# IB DP Maths Topic 8.11 Use and proof of the result that the order of a finite group is divisible by the order of any element HL Paper 3

## 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 $$y \in S$$.


a.

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

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


b.

Determine whether or not $$e$$ is an identity element.


c.

## Markscheme

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

the only solution is $$e = 5$$     A1

[2 marks]

a.

(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]

b.

($$5$$ 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]

c.

## 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.

a.

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.

b.

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.

c.

## Question

Consider the group $$\{ G,{\text{ }}{ \times _{18}}\}$$ defined on the set $$\{ 1,{\text{ }}5,{\text{ }}7,{\text{ }}11,{\text{ }}13,{\text{ }}17\}$$ where $${ \times _{18}}$$ denotes multiplication modulo 18. The group $$\{ G,{\text{ }}{ \times _{18}}\}$$ is shown in the following Cayley table. The subgroup of $$\{ G,{\text{ }}{ \times _{18}}\}$$ of order two is denoted by $$\{ K,{\text{ }}{ \times _{18}}\}$$.

Find the order of elements 5, 7 and 17 in $$\{ G,{\text{ }}{ \times _{18}}\}$$.


a.i.

State whether or not $$\{ G,{\text{ }}{ \times _{18}}\}$$ is cyclic, justifying your answer.


a.ii.

Write down the elements in set $$K$$.


b.

Find the left cosets of $$K$$ in $$\{ G,{\text{ }}{ \times _{18}}\}$$.


c.

## Markscheme

considering powers of elements     (M1)

5 has order 6     A1

7 has order 3     A1

17 has order 2     A1

[4 marks]

a.i.

$$G$$ is cyclic     A1

because there is an element (are elements) of order 6     R1

Note:     Accept “there is a generator”; allow A1R0.

[3 marks]

a.ii.

$$\{ 1,{\text{ }}17\}$$     A1

[1 mark]

b.

multiplying $$\{ 1,{\text{ }}17\}$$ by each element of $$G$$     (M1)

$$\{ 1,{\text{ }}17\} ,{\text{ }}\{ 5,{\text{ }}13\} ,{\text{ }}\{ 7,{\text{ }}11\}$$     A1A1A1

[4 marks]

c.

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.

[N/A]

c.