## 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) Consider the set *A* = {1, 3, 5, 7} under the binary operation \( * \), where \( * \) denotes multiplication modulo 8.

(i) Write down the Cayley table for \(\{ A,{\text{ }} * \} \).

(ii) Show that \(\{ A,{\text{ }} * \} \) is a group.

(iii) Find all solutions to the equation \(3 * x * 7 = y\). Give your answers in the form \((x,{\text{ }}y)\).

(b) Now consider the set *B* = {1, 3, 5, 7, 9} under the binary operation \( \otimes \), where \( \otimes \) denotes multiplication modulo 10. Show that \(\{ B,{\text{ }} \otimes \} \) is not a group.

(c) Another set *C* can be formed by removing an element from *B* so that \(\{ C,{\text{ }} \otimes \} \) is a group.

(i) State which element has to be removed.

(ii) Determine whether or not \(\{ A,{\text{ }} * \} \) and \(\{ C,{\text{ }} \otimes \} \) are isomorphic.

**Answer/Explanation**

## Markscheme

(a) (i) *A3*

**Note:** Award ** A2** for 15 correct,

**for 14 correct and**

*A1***otherwise.**

*A0*(ii) it is a group because:

the table shows closure *A1*

multiplication is associative *A1*

it possesses an identity 1 *A1*

justifying that every element has an inverse *e.g.* all self-inverse *A1*

(iii) (since \( * \) is commutative, \(5 * x = y\))

so solutions are (1, 5), (3, 7), (5, 1), (7, 3) *A2*

**Notes:** Award ** A1** for 3 correct and

**otherwise.**

*A0*Do not penalize extra incorrect solutions.

*[9 marks]*

(b)

**Note:** It is not necessary to see the Cayley table.

a valid reason *R2*

*e.g.* from the Cayley table the 5 row does not give a Latin square, or 5 does not have an inverse, so it cannot be a group

*[2 marks]*

(c) (i) remove the 5 *A1*

(ii) they are not isomorphic because all elements in *A* are self-inverse this is not the case in *C*, (e.g. \(3 \otimes 3 = 9 \ne 1\)) *R2*

**Note:** Accept any valid reason.

*[3 marks]*

*Total [14 marks]*

## Examiners report

Candidates are generally confident when dealing with a specific group and that was the situation again this year. Some candidates lost marks in (a)(ii) by not giving an adequate explanation for the truth of some of the group axioms, eg some wrote ‘every element has an inverse’. Since the question told the candidates that \(\{ A,{\text{ }} * )\) was a group, this had to be the case and the candidates were expected to justify their statement by noting that every element was self-inverse. Solutions to (c)(ii) were reasonably good in general, certainly better than solutions to questions involving isomorphisms set in previous years.

## Question

The binary operator multiplication modulo 14, denoted by \( * \), is defined on the set *S* = {2, 4, 6, 8, 10, 12}.

Copy and complete the following operation table.

(i) Show that {*S* , \( * \)} is a group.

(ii) Find the order of each element of {*S* , \( * \)}.

(iii) Hence show that {*S* , \( * \)} is cyclic and find all the generators.

The set *T* is defined by \(\{ x * x:x \in S\} \). Show that {*T* , \( * \)} is a subgroup of {*S* , \( * \)}.

**Answer/Explanation**

## Markscheme

*A4 *

**Note:** Award ** A4** for all correct,

**for one error,**

*A3***for two errors,**

*A2***for three errors and**

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

*A0**[4 marks]*

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

identity: 8 is the identity element *A1*

inverse: every element has an inverse because there is an 8 in every row and column *A1*

associativity: (modulo) multiplication is associative *A1*

therefore {*S *, \( * \)} is a group *AG*

(ii) the orders of the elements are as follows

*A4*

**Note:** Award ** A4** for all correct,

**for one error,**

*A3***for two errors,**

*A2***for three errors and**

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

*A0*(iii) **EITHER**

the group is cyclic because there are elements of order 6 *R1*

**OR**

the group is cyclic because there are generators *R1*

**THEN**

10 and 12 are the generators *A1A1** *

*[11 marks]*

looking at the Cayley table, we see that

*T* = {2, 4, 8} *A1*

this is a subgroup because it contains the identity element 8, no new elements are formed and 2 and 4 form an inverse pair *R2*

**Note:** Award ** R1** for any two conditions

*[3 marks]*

## Examiners report

Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of *T*. This approach was invariably unsuccessful.

Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of *T*. This approach was invariably unsuccessful.

Parts (a) and (b) were well done in general. Some candidates, however, when considering closure and associativity simply wrote ‘closed’ and ‘associativity’ without justification. Here, candidates were expected to make reference to their Cayley table to justify closure and to state that multiplication is associative to justify associativity. In (c), some candidates tried to show the required result without actually identifying the elements of *T*. This approach was invariably unsuccessful.

## Question

Consider the following Cayley table for the set *G* = {1, 3, 5, 7, 9, 11, 13, 15} under the operation \({ \times _{16}}\), where \({ \times _{16}}\) denotes multiplication modulo 16.

(i) Find the values of *a*, *b*, *c*, *d*, *e*, *f*, *g*, *h*, *i* and *j*.

(ii) Given that \({ \times _{16}}\) is associative, show that the set *G*, together with the operation \({ \times _{16}}\), forms a group.

The Cayley table for the set \(H = \{ e,{\text{ }}{a_1},{\text{ }}{a_2},{\text{ }}{a_3},{\text{ }}{b_1},{\text{ }}{b_2},{\text{ }}{b_3},{\text{ }}{b_4}\} \) under the operation \( * \), is shown below.

(i) Given that \( * \) is associative, show that *H* together with the operation \( * \) forms a group.

(ii) Find two subgroups of order 4.

Show that \(\{ G,{\text{ }}{ \times _{16}}\} \) and \(\{ H,{\text{ }} * \} \) are not isomorphic.

Show that \(\{ H,{\text{ }} * \} \) is not cyclic.

**Answer/Explanation**

## Markscheme

(i) \(a = 9,{\text{ }}b = 1,{\text{ }}c = 13,{\text{ }}d = 5,{\text{ }}e = 15,{\text{ }}f = 11,{\text{ }}g = 15,{\text{ }}h = 1,{\text{ }}i = 15,{\text{ }}j = 15\) *A3*

**Note:** Award ** A2** for one or two errors,

** A1** for three or four errors,

** A0** for five or more errors.

(ii) since the Cayley table only contains elements of the set *G*, then it is closed *A1*

there is an identity element which is 1 *A1*

{3, 11} and {5, 13} are inverse pairs and all other elements are self inverse *A1*

hence every element has an inverse *R1*

**Note:** Award ** A0R0** if no justification given for every element having an inverse.

since the set is closed, has an identity element, every element has an inverse and it is associative, it is a group *AG*

*[7 marks]*

(i) since the Cayley table only contains elements of the set *H*, then it is closed *A1*

there is an identity element which is *e* *A1*

\(\{ {a_1},{\text{ }}{a_3}\} \) form an inverse pair and all other elements are self inverse *A1*

hence every element has an inverse *R1*

**Note:** Award ** A0R0** if no justification given for every element having an inverse.

since the set is closed, has an identity element, every element has an inverse and it is associative, it is a group *AG*

(ii) any 2 of \(\{ e,{\text{ }}{a_1},{\text{ }}{a_2},{\text{ }}{a_3}\} ,{\text{ }}\{ e,{\text{ }}{a_2},{\text{ }}{b_1},{\text{ }}{b_2}\} ,{\text{ }}\{ e,{\text{ }}{a_2},{\text{ }}{b_3},{\text{ }}{b_4}\} \) *A2A2** *

*[8 marks]*

the groups are not isomorphic because \(\{ H,{\text{ }} * \} \) has one inverse pair whereas \(\{ G,{\text{ }}{ \times _{16}}\} \) has two inverse pairs *A2*

**Note:** Accept any other valid reason:

*e.g.* the fact that \(\{ G,{\text{ }}{ \times _{16}}\} \) is commutative and \(\{ H,{\text{ }} * \} \) is not.

*[2 marks]*

**EITHER**

a group is not cyclic if it has no generators *R1*

for the group to have a generator there must be an element in the group of order eight *A1*

since there is no element of order eight in the group, it is not cyclic *A1*

**OR**

a group is not cyclic if it has no generators *R1*

only possibilities are \({a_1}\), \({a_3}\) since all other elements are self inverse *A1*

this is not possible since it is not possible to generate any of the “*b*” elements from the “*a*” elements – the elements \({a_1},{\text{ }}{a_2},{\text{ }}{a_3},{\text{ }}{a_4}\) form a closed set *A1*

*[3 marks]*

## Examiners report

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.

