## Question

The binary operation \( * \) is defined for \(a{\text{, }}b \in {\mathbb{Z}^ + }\) by

\[a * b = a + b – 2.\]

(a) Determine whether or not \( * \) is

(i) closed,

(ii) commutative,

(iii) associative.

(b) (i) Find the identity element.

(ii) Find the set of positive integers having an inverse under \( * \).

**Answer/Explanation**

## Markscheme

(a) (i) It is not closed because

\(1 * 1 = 0 \notin {\mathbb{Z}^ + }\) . *R2*

* *

(ii) \(a * b = a + b – 2\)

\(b * a = b + a – 2 = a * b\) *M1*

It is commutative. *A1*

* *

(iii) It is not associative. *A1*

Consider \((1 * 1) * 5\) and \(1 * (1 * 5)\) .

The first is undefined because \(1 * 1 \notin {\mathbb{Z}^ + }\) .

The second equals 3. *R2*

**Notes: **Award ** A1R2 **for stating that non-closure implies non-associative.

Award ** A1R1 **to candidates who show that \(a * (b * c) = (a * b) * c = a + b + c – 4\) and therefore conclude that it is associative, ignoring the non-closure.

*[7 marks]*

* *

(b) (i) The identity *e *satisfies

\(a * e = a + e – 2 = a\) *M1*

\(e = 2\,\,\,\,\,({\text{and }}2 \in {\mathbb{Z}^ + })\) *A1*

* *

(ii) \(a * {a^{ – 1}} = a + {a^{ – 1}} – 2 = 2\) *M1*

\(a + {a^{ – 1}} = 4\) *A1*

So the only elements having an inverse are 1, 2 and 3. *A1*

**Note: **Due to commutativity there is no need to check two sidedness of identity and inverse.

* *

*[5 marks]*

*Total [12 marks]*

## Examiners report

Almost all the candidates thought that the binary operation was associative, not realising that the non-closure prevented this from being the case. In the circumstances, however, partial credit was given to candidates who ‘proved’ associativity. Part (b) was well done by many candidates.

## 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

The binary operation \( * \) is defined on \(\mathbb{R}\) as follows. For any elements *a* , \(b \in \mathbb{R}\)

\[a * b = a + b + 1.\]

(i) Show that \( * \) is commutative.

(ii) Find the identity element.

(iii) Find the inverse of the element *a* .

The binary operation \( \cdot \) is defined on \(\mathbb{R}\) as follows. For any elements *a* , \(b \in \mathbb{R}\)

\(a \cdot b = 3ab\) . The set *S* is the set of all ordered pairs \((x,{\text{ }}y)\) of real numbers and the binary operation \( \odot \) is defined on the set *S* as

\(({x_1},{\text{ }}{y_1}) \odot ({x_2},{\text{ }}{y_2}) = ({x_1} * {x_2},{\text{ }}{y_1} \cdot {y_2}){\text{ }}.\)

Determine whether or not \( \odot \) is associative.

**Answer/Explanation**

## Markscheme

(i) if \( * \) is commutative \(a * b = b * a\)

since \(a + b + 1 = b + a + 1\) , \( * \) is commutative *R1*

* *

(ii) let *e* be the identity element

\(a * e = a + e + 1 = a\) *M1*

\( \Rightarrow e = – 1\) *A1*

* *

(iii) let *a* have an inverse, \({a^{ – 1}}\)

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

\( \Rightarrow {a^{ – 1}} = – 2 – a\) *A1*

*[5 marks]*

\(({x_1},{\text{ }}{y_1}) \odot \left( {({x_2},{\text{ }}{y_2}) \odot ({x_3},{\text{ }}{y_3})} \right) = ({x_1},{\text{ }}{y_1}) \odot ({x_2} + {x_3} + 1,{\text{ }}3{y_2}{y_3})\) *M1*

\( = ({x_1} + {x_2} + {x_3} + 2,{\text{ }}9{y_1}{y_2}{y_3})\) *A1A1*

\(\left( {({x_1},{\text{ }}{y_1}) \odot ({x_2},{\text{ }}{y_2})} \right) \odot ({x_3},{\text{ }}{y_3}) = ({x_1} + {x_2} + 1,{\text{ }}3{y_1}{y_2}) \odot ({x_3},{\text{ }}{y_3})\) *M1*

\( = ({x_1} + {x_2} + {x_3} + 2,{\text{ }}9{y_1}{y_2}{y_3})\) *A1*

hence \( \odot \) is associative *R1*

*[6 marks]*

## Examiners report

Part (a) of this question was the most accessible on the paper and was completed correctly by the majority of candidates.

Part (b) was completed by many candidates, but a significant number either did not understand what was meant by associative, confused associative with commutative, or were unable to complete the algebra.

