## 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

\(\{ G,{\text{ }} * \} \) is a group with identity element \(e\). Let \(a,{\text{ }}b \in G\).

State Lagrange’s theorem.

Verify that the inverse of \(a * {b^{ – 1}}\) is equal to \(b * {a^{ – 1}}\).

Let \(\{ H,{\rm{ }} * {\rm{\} }}\) be a subgroup of \(\{ G,{\rm{ }} * {\rm{\} }}\). Let \(R\) be a relation defined on \(G\) by

\[aRb \Leftrightarrow a * {b^{ – 1}} \in H.\]

Prove that \(R\) is an equivalence relation, indicating clearly whenever you are using one of the four properties required of a group.

Let \(\{ H,{\rm{ }} * {\rm{\} }}\) be a subgroup of \(\{ G,{\rm{ }} * {\rm{\} }}\) .Let \(R\) be a relation defined on \(G\) by

\[aRb \Leftrightarrow a * {b^{ – 1}} \in H.\]

Show that \(aRb \Leftrightarrow a \in Hb\), where \(Hb\) is the right coset of \(H\) containing \(b\).

Let \(\{ H,{\rm{ }} * {\rm{\} }}\) be a subgroup of \(\{ G,{\rm{ }} * {\rm{\} }}\) .Let \(R\) be a relation defined on \(G\) by

\[aRb \Leftrightarrow a * {b^{ – 1}} \in H.\]

It is given that the number of elements in any right coset of \(H\) is equal to the order of \(H\).

Explain how this fact together with parts (c) and (d) prove Lagrange’s theorem.

**Answer/Explanation**

## Markscheme

in a **finite **group the order of any subgroup (exactly) divides the order of the group *A1A1*

*[2 marks]*

**METHOD 1**

\((a * {b^{ – 1}}) * (b * {a^{ – 1}}) = a * {b^{ – 1}} * b * {a^{ – 1}} = a * e * {a^{ – 1}} = a * {a^{ – 1}} = e\) *M1A1A1*

**Note: M1 **for multiplying,

**for at least one of the next 3 expressions,**

*A1*** A1 **for \(e\).

Allow \((b * {a^{ – 1}}) * (a * {b^{ – 1}}) = b * {a^{ – 1}} * a * {b^{ – 1}} = b * e * {b^{ – 1}} = b * {b^{ – 1}} = e\).

**METHOD 2**

\({(a * {b^{ – 1}})^{ – 1}} = {({b^{ – 1}})^{ – 1}} * {a^{ – 1}}\) *M1A1*

\( = b * {a^{ – 1}}\)*A1*

*[3 marks]*

\(a * {a^{ – 1}} = e \in H\;\;\;\)(as \(H\) is a subgroup) *M1*

so \(aRa\) and hence \(R\) is reflexive

\(aRb \Leftrightarrow a * {b^{ – 1}} \in H\). \(H\) is a subgroup so every element has an inverse in \(H\) so

\({(a * {b^{ – 1}})^{ – 1}} \in H\) *R1*

\( \Leftrightarrow b * {a^{ – 1}} \in H \Leftrightarrow bRa\) *M1*

so \(R\) is symmetric

\(aRb,{\text{ }}bRc \Leftrightarrow a * {b^{ – 1}} \in H,{\text{ }}b * {c^{ – 1}} \in H\) *M1*

as \(H\) is closed \((a * {b^{ – 1}}) * {\text{(}}b * {c^{ – 1}}) \in H\) *R1*

and using associativity *R1*

\((a * {b^{ – 1}}) * {\text{(}}b * {c^{ – 1}}) = a * ({b^{ – 1}} * b) * {c^{ – 1}} = a * {c^{ – 1}} \in H \Leftrightarrow aRc\) *A1*

therefore \(R\) is transitive

\(R\) is reflexive, symmetric and transitive

**Note: **Can be said separately at the end of each part.

hence it is an equivalence relation *AG*

*[8 marks]*

\(aRb \Leftrightarrow a * {b^{ – 1}} \in H \Leftrightarrow a * {b^{ – 1}} = h \in H\) *A1*

\( \Leftrightarrow a = h * b \Leftrightarrow a \in Hb\) *M1R1*

*[3 marks]*

(d) implies that the right cosets of \(H\) are equal to the equivalence classes of the relation in (c) *R1*

hence the cosets partition \(G\) *R1*

all the cosets are of the same size as the subgroup \(H\) so the order of \(G\) must be a multiple of \(\left| H \right|\) *R1*

*[3 marks]*

*Total [19 marks]*

## Examiners report

Many students obtained just half marks in (a) for not stating the requirement of the order to be finite.

Part (b) should have been more straightforward than many found.

In part (c) it was evident that most candidates knew what to do, but being a more difficult question fell down on a lack of rigour. Nonetheless, many candidates obtained full or partial marks on this question part.

Part (d) enabled many candidates to obtain, at least partial marks, but there were few students with the insight to be able to answer part (e) satisfactorily.

Part (d) enabled many candidates to obtain, at least partial marks, but there were few students with the insight to be able to answer part (e) satisfactorily.

## 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 set of all permutations of the elements \(1,{\text{ }}2,{\text{ }} \ldots 10\) is denoted by \(H\) and the binary operation \( \circ \) represents the composition of permutations.

The permutation \(p = (1{\text{ }}2{\text{ }}3{\text{ }}4{\text{ }}5{\text{ }}6)(7{\text{ }}8{\text{ }}9{\text{ }}10)\) generates the subgroup \(\{ G,{\text{ }} \circ \} \) of the group \(\{ H,{\text{ }} \circ \} \).