Most candidates were aware of the group axioms and the properties of a group, but they were not always explained clearly. Surprisingly, a number of candidates tried to show the non-isomorphic nature of the two groups by stating that elements of different groups were not in the same position rather than considering general group properties. Many candidates understood the conditions for a group to be cyclic, but again explanations were sometimes incomplete. Overall, a good number of substantially correct solutions to this question were seen.

## Question

The group *G* has a subgroup *H*. The relation *R* is defined on *G* by *xRy* if and only if \(x{y^{ – 1}} \in H\), for \(x,{\text{ }}y \in G\).

Show that *R* is an equivalence relation.

The Cayley table for *G* is shown below.

The subgroup *H* is given as \(H = \{ e,{\text{ }}{a^2}b\} \).

(i) Find the equivalence class with respect to *R* which contains *ab*.

(ii) Another equivalence relation \(\rho \) is defined on *G* by \(x\rho y\) if and only if \({x^{ – 1}}y \in H\), for \(x,{\text{ }}y \in G\). Find the equivalence class with respect to \(\rho \) which contains *ab*.

**Answer/Explanation**

## Markscheme

\(x{x^{ – 1}} = e \in H\) *M1*

\( \Rightarrow xRx\)

hence *R* is reflexive *A1*

if *xRy* then \(x{y^{ – 1}} \in H\)

\( \Rightarrow {(x{y^{ – 1}})^{ – 1}} \in H\) *M1*

now \((x{y^{ – 1}}){(x{y^{ – 1}})^{ – 1}} = e\) and \(x{y^{ – 1}}y{x^{ – 1}} = e\)

\( \Rightarrow {(x{y^{ – 1}})^{ – 1}} = y{x^{ – 1}}\) *A1*

hence \(y{x^{ – 1}} \in H \Rightarrow yRx\)

hence *R* is symmetric *A1*

if *xRy*, *yRz* then \(x{y^{ – 1}} \in H,{\text{ }}y{z^{ – 1}} \in H\) *M1*

\( \Rightarrow (x{y^{ – 1}})(y{z^{ – 1}}) \in H\) *M1*

\( \Rightarrow x({y^{ – 1}}y){z^{ – 1}} \in H\)

\( \Rightarrow {x^{ – 1}}z \in H\)

hence *R* is transitive *A1*

hence *R* is an equivalence relation *AG*

*[8 marks]*

(i) for the equivalence class, solving:

**EITHER**

\(x{(ab)^{ – 1}} = e{\text{ or }}x{(ab)^{ – 1}} = {a^2}b\) *(M1)*

\(\{ ab,{\text{ }}a\} \) *A2*

**OR**

\(ab{(x)^{ – 1}} = e{\text{ or }}ab{(x)^{ – 1}} = {a^2}b\) *(M1)*

\(\{ ab,{\text{ }}a\} \) *A2*

* *

(ii) for the equivalence class, solving:

**EITHER**

\({x^{ – 1}}(ab) = e{\text{ or }}{x^{ – 1}}(ab) = {a^2}b\) *(M1)*

\(\{ ab,{\text{ }}{a^2}\} \) *A2*

**OR**

\({(ab)^{ – 1}}x = e{\text{ or }}{(ab)^{ – 1}}x = {a^2}b\) *(M1)*

\(\{ ab,{\text{ }}{a^2}\} \) *A2** *

*[6 marks]*

## Examiners report

Stronger candidates made a reasonable start to (a), and many were able to demonstrate that the relation was reflexive and transitive. However, the majority of candidates struggled to make a meaningful attempt to show the relation was symmetric, with many making unfounded assumptions. Equivalence classes still cause major problems and few fully correct answers were seen to (b).

Stronger candidates made a reasonable start to (a), and many were able to demonstrate that the relation was reflexive and transitive. However, the majority of candidates struggled to make a meaningful attempt to show the relation was symmetric, with many making unfounded assumptions. Equivalence classes still cause major problems and few fully correct answers were seen to (b).

## 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

Let \(A = \left\{ {a,{\text{ }}b} \right\}\).

Let the set of all these subsets be denoted by \(P(A)\) . The binary operation symmetric difference, \(\Delta\) , is defined on \(P(A)\) by \(X\Delta Y = (X\backslash Y) \cup (Y\backslash X)\) where \(X\) , \(Y \in P(A)\).