## 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

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 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

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

The binary operation \( * \) is defined on \(\mathbb{N}\) by \(a * b = 1 + ab\).

Determine whether or not \( * \)

is closed;

is commutative;

is associative;

has an identity element.

**Answer/Explanation**

## Markscheme

\( * \) is closed *A1*

because \(1 + ab \in \mathbb{N}\) (when \(a,b \in \mathbb{N}\)) *R1*

*[2 marks]*

consider

\(a * b = 1 + ab = 1 + ba = b * a\) *M1A1*

therefore \( * \) is commutative

*[2 marks]*

**EITHER**

\(a * (b * c) = a * (1 + bc) = 1 + a(1 + bc){\text{ }}( = 1 + a + abc)\) *A1*

\((a * b) * c = (1 + ab) * c = 1 + c(1 + ab){\text{ }}( = 1 + c + abc)\) *A1*

(these two expressions are unequal when \(a \ne c\)) so \( * \) is not associative *R1*

**OR**

proof by counter example, for example

\(1 * (2 * 3) = 1 * 7 = 8\) *A1*

\((1 * 2) * 3 = 3 * 3 = 10\) *A1*

(these two numbers are unequal) so \( * \) is not associative *R1*

*[3 marks]*

let *e* denote the identity element; so that

\(a * e = 1 + ae = a\) gives \(e = \frac{{a – 1}}{a}\) (where \(a \ne 0\)) *M1*

then any valid statement such as: \(\frac{{a – 1}}{a} \notin \mathbb{N}\) or *e* is not unique *R1*

there is therefore no identity element *A1*

**Note:** Award the final ** A1** only if the previous

**is awarded.**

*R1** *

*[3 marks]*

## Examiners report

For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

For the commutative property some candidates began by setting \(a * b = b * a\) . For the identity element some candidates confused \(e * a\) and \(ea\) stating \(ea = a\) . Others found an expression for an inverse element but then neglected to state that it did not belong to the set of natural numbers or that it was not unique.

## 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

The binary operation \(\Delta\) is defined on the set \(S =\) {1, 2, 3, 4, 5} by the following Cayley table.

(a) State whether *S *is closed under the operation Δ and justify your answer.

(b) State whether Δ is commutative and justify your answer.

(c) State whether there is an identity element and justify your answer.

(d) Determine whether Δ is associative and justify your answer.

(e) Find the solutions of the equation \(a\Delta b = 4\Delta b\), for \(a \ne 4\).

**Answer/Explanation**

## Markscheme

(a) yes *A1*

because the Cayley table only contains elements of *S **R1*

*[2 marks]*

* *

(b) yes *A1*

because the Cayley table is symmetric *R1*

*[2 marks]*

* *

(c) no *A1*

because there is no row (and column) with 1, 2, 3, 4, 5 *R1*

*[2 marks]*

* *

(d) attempt to calculate \((a\Delta b)\Delta c\) and \(a\Delta (b\Delta c)\) for some \(a,{\text{ }}b,{\text{ }}c \in S\) *M1*

counterexample: for example, \((1\Delta 2)\Delta 3 = 2\)

\(1\Delta (2\Delta 3) = 1\) *A1*

Δ is not associative *A1*

**Note: **Accept a correct evaluation of \((a\Delta b)\Delta c\) and \(a\Delta (b\Delta c)\) for some \(a,{\text{ }}b,{\text{ }}c \in S\) for the ** M1**.

**[3 marks]**

** **

(e) for example, attempt to enumerate \(4\Delta b\) for *b* = 1, 2, 3, 4, 5 and obtain (3, 2, 1, 4, 1) *(M1)*

find \((a,{\text{ }}b) \in \left\{ {{\text{(2, 2), (2, 3)}}} \right\}\) for \(a \ne 4\) (or equivalent) *A1A1*

**Note:** Award ** M1A1A0 **if extra ‘solutions’ are listed.

*[3 marks]*

* *

*Total [12 marks]*

## Examiners report

## 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 for \(x,{\text{ }}y \in S = \{ 0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}6\} \) by

\[x * y = ({x^3}y – xy)\bmod 7.\]

Find the element \(e\) such that \(e * y = y\), for all \(y \in S\).

(i) Find the least solution of \(x * x = e\).

(ii) Deduce that \((S,{\text{ }} * )\) is not a group.

Determine whether or not \(e\) is an identity element.

