## Question

A binary operation is defined on {−1, 0, 1} by

\[A \odot B = \left\{ {\begin{array}{*{20}{c}}

{ – 1,}&{{\text{if }}\left| A \right| < \left| B \right|} \\

{0,}&{{\text{if }}\left| A \right| = \left| B \right|} \\

{1,}&{{\text{if }}\left| A \right| > \left| B \right|{\text{.}}}

\end{array}} \right.\]

(a) Construct the Cayley table for this operation.

(b) Giving reasons, determine whether the operation is

(i) closed;

(ii) commutative;

(iii) associative.

**Answer/Explanation**

## Markscheme

(a) the Cayley table is

\(\begin{gathered}

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

{}&{ – 1}&0&1

\end{array} \\

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

{ – 1} \\

0 \\

1

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

0&1&0 \\

{ – 1}&0&{ – 1} \\

0&1&0

\end{array}} \right) \\

\end{gathered} \) *M1A2*

**Notes:** Award ** M1** for setting up a Cayley table with labels.

Deduct ** A1** for each error or omission.

*[3 marks]*

* *

(b) (i) closed *A1*

because all entries in table belong to {–1, 0, 1} *R1*

* *

(ii) not commutative *A1*

because the Cayley table is not symmetric, or counter-example given *R1*

* *

(iii) not associative *A1*

for example because *M1*

\(0 \odot ( – 1 \odot 0) = 0 \odot 1 = – 1\)

but

\((0 \odot – 1) \odot 0 = – 1 \odot 0 = 1\) *A1*

or alternative counter-example

*[7 marks]*

*Total [10 marks]*

## Examiners report

This question was generally well done, with the exception of part(b)(iii), showing that the operation is non-associative.

## 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

The binary operation \( * \) is defined on the set *S* = {0, 1, 2, 3} by

\[a * b = a + 2b + ab(\bmod 4){\text{ .}}\]

(a) (i) Construct the Cayley table.

(ii) Write down, with a reason, whether or not your table is a Latin square.

(b) (i) Write down, with a reason, whether or not \( * \) is commutative.

(ii) Determine whether or not \( * \) is associative, justifying your answer.

(c) Find all solutions to the equation \(x * 1 = 2 * x\) , for \(x \in S\) .

**Answer/Explanation**

## Markscheme

(a) (i)

*A3*

**Note:** Award ** A3** for no errors,

**for one error,**

*A2***for two errors and**

*A1***for three or more errors.**

*A0*

(ii) it is not a Latin square because some rows/columns contain the same digit more than once *A1*

*[4 marks]*

* *

(b) (i) **EITHER**

it is not commutative because the table is not symmetric about the leading diagonal *R2*

**OR**

it is not commutative because \(a + 2b + ab \ne 2a + b + ab\) in general *R2*

**Note:** Accept a counter example *e.g.* \(1 * 2 = 3\) whereas \(2 * 1 = 2\) .

(ii) **EITHER**

for example \((0 * 1) * 1 = 2 * 1 = 2\) *M1*

and \(0 * (1 * 1) = 0 * 0 = 0\) *A1*

so \( * \) is not associative *A1*

**OR**

associative if and only if \(a * (b * c) = (a * b) * c\) *M1*

which gives

\(a + 2b + 4c + 2bc + ab + 2ac + abc = a + 2b + ab + 2c + ac + 2bc + abc\) *A1*

so \( * \) is not associative as \(2ac \ne 2c + ac\) , in general *A1*

*[5 marks]*

* *

(c) *x* = 0 is a solution *A2*

*x* = 2 is a solution *A2*

*[4 marks]*

*Total [13 marks]*

## Examiners report

This question was generally well answered.

## 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

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

A group with the binary operation of multiplication modulo 15 is shown in the following Cayley table.

Find the values represented by each of the letters in the table.

Find the order of each of the elements of the group.

Write down the three sets that form subgroups of order 2.

Find the three sets that form subgroups of order 4.