Let \({\mathbb{Z}_4} = \left\{ {0,{\text{ }}1,{\text{ }}2,{\text{ }}3} \right\}\) and \({ + _4}\) denote addition modulo \(4\).

Let \(S\) be any non-empty set. Let \(P(S)\) be the set of all subsets of \(S\) . For the following parts, you are allowed to assume that \(\Delta\), \( \cup \) and \( \cap \) are associative.

Write down all four subsets of *A *.

Construct the Cayley table for \(P(A)\) under \(\Delta \) .

Prove that \(\left\{ {P(A),{\text{ }}\Delta } \right\}\) is a group. You are allowed to assume that \(\Delta \) is associative.

Is \(\{ P(A){\text{, }}\Delta \} \) isomorphic to \(\{ {\mathbb{Z}_4},{\text{ }}{ + _4}\} \) ? Justify your answer.

(i) State the identity element for \(\{ P(S){\text{, }}\Delta \} \).

(ii) Write down \({X^{ – 1}}\) for \(X \in P(S)\) .

(iii) Hence prove that \(\{ P(S){\text{, }}\Delta \} \) is a group.

Explain why \(\{ P(S){\text{, }} \cup \} \) is not a group.

Explain why \(\{ P(S){\text{, }} \cap \} \) is not a group.

**Answer/Explanation**

## Markscheme

\(\emptyset {\text{, \{ a\} , \{ b\} , \{ a, b\} }}\) *A1*

*[1 mark]*

**A3**

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

**for two errors,**

*A1***for more than two errors.**

*A0** *

*[3 marks]*

closure is seen from the table above *A1*

\(\emptyset \) is the identity *A1*

each element is self-inverse *A1*

**Note: **Showing each element has an inverse is sufficient.

associativity is assumed so we have a group *AG*

*[3 marks]*

not isomorphic as in the above group all elements are self-inverse whereas in \(({\mathbb{Z}_4},{\text{ }}{ + _4})\) there is an element of order 4 (*e.g. *1) *R2*

*[2 marks]*

(i) \(\emptyset \) is the identity *A1*

* *

(ii) \({X^{ – 1}} = X\) *A1*

(iii) if *X *and *Y *are subsets of *S *then \(X\Delta Y\) (the set of elements that belong to *X *or *Y *but not both) is also a subset of *S*, hence closure is proved *R1*

\(\{ P(S){\text{, }}\Delta \} \) is a group because it is closed, has an identity, all elements have inverses (and \(\Delta \) is associative) *R1AG*

*[4 marks]*

not a group because although the identity is \(\emptyset {\text{, if }}X \ne \emptyset \) it is impossible to find a set *Y *such that \(X \cup Y = \emptyset \), so there are elements without an inverse *R1AG*

*[1 mark]*

not a group because although the identity is *S*, if \(X \ne S\) is impossible to find a set *Y *such that \(X \cap Y = S\), so there are elements without an inverse *R1AG*

*[1 mark]*

## Examiners report

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

## Question

Let *c *be a positive, real constant. Let *G *be the set \(\{ \left. {x \in \mathbb{R}} \right| – c < x < c\} \) . The binary operation \( * \) is defined on the set *G *by \(x * y = \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}}\).

Simplify \(\frac{c}{2} * \frac{{3c}}{4}\) .

State the identity element for *G *under \( * \).

For \(x \in G\) find an expression for \({x^{ – 1}}\) (the inverse of *x *under \( * \)).

Show that the binary operation \( * \) is commutative on *G *.

Show that the binary operation \( * \) is associative on *G *.

(i) If \(x,{\text{ }}y \in G\) explain why \((c – x)(c – y) > 0\) .

(ii) Hence show that \(x + y < c + \frac{{xy}}{c}\) .

Show that *G *is closed under \( * \).

Explain why \(\{ G, * \} \) is an Abelian group.

