end{array}\)

*[8 marks]*

(i) \(H\) has order 6 *(R1)*

generator is \({a^2} * b\) or \({a^2} * {b^2}\) *A1*

(ii) \(H = \left\{ {e,{\text{ }}{a^2} * b,{\text{ }}{b^2},{\text{ }}{a^2},{\text{ }}b,{\text{ }}{a^2} * {b^2}} \right\}\) *A3*

**Note: A2 **for 4 or 5 correct.

**for 2 or 3 correct.**

*A1*(iii) required coset is \(Ha\) (or \(aH\)) *(R1)*

\(Ha = \left\{ {a,{\text{ }}{a^3} * b,{\text{ }}a * {b^2},{\text{ }}{a^3},{\text{ }}a * b,{\text{ }}{a^3} * {b^2}} \right\}\) *A1*

*[7 marks]*

## Examiners report

[N/A]

[N/A]

## Question

An Abelian group, \(\{ G,{\text{ }} * \} \), has 12 different elements which are of the form \({a^i} * {b^j}\) where \(i \in \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }}4\} \) and \(j \in \{ 1,{\text{ }}2,{\text{ }}3\} \). The elements \(a\) and \(b\) satisfy \({a^4} = e\) and \({b^3} = e\) where \(e\) is the identity.

Let \(\{ H,{\text{ }} * \} \) be the proper subgroup of \(\{ G,{\text{ }} * \} \) having the maximum possible order.

State the possible orders of an element of \(\{ G,{\text{ }} * \} \) and for each order give an example of an element of that order.

(i) State a generator for \(\{ H,{\text{ }} * \} \).

(ii) Write down the elements of \(\{ H,{\text{ }} * \} \).

(iii) Write down the elements of the coset of \(H\) containing \(a\).

**Answer/Explanation**

## Markscheme

orders are 1 2 3 4 6 12 *A2*

**Note: A1 **for four or five correct orders.

**Note: **For the rest of this question condone absence of xxx and accept equivalent expressions.

\(\begin{array}{*{20}{l}} {{\text{order:}}}&1&{{\text{element:}}}&2&{A1} \\ {}&2&{}&{{a^2}}&{A1} \\ {}&3&{}&{b{\text{ or }}{{\text{b}}^2}}&{A1} \\ {}&4&{}&{a{\text{ or }}{a^3}}&{A1} \\ {}&6&{}&{{a^2} * b{\text{ or }}{a^2} * {b^2}}&{A1} \\ {}&{12}&{}&{a * b{\text{ or }}a * {b^2}{\text{ or }}{a^3} * b{\text{ or }}{a^3} * {b^2}}&{A1} \end{array}\)

*[8 marks]*

(i) \(H\) has order 6 *(R1)*

generator is \({a^2} * b\) or \({a^2} * {b^2}\) *A1*

(ii) \(H = \left\{ {e,{\text{ }}{a^2} * b,{\text{ }}{b^2},{\text{ }}{a^2},{\text{ }}b,{\text{ }}{a^2} * {b^2}} \right\}\) *A3*

**Note: A2 **for 4 or 5 correct.

**for 2 or 3 correct.**

*A1*(iii) required coset is \(Ha\) (or \(aH\)) *(R1)*

\(Ha = \left\{ {a,{\text{ }}{a^3} * b,{\text{ }}a * {b^2},{\text{ }}{a^3},{\text{ }}a * b,{\text{ }}{a^3} * {b^2}} \right\}\) *A1*

*[7 marks]*

## Examiners report

[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

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