**Answer/Explanation**

## Markscheme

\(a = 1\;\;\;b = 8\;\;\;c = 4\)

\(d = 8\;\;\;e = 4\;\;\;f = 2\)

\(g = 4\;\;\;h = 2\;\;\;i = 1\) *A3*

**Note: **Award ** A3 **for 9 correct answers,

**for 6 or more, and**

*A2***for 3 or more.**

*A1***[3 marks]**

*A3*

**Note: **Award ** A3 **for 8 correct answers,

**for 6 or more, and**

*A2***for 4 or more.**

*A1***[3 marks]**

\(\{ 1,{\text{ }}4\} ,{\text{ }}\{ 1,{\text{ }}11\} ,{\text{ }}\{ 1,{\text{ }}14\} \) *A1A1*

**Note: **Award ** A1 **for 1 correct answer and

**for all 3 (and no extras).**

*A2***[2 marks]**

\(\{ 1,{\text{ }}2,{\text{ }}4,{\text{ }}8\} ,{\text{ }}\{ 1,{\text{ }}4,{\text{ }}7,{\text{ }}13\} ,\) *A1A1*

\(\{ 1,{\text{ }}4,{\text{ }}11,{\text{ }}14\} \) *A2*

*[4 marks]*

*Total [12 marks]*

## Examiners report

The first two parts of this question were generally well done. It was surprising to see how many difficulties there were with parts (c) and (d) with many answers given as {4}, {11} and {14} for example.

The first two parts of this question were generally well done. It was surprising to see how many difficulties there were with parts (c) and (d) with many answers given as {4}, {11} and {14} for example.

The first two parts of this question were generally well done. It was surprising to see how many difficulties there were with parts (c) and (d) with many answers given as {4}, {11} and {14} for example.

## Question

Consider the set \({S_3} = \{ {\text{ }}p,{\text{ }}q,{\text{ }}r,{\text{ }}s,{\text{ }}t,{\text{ }}u\} \) of permutations of the elements of the set \(\{ 1,{\text{ }}2,{\text{ }}3\} \), defined by

\(p = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 1&2&3 \end{array}} \right),{\text{ }}q = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 1&3&2 \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 3&2&1 \end{array}} \right),{\text{ }}s = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 2&1&3 \end{array}} \right),{\text{ }}t = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 2&3&1 \end{array}} \right),{\text{ }}u = \left( {\begin{array}{*{20}{c}} 1&2&3 \\ 3&1&2 \end{array}} \right).\)

Let \( \circ \) denote composition of permutations, so \(a \circ b\) means \(b\) followed by \(a\). You may assume that \(({S_3},{\text{ }} \circ )\) forms a group.

Complete the following Cayley table

**[5 marks]**

(i) State the inverse of each element.

(ii) Determine the order of each element.

Write down the subgroups containing

(i) \(r\),

(ii) \(u\).

**Answer/Explanation**

## Markscheme

**(M1)A4**

**Note:** Award ** M1** for use of Latin square property and/or attempted multiplication,

**for the first row or column,**

*A1***for the squares of \(q\), \(r\) and \(s\), then**

*A1***for all correct.**

*A2*(i) \({p^{ – 1}} = p,{\text{ }}{q^{ – 1}} = q,{\text{ }}{r^{ – 1}} = r,{\text{ }}{s^{ – 1}} = s\) *A1*

\({t^{ – 1}} = u,{\text{ }}{u^{ – 1}} = t\) *A1*

**Note:** Allow FT from part (a) unless the working becomes simpler.

(ii) using the table or direct multiplication *(M1)*

the orders of \(\{ p,{\text{ }}q,{\text{ }}r,{\text{ }}s,{\text{ }}t,{\text{ }}u\} \) are \(\{ 1,{\text{ }}2,{\text{ }}2,{\text{ }}2,{\text{ }}3,{\text{ }}3\} \) *A3*

**Note:** Award ** A1** for two, three or four correct,

**for five correct.**

*A2***[6 marks]**

(i) \(\{ p,{\text{ }}r\} {\text{ }}\left( {{\text{and }}({S_3},{\text{ }} \circ )} \right)\) *A1*

(ii) \(\{ p,{\text{ }}u,{\text{ }}t\} {\text{ }}\left( {{\text{and }}({S_3},{\text{ }} \circ )} \right)\) *A1*

**Note:** Award ** A0A1** if the identity has been omitted.

Award ** A0** in (i) or (ii) if an extra incorrect “subgroup” has been included.

**[2 marks]**

**Total [13 marks]**

## Examiners report

The majority of candidates were able to complete the Cayley table correctly. Unfortunately, many wasted time and space, laboriously working out the missing entries in the table – the identity is \(p\) and the elements \(q\), \(r\) and \(s\) are clearly of order two, so 14 entries can be filled in without any calculation. A few candidates thought \(t\) and \(u\) had order two.

Generally well done. A few candidates were unaware of the definition of the order of an element.

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

## 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 \(x = 7\).