**Answer/Explanation**

## Markscheme

\(\frac{c}{2} * \frac{{3c}}{4} = \frac{{\frac{c}{2} + \frac{{3c}}{4}}}{{1 + \frac{1}{2} \cdot \frac{3}{4}}}\) *M1*

\( = \frac{{\frac{{5c}}{4}}}{{\frac{{11}}{8}}} = \frac{{10c}}{{11}}\) *A1*

*[2 marks]*

identity is 0 *A1*

*[1 mark]*

inverse is –*x* *A1*

*[1 mark]*

\(x * y = \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}},{\text{ }}y * x = \frac{{y + x}}{{1 + \frac{{yx}}{{{c^2}}}}}\) *M1*

(since ordinary addition and multiplication are commutative)

\(x * y = y * x{\text{ so }} * \) is commutative *R1*

**Note: **Accept arguments using symmetry.

* *

*[2 marks]*

\((x * y) * z = \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}} * z = \frac{{\left( {\frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}}} \right) + z}}{{1 + \left( {\frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}}} \right)\frac{z}{{{c^2}}}}}\) *M1*

\( = \frac{{\frac{{\left( {x + y + z + \frac{{xyz}}{{{c^2}}}} \right)}}{{\left( {1 + \frac{{xy}}{{{c^2}}}} \right)}}}}{{\frac{{\left( {1 + \frac{{xy}}{{{c^2}}} + \frac{{xz}}{{{c^2}}} + \frac{{yz}}{{{c^2}}}} \right)}}{{\left( {1 + \frac{{xy}}{{{c^2}}}} \right)}}}} = \frac{{\left( {x + y + z + \frac{{xyz}}{{{c^2}}}} \right)}}{{\left( {1 + \left( {\frac{{xy + xz + yz}}{{{c^2}}}} \right)} \right)}}\) *A1*

\(x * (y * z) = x * \left( {\frac{{y + z}}{{1 + \frac{{yz}}{{{c^2}}}}}} \right) = \frac{{x + \left( {\frac{{y + z}}{{1 + \frac{{yz}}{{{c^2}}}}}} \right)}}{{1 + \frac{x}{{{c^2}}}\left( {\frac{{y + z}}{{1 + \frac{{yz}}{{{c^2}}}}}} \right)}}\)

\( = \frac{{\frac{{\left( {x + \frac{{xyz}}{{{c^2}}} + y + z} \right)}}{{\left( {1 + \frac{{yz}}{{{c^2}}}} \right)}}}}{{\frac{{\left( {1 + \frac{{yz}}{{{c^2}}} + \frac{{xy}}{{{c^2}}} + \frac{{xz}}{{{c^2}}}} \right)}}{{\left( {1 + \frac{{yz}}{{{c^2}}}} \right)}}}} = \frac{{\left( {x + y + z + \frac{{xyz}}{{{c^2}}}} \right)}}{{\left( {1 + \left( {\frac{{xy + xz + yz}}{{{c^2}}}} \right)} \right)}}\) **A1**

since both expressions are the same \( * \) is associative *R1*

**Note**: After the initial ** M1A1**, correct arguments using symmetry also gain full marks.

* *

*[4 marks]*

(i) \(c > x{\text{ and }}c > y \Rightarrow c – x > 0{\text{ and }}c – y > 0 \Rightarrow (c – x)(c – y) > 0\) *R1AG*

* *

(ii) \({c^2} – cx – cy + xy > 0 \Rightarrow {c^2} + xy > cx + cy \Rightarrow c + \frac{{xy}}{c} > x + y{\text{ (as }}c > 0)\)

so \(x + y < c + \frac{{xy}}{c}\) *M1AG*

*[2 marks]*

if \(x,{\text{ }}y \in G{\text{ then }} – c – \frac{{xy}}{c} < x + y < c + \frac{{xy}}{c}\)

thus \( – c\left( {1 + \frac{{xy}}{{{c^2}}}} \right) < x + y < c\left( {1 + \frac{{xy}}{{{c^2}}}} \right){\text{ and }} – c < \frac{{x + y}}{{1 + \frac{{xy}}{{{c^2}}}}} < c\) *M1*

\(({\text{as }}1 + \frac{{xy}}{{{c^2}}} > 0){\text{ so }} – c < x * y < c\) *A1*

proving that *G *is closed under \( * \) *AG*

*[2 marks]*

as \(\{ G, * \} \) is closed, is associative, has an identity and all elements have an inverse *R1*

it is a group *AG*

as \( * \) is commutative *R1*

it is an Abelian group *AG*

*[2 marks]*

## Examiners report

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully – with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully – with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

Most candidates were able to answer part (a) indicating preparation in such questions. Many students failed to identify the command term “state” in parts (b) and (c) and spent a lot of time – usually unsuccessfully – with algebraic methods. Most students were able to offer satisfactory solutions to part (d) and although most showed that they knew what to do in part (e), few were able to complete the proof of associativity. Surprisingly few managed to answer parts (f) and (g) although many who continued to this stage, were able to pick up at least one of the marks for part (h), regardless of what they had done before. Many candidates interpreted the question as asking to prove that the group was Abelian, rather than proving that it was an Abelian group. Few were able to fully appreciate the significance in part (i) although there were a number of reasonable solutions.