**Answer/Explanation**

## Markscheme

attempt to solve \({e^3}y – ey \equiv y\bmod 7\) *(M1)*

the only solution is \(e = 5\) *A1*

*[2 marks]*

(i) attempt to solve \({x^4} – {x^2} \equiv 5\bmod 7\) *(M1)*

least solution is \(x = 2\) *A1*

(ii) suppose \((S,{\text{ }} * )\) is a group with order 7 *A1*

\(2\) has order \(2\) *A1*

since \(2\) does not divide \(7\), Lagrange’s Theorem is contradicted *R1*

hence, \((S,{\text{ }} * )\) is not a group *AG*

*[5 marks]*

(\(5\) is a left-identity), so need to test if it is a right-identity:

ie, is \(y * 5 = y\)? *M1*

\(1 * 5 = 0 \ne 1\) *A1*

so \(5\) is not an identity *A1*

*[3 marks]*

*Total [10 marks]*

## Examiners report

Many candidates were not sufficiently familiar with modular arithmetic to complete this question satisfactorily. In particular, some candidates completely ignored the requirement that solutions were required to be found modulo 7, and returned decimal answers to parts (a) and (b). Very few candidates invoked Lagrange’s theorem in part (b)(ii). Some candidates were under the misapprehension that a group had to be Abelian, so tested for commutativity in part (b)(ii). It was pleasing that many candidates realised that an identity had to be both a left and right identity.

Many candidates were not sufficiently familiar with modular arithmetic to complete this question satisfactorily. In particular, some candidates completely ignored the requirement that solutions were required to be found modulo 7, and returned decimal answers to parts (a) and (b). Very few candidates invoked Lagrange’s theorem in part (b)(ii). Some candidates were under the misapprehension that a group had to be Abelian, so tested for commutativity in part (b)(ii). It was pleasing that many candidates realised that an identity had to be both a left and right identity.

Many candidates were not sufficiently familiar with modular arithmetic to complete this question satisfactorily. In particular, some candidates completely ignored the requirement that solutions were required to be found modulo 7, and returned decimal answers to parts (a) and (b). Very few candidates invoked Lagrange’s theorem in part (b)(ii). Some candidates were under the misapprehension that a group had to be Abelian, so tested for commutativity in part (b)(ii). It was pleasing that many candidates realised that an identity had to be both a left and right identity.

## 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 set \(S\) is defined as the set of real numbers greater than 1.

The binary operation \( * \) is defined on \(S\) by \(x * y = (x – 1)(y – 1) + 1\) for all \(x,{\text{ }}y \in S\).

Let \(a \in S\).

Show that \(x * y \in S\) for all \(x,{\text{ }}y \in S\).

Show that the operation \( * \) on the set \(S\) is commutative.

Show that the operation \( * \) on the set \(S\) is associative.

Show that 2 is the identity element.

Show that each element \(a \in S\) has an inverse.

**Answer/Explanation**

## Markscheme

\(x,{\text{ }}y > 1 \Rightarrow (x – 1)(y – 1) > 0\) *M1*

\((x – 1)(y – 1) + 1 > 1\) *A1*

so \(x * y \in S\) for all \(x,{\text{ }}y \in S\) *AG*

*[2 marks]*

\(x * y = (x – 1)(y – 1) + 1 = (y – 1)(x – 1) + 1 = y * x\) *M1A1*

so \( * \) is commutative *AG*

*[2 marks]*

\(x * (y * z) = x * \left( {(y – 1)(z – 1) + 1} \right)\) *M1*

\( = (x – 1)\left( {(y – 1)(z – 1) + 1 – 1} \right) + 1\) *(A1)*

\( = (x – 1)(y – 1)(z – 1) + 1\) *A1*

\((x * y) * z = \left( {(x – 1)(y – 1) + 1} \right) * z\) *M1*

\( = \left( {(x – 1)(y – 1) + 1 – 1} \right)(z – 1) + 1\)

\( = (x – 1)(y – 1)(z – 1) + 1\) *A1*

so \( * \) is associative *AG*

*[5 marks]*

\(2 * x = (2 – 1)(x – 1) + 1 = x,{\text{ }}x * 2 = (x – 1)(2 – 1) + 1 = x\) *M1*

\(2 * x = x * 2 = 2{\text{ }}(2 \in S)\) *R1*

**Note:** Accept reference to commutativity instead of explicit expressions.

so 2 is the identity element *AG*

*[2 marks]*

\(a * {a^{ – 1}} = 2 \Rightarrow (a – 1)({a^{ – 1}} – 1) + 1 = 2\) *M1*

so \({a^{ – 1}} = 1 + \frac{1}{{a – 1}}\) *A1*

since \(a – 1 > 0 \Rightarrow {a^{ – 1}} > 1{\text{ }}({a^{ – 1}} * a = a * {a^{ – 1}})\) *R1*

**Note:** ** R1 **dependent on

**.**

*M1*so each element, \(a \in S\), has an inverse *AG*

*[3 marks]*

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