## Question

Given that *p* , *q* and *r* are elements of a group, prove the left-cancellation rule, *i.e.* \(pq = pr \Rightarrow q = r\) .

Your solution should indicate which group axiom is used at each stage of the proof.

Consider the group *G* , of order 4, which has distinct elements *a* , *b* and *c* and the identity element *e* .

(i) Giving a reason in each case, explain why *ab* cannot equal *a* or *b* .

(ii) Given that *c* is self inverse, determine the two possible Cayley tables for *G* .

(iii) Determine which one of the groups defined by your two Cayley tables is isomorphic to the group defined by the set {1, −1, i, −i} under multiplication of complex numbers. Your solution should include a correspondence between *a*, *b*, *c*, *e* and 1, −1, i, −i .

**Answer/Explanation**

## Markscheme

\(pq = pr\)

\({p^{ – 1}}(pq) = {p^{ – 1}}(pr)\) , every element has an inverse *A1*

\(({p^{ – 1}}p)q = ({p^{ – 1}}p)r\) , Associativity *A1*

**Note:** Brackets in lines 2 and 3 must be seen.

\(eq = er\), \({p^{ – 1}}p = e\), the identity *A1*

\(q = r\), \(ea = a\) for all elements *a* of the group *A1*

*[4 marks]*

(i) let *ab* = *a* so *b* = *e* be which is a contradiction *R1*

let *ab* = *b* so *a* = *e* which is a contradiction *R1*

therefore *ab* cannot equal either *a* or *b* *AG*

(ii) the two possible Cayley tables are

table 1

*A2*

table 2

*A2*

(iii) the group defined by table 1 is isomorphic to the given group *R1*

because

**EITHER**

both contain one self-inverse element (other than the identity) *R1*

**OR**

both contain an inverse pair *R1*

**OR**

both are cyclic *R1*

**THEN**

the correspondence is \(e \to 1\), \(c \to – 1\), \(a \to i\), \(b \to – i\)

(or vice versa for the last two) *A2*

**Note:** Award the final ** A2** only if the correct group table has been identified.

*[10 marks]*

## Examiners report

Solutions to (a) were often poor with inadequate explanations often seen. It was not uncommon to see \(pq = pr\)

\({p^{ – 1}}pq = {p^{ – 1}}pr\)

\(q = r\)

without any mention of associativity. Many candidates understood what was required in (b)(i), but solutions to (b)(ii) were often poor with the tables containing elements such as *ab* and *bc* without simplification. In (b)(iii), candidates were expected to determine the isomorphism by noting that the group defined by {1, –1, i, –i} under multiplication is cyclic or that –1 is the only self-inverse element apart from the identity, without necessarily writing down the Cayley table in full which many candidates did. Many candidates just stated that there was a bijection between the two groups without giving any justification for this.

Solutions to (a) were often poor with inadequate explanations often seen. It was not uncommon to see \(pq = pr\)

\({p^{ – 1}}pq = {p^{ – 1}}pr\)

\(q = r\)

without any mention of associativity. Many candidates understood what was required in (b)(i), but solutions to (b)(ii) were often poor with the tables containing elements such as *ab* and *bc* without simplification. In (b)(iii), candidates were expected to determine the isomorphism by noting that the group defined by {1, –1, i, –i} under multiplication is cyclic or that –1 is the only self-inverse element apart from the identity, without necessarily writing down the Cayley table in full which many candidates did. Many candidates just stated that there was a bijection between the two groups without giving any justification for this.

## 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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