## Question

Consider the set *S* = {1, 3, 5, 7, 9, 11, 13} under the binary operation multiplication modulo 14 denoted by \({ \times _{14}}\).

Copy and complete the following Cayley table for this binary operation.

Give one reason why \(\{ S,{\text{ }}{ \times _{14}}\} \) is not a group.

Show that a new set *G* can be formed by removing one of the elements of *S* such that \(\{ G,{\text{ }}{ \times _{14}}\} \) is a group.

Determine the order of each element of \(\{ G,{\text{ }}{ \times _{14}}\} \).

Find the proper subgroups of \(\{ G,{\text{ }}{ \times _{14}}\} \).

**Answer/Explanation**

## Markscheme

*A4*

**Note:** Award ** A3** for one error,

**for two errors,**

*A2***for three errors,**

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

*A0**[4 marks]*

any valid reason, for example *R1*

not a Latin square

7 has no inverse

*[1 mark]*

delete 7 (so that *G* = {1, 3, 5, 9, 11, 13}) *A1*

closure – evident from the table *A1*

associative because multiplication is associative *A1*

the identity is 1 *A1*

13 is self-inverse, 3 and 5 form an inverse

pair and 9 and 11 form an inverse pair *A1*

the four conditions are satisfied so that \(\{ G,{\text{ }}{ \times _{14}}\} \) is a group *AG*

*[5 marks]*

*A4*

**Note:** Award ** A3** for one error,

**for two errors,**

*A2***for three errors,**

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

*A0*

*[4 marks]*

{1}

{1, 13}\(\,\,\,\,\,\){1, 9, 11} *A1A1*

*[2 marks]*

## Examiners report

There were no problems with parts (a), (b) and (d).

There were no problems with parts (a), (b) and (d).

There were no problems with parts (a), (b) and (d) but in part (c) candidates often failed to state that the set was associative under the operation because multiplication is associative. Likewise they often failed to list the inverses of each element simply stating that the identity was present in each row and column of the Cayley table.

The majority of candidates did not answer part (d) correctly and often simply listed all subsets of order 2 and 3 as subgroups.

[N/A]

## Question

Consider the set *S *defined by \(S = \{ s \in \mathbb{Q}:2s \in \mathbb{Z}\} \).

You may assume that \( + \) (addition) and \( \times \) (multiplication) are associative binary operations

on \(\mathbb{Q}\).

(i) Write down the six smallest non-negative elements of \(S\).

(ii) Show that \(\{ S,{\text{ }} + \} \) is a group.

(iii) Give a reason why \(\{ S,{\text{ }} \times \} \) is not a group. Justify your answer.

The relation \(R\) is defined on \(S\) by \({s_1}R{s_2}\) if \(3{s_1} + 5{s_2} \in \mathbb{Z}\).

(i) Show that \(R\) is an equivalence relation.

(ii) Determine the equivalence classes.

**Answer/Explanation**

## Markscheme

(i) \({\text{0, }}\frac{{\text{1}}}{{\text{2}}}{\text{, 1, }}\frac{{\text{3}}}{{\text{2}}}{\text{, 2, }}\frac{{\text{5}}}{{\text{2}}}\) *A2*

**Notes: A2 **for all correct,

**for three to five correct.**

*A1*

(ii) **EITHER**

closure: if \({s_1},{\text{ }}{s_2} \in S\), then \({s_1} = \frac{m}{2}\) and \({s_2} = \frac{n}{2}\) for some \(m,{\text{ }}n \in {\text{¢}}\). *M1*

**Note: **Accept two distinct examples (*eg*, \(\frac{1}{2} + \frac{1}{2} = 1;{\text{ }}\frac{1}{2} + 1 = \frac{3}{2}\)) for the ** M1**.