Find the order of \(\{ G,{\text{ }} \circ \} \).

State the identity element in \(\{ G,{\text{ }} \circ \} \).

Find

(i) \(p \circ p\);

(ii) the inverse of \(p \circ p\).

(i) Find the maximum possible order of an element in \(\{ H,{\text{ }} \circ \} \).

(ii) Give an example of an element with this order.

**Answer/Explanation**

## Markscheme

the order of \((G,{\text{ }} \circ )\) is \({\text{lcm}}(6,{\text{ }}4)\) *(M1)*

\( = 12\) *A1*

*[2 marks]*

\(\left( 1 \right){\rm{ }}\left( 2 \right){\rm{ }}\left( 3 \right){\rm{ }}\left( 4 \right){\rm{ }}\left( 5 \right){\rm{ }}\left( 6 \right){\rm{ }}\left( 7 \right){\rm{ }}\left( 8 \right){\rm{ }}\left( 9 \right){\rm{ }}\left( {10} \right)\) *A1*

**Note: **Accept ( ) or a word description.

**[1 mark]**

(i) \(p \circ p = (1{\text{ }}3{\text{ }}5)(2{\text{ }}4{\text{ }}6)(7{\text{ }}9)(810)\) *(M1)A1*

(ii) its inverse \( = (1{\text{ }}5{\text{ }}3)(2{\text{ }}6{\text{ }}4)(7{\text{ }}9)(810)\) *A1A1*

**Note: **Award ** A1 **for cycles of 2,

**for cycles of 3.**

*A1***[4 marks]**

(i) considering LCM of length of cycles with length \(2\), \(3\) and \(5\) *(M1)*

\(30\) *A1*

(ii) *eg*\(\;\;\;(1{\text{ }}2)(3{\text{ }}4{\text{ }}5)(6{\text{ }}7{\text{ }}8{\text{ }}9{\text{ }}10)\) *A1*

**Note: **allow FT as long as the length of cycles adds to \(10\) and their LCM is consistent with answer to part (i).

**Note: **Accept alternative notation for each part

**[3 marks]**

**Total [10 marks]**

## Examiners report

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## 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]*

## Examiners report

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## Question

The function \(f\,{\text{: }}\mathbb{Z} \to \mathbb{Z}\) is defined by \(f\left( n \right) = n + {\left( { – 1} \right)^n}\).

Prove that \(f \circ f\) is the identity function.

Show that \(f\) is injective.

Show that \(f\) is surjective.

**Answer/Explanation**

## Markscheme

**METHOD 1**

\(\left( {f \circ f} \right)\left( n \right) = n + {\left( { – 1} \right)^n} + {\left( { – 1} \right)^{n + {{\left( { – 1} \right)}^n}}}\) **M1A1**

\( = n + {\left( { – 1} \right)^n} + {\left( { – 1} \right)^n} \times {\left( { – 1} \right)^{{{\left( { – 1} \right)}^n}}}\) **(A1)**

considering \({\left( { – 1} \right)^n}\) for even and odd \(n\) **M1**

if \(n\) is odd, \({\left( { – 1} \right)^n} = – 1\) and if \(n\) is even, \({\left( { – 1} \right)^n} = 1\) and so \({\left( { – 1} \right)^{ \pm 1}} = – 1\) **A1**

\( = n + {\left( { – 1} \right)^n} – {\left( { – 1} \right)^n}\) **A1**

= \(n\) and so \(f \circ f\) is the identity function **AG**

**METHOD 2**

\(\left( {f \circ f} \right)\left( n \right) = n + {\left( { – 1} \right)^n} + {\left( { – 1} \right)^{n + {{\left( { – 1} \right)}^n}}}\) **M1A1**

\( = n + {\left( { – 1} \right)^n} + {\left( { – 1} \right)^n} \times {\left( { – 1} \right)^{{{\left( { – 1} \right)}^n}}}\) **(A1)**

\( = n + {\left( { – 1} \right)^n} \times \left( {1 + {{\left( { – 1} \right)}^{{{\left( { – 1} \right)}^n}}}} \right)\) **M1**

\({\left( { – 1} \right)^{ \pm 1}} = – 1\) **R1**

\(1 + {\left( { – 1} \right)^{{{\left( { – 1} \right)}^n}}} = 0\) **A1**

\(\left( {f \circ f} \right)\left( n \right) = n\) and so \(f \circ f\) is the identity function **AG**

**METHOD 3**

\(\left( {f \circ f} \right)\left( n \right) = f\left( {n + {{\left( { – 1} \right)}^n}} \right)\) **M1**

considering even and odd \(n\) **M1**

if \(n\) is even, \(f\left( n \right) = n + 1\) which is odd **A1**

so \(\left( {f \circ f} \right)\left( n \right) = f\left( {n + 1} \right) = \left( {n + 1} \right) – 1 = n\) **A1**

if \(n\) is odd, \(f\left( n \right) = n – 1\) which is even **A1**

so \(\left( {f \circ f} \right)\left( n \right) = f\left( {n – 1} \right) = \left( {n – 1} \right) + 1 = n\) **A1**

\(\left( {f \circ f} \right)\left( n \right) = n\) in both cases

hence \(f \circ f\) is the identity function ** AG**

**[6 marks]**

suppose \(f\left( n \right) = f\left( m \right)\) **M1**

applying \(f\) to both sides \( \Rightarrow n = m\) **R1**

hence \(f\) is injective **AG**

**[2 marks]**

\(m = f\left( n \right)\) has solution \(n = f\left( m \right)\) **R1**

hence surjective **AG**

**[1 mark]**

## Examiners report

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