\({s_1} + {s_2} = \frac{{m + n}}{2} \in S\) *A1*

**OR**

the sum of two half-integers *A1*

is a half-integer *R1*

**THEN**

identity: 0 is the (additive) identity *A1*

inverse: \(s + ( – s) = 0\), where \( – s \in S\) *A1*

it is associative (since \(S \subset \S\)) *A1*

the group axioms are satisfied *AG*

(iii) **EITHER**

the set is not closed under multiplication, *A1*

for example, \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\), but \(\frac{1}{4} \notin S\) *R1*

**OR**

not every element has an inverse, *A1*

for example, 3 does not have an inverse *R1*

*[9 marks]*

(i) reflexive: consider \(3s + 5s\) *M1*

\( = 8s \in {\text{¢}} \Rightarrow \) reflexive *A1*

symmetric: if \({s_1}R{s_2}\), consider \(3{s_2} + 5{s_1}\) *M1*

for example, \( = 3{s_1} + 5{s_2} + (2{s_1} – 2{s_2}) \in {\text{¢}} \Rightarrow \)symmetric *A1*

transitive: if \({s_1}R{s_2}\) and \({s_2}R{s_3}\), consider *(M1)*

\(3{s_1} + 5{s_3} = (3{s_1} + 5{s_2}) + (3{s_2} + 5{s_3}) – 8{s_2}\) **M1**

\( \in {\text{¢}} \Rightarrow \)transitive *A1*

so *R *is an equivalence relation *AG*

(ii) \({C_1} = {\text{¢}}\) **A1**

\({C_2} = \left\{ { \pm \frac{1}{2},{\text{ }} \pm \frac{3}{2},{\text{ }} \pm \frac{5}{2},{\text{ }} \ldots } \right\}\) **A1A1**

** **

**Note: A1 **for half odd integers and

**for ±.**

*A1**[10 marks]*

## Examiners report

[N/A]

[N/A]

## Question

Consider the sets

\[G = \left\{ {\frac{n}{{{6^i}}}|n \in \mathbb{Z},{\text{ }}i \in \mathbb{N}} \right\},{\text{ }}H = \left\{ {\frac{m}{{{3^j}}}|m \in \mathbb{Z},{\text{ }}j \in \mathbb{N}} \right\}.\]

Show that \((G,{\text{ }} + )\) forms a group where \( + \) denotes addition on \(\mathbb{Q}\). Associativity may be assumed.

Assuming that \((H,{\text{ }} + )\) forms a group, show that it is a proper subgroup of \((G,{\text{ }} + )\).

The mapping \(\phi :G \to G\) is given by \(\phi (g) = g + g\), for \(g \in G\).

Prove that \(\phi \) is an isomorphism.

**Answer/Explanation**

## Markscheme

closure: \(\frac{{{n_1}}}{{{6^{{i_1}}}}} + \frac{{{n_2}}}{{{6^{{i_2}}}}} = \frac{{{6^{{i_2}}}{n_1} + {6^{{i_1}}}{n_2}}}{{{6^{{i_1} + {i_2}}}}} \in G\) *A1R1*

**Note:** Award ** A1** for RHS of equation.

**is for the use of two different, but not necessarily most general elements, and the result \( \in G\) or equivalent.**

*R1*identity: \(0\) *A1*

inverse: \(\frac{{ – n}}{{{6^i}}}\) **A1**

since associativity is given, \((G,{\text{ }} + )\) forms a group *R1AG*

**Note:** The * R1* is for considering closure, the identity, inverses and associativity.

**[5 marks]**

it is required to show that \(H\) is a proper subset of \(G\) *(M1)*

let \(\frac{n}{{{3^i}}} \in H\) *M1*

then \(\frac{n}{{{3^i}}} = \frac{{{2^i}n}}{{{6^i}}} \in G\) hence \(H\) is a subgroup of \(G\) *A1*

\(H \ne G\) since \(\frac{1}{6} \in G\) but \(\frac{1}{6} \notin H\) *A1*

**Note:** The final ** A1** is only dependent on the first

**.**

*M1*hence, \(H\) is a proper subgroup of \(G\) *AG*

*[4 marks]*

consider \(\phi ({g_1} + {g_2}) = ({g_1} + {g_2}) + ({g_1} + {g_2})\) *M1*

\( = ({g_1} + {g_1}) + ({g_2} + {g_2}) = \phi ({g_1}) + \phi ({g_2})\) *A1*

(hence \(\phi \) is a homomorphism)

injectivity: let \(\phi ({g_1}) = \phi ({g_2})\) *M1*

working within \(\mathbb{Q}\) we have \(2{g_1} = 2{g_2} \Rightarrow {g_1} = {g_2}\) *A1*

surjectivity: considering even and odd numerators *M1*

\(\phi \left( {\frac{n}{{{6^i}}}} \right) = \frac{{2n}}{{{6^i}}}\) and \(\phi \left( {\frac{{3(2n + 1)}}{{{6^{i + 1}}}}} \right) = \frac{{2n + 1}}{{{6^i}}}\) *A1A1*

hence \(\phi \) is an isomorphism *AG*

*[7 marks]*

*Total [16 marks]*

## Examiners report

This part was generally well done. Where marks were lost, it was usually because a candidate failed to choose two different elements in the proof of closure.

Only a few candidates realised that they did not have to prove that \(H\) is a group – that was stated in the question. Some candidates tried to invoke Lagrange’s theorem, even though \(G\) is an infinite group.

Many candidates showed that the mapping is injective. Most attempts at proving surjectivity were unconvincing. Those candidates who attempted to establish the homomorphism property sometimes failed to use two different elements.

## Question

The binary operation \( * \) is defined on the set \(T = \{ 0,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6\} \) by \(a * b = (a + b – ab)(\bmod 7),{\text{ }}a,{\text{ }}b \in T\).

Copy and complete the following Cayley table for \(\{ T,{\text{ }} * \} \).

Prove that \(\{ T,{\text{ }} * \} \) forms an Abelian group.

Find the order of each element in \(T\).

Given that \(\{ H,{\text{ }} * \} \) is the subgroup of \(\{ T,{\text{ }} * \} \) of order \(2\), partition \(T\) into the left cosets with respect to \(H\).

**Answer/Explanation**

## Markscheme

Cayley table is

*A4*

award ** A4 **for all 16 correct,

**for up to 2 errors,**

*A3***for up to 4 errors,**

*A2***for up to 6 errors**

*A1***[4 marks]**

closed as no other element appears in the Cayley table *A1*

symmetrical about the leading diagonal so commutative *R1*

hence it is Abelian

\(0\) is the identity

as \(x * 0( = 0 * x) = x + 0 – 0 = x\) *A1*

\(0\) and \(2\) are self inverse, \(3\) and \(5\) is an inverse pair, \(4\) and \(6\) is an inverse pair *A1*

**Note: **Accept “Every row and every column has a \(0\) so each element has an inverse”.

\((a * b) * c = (a + b – ab) * c = a + b – ab + c – (a + b – ab)c\) *M1*

\( = a + b + c – ab – ac – bc + abc\) *A1*

\(a * (b * c) = a * (b + c – bc) = a + b + c – bc – a(b + c – bc)\) *A1*

\( = a + b + c – ab – ac – bc + abc\)

so \((a * b) * c = a * (b * c)\) and \( * \) is associative

**Note: **Inclusion of mod 7 may be included at any stage.

**[7 marks]**

\(0\) has order \(1\) and \(2\) has order \(2\) *A1*

\({3^2} = 4,{\text{ }}{3^3} = 2,{\text{ }}{3^4} = 6,{\text{ }}{3^5} = 5,{\text{ }}{3^6} = 0\) so \(3\) has order \(6\) *A1*

\({4^2} = 6,{\text{ }}{4^3} = 0\) so \(4\) has order \(3\) *A1*

\(5\) has order \(6\) and \(6\) has order \(3\) *A1*

*[4 marks]*

\(H = \{ 0,{\text{ }}2\} \) *A1*

\(0 * \{ 0,{\text{ }}2\} = \{ 0,{\text{ }}2\} ,{\text{ }}2 * \{ 0,{\text{ }}2\} = \{ 2,{\text{ }}0\} ,{\text{ }}3 * \{ 0,{\text{ }}2\} = \{ 3,{\text{ }}6\} ,{\text{ }}4 * \{ 0,{\text{ }}2\} = \{ 4,{\text{ }}5\} ,\)

\(5 * \{ 0,{\text{ }}2\} = \{ 5,{\text{ }}4\} ,{\text{ }}6 * \{ 0,{\text{ }}2\} = \{ 6,{\text{ }}3\} \) *M1*

**Note: **Award the ** M1 **if sufficient examples are used to find at least two of the cosets.

so the left cosets are \(\{ 0,{\text{ }}2\} ,{\text{ }}\{ 3,{\text{ }}6\} ,{\text{ }}\{ 4,{\text{ }}5\} \) *A1*

*[3 marks]*

*Total [18 marks]*

## Examiners report

[N/A]

[N/A]

[N/A]

## 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]

[N/A]

[N/A]

[N/A]