# IB DP Maths Topic 8.11 Use and proof of subgroup tests HL Paper 3

## Question

A group $$\{ D,{\text{ }}{ \times _3}\}$$ is defined so that $$D = \{ 1,{\text{ }}2\}$$ and $${ \times _3}$$ is multiplication modulo $$3$$.

A function $$f:\mathbb{Z} \to D$$ is defined as $$f:x \mapsto \left\{ {\begin{array}{*{20}{c}} {1,{\text{ }}x{\text{ is even}}} \\ {2,{\text{ }}x{\text{ is odd}}} \end{array}} \right.$$.

Prove that the function $$f$$ is a homomorphism from the group $$\{ \mathbb{Z},{\text{ }} + \} {\text{ to }}\{ D,{\text{ }}{ \times _3}\}$$.

[6]
a.

Find the kernel of $$f$$.

[3]
b.

Prove that $$\{ {\text{Ker}}(f),{\text{ }} + \}$$ is a subgroup of $$\{ \mathbb{Z},{\text{ }} + \}$$.

[4]
c.

## Markscheme

consider the cases, $$a$$ and $$b$$ both even, one is even and one is odd and $$a$$ and $$b$$ are both odd     (M1)

calculating $$f(a + b)$$ and $$f(a){ \times _3}f(b)$$ in at least one case     M1

if $$a$$ is even and $$b$$  is even, then $$a + b$$ is even

so$$\;\;\;f(a + b) = 1.\;\;\;f(a){ \times _3}f(b) = 1{ \times _3}1 = 1$$     A1

so$$\;\;\;f(a + b) = f(a){ \times _3}f(b)$$

if one is even and the other is odd, then $$a + b$$ is odd

so$$\;\;\;f(a + b) = 2.\;\;\;f(a){ \times _3}f(b) = 1{ \times _3}2 = 2$$     A1

so$$\;\;\;f(a + b) = f(a){ \times _3}f(b)$$

if $$a$$ is odd and $$b$$ is odd, then $$a + b$$ is even

so$$\;\;\;f(a + b) = 1.\;\;\;f(a){ \times _3}f(b) = 2{ \times _3}2 = 1$$     A1

so$$\;\;\;f(a + b) = f(a){ \times _3}f(b)$$

as$$\;\;\;f(a + b) = f(a){ \times _3}f(b)\;\;\;$$in all cases, so$$\;\;\;f:\mathbb{Z} \to D$$ is a homomorphism     R1AG

[6 marks]

a.

$$1$$ is the identity of $$\{ D,{\text{ }}{ \times _3}\}$$     (M1)(A1)

so$$\;\;\;{\text{Ker}}(f)$$ is all even numbers     A1

[3 marks]

b.

METHOD 1

sum of any two even numbers is even so closure applies     A1

associative as it is a subset of $$\{ \mathbb{Z},{\text{ }} + \}$$     A1

identity is $$0$$, which is in the kernel     A1

the inverse of any even number is also even     A1

METHOD 2

$${\text{ker}}(f) \ne \emptyset$$

$${b^{ – 1}} \in {\text{ker}}(f)$$ for any $$b$$

$$a{b^{ – 1}} \in {\text{ker}}(f)$$ for any $$a$$ and $$b$$

Note:     Allow a general proof that the Kernel is always a subgroup.

[4 marks]

Total [13 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.

## 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.

[3]
a.

Show that $$\{ S,{\text{ }} * \}$$ is an Abelian group.

[5]
b.

Determine the orders of all the elements of $$\{ S,{\text{ }} * \}$$.

[3]
c.

(i)     Find the two proper subgroups of $$\{ S,{\text{ }} * \}$$.

(ii)     Find the coset of each of these subgroups with respect to the element 5.

[4]
d.

Solve the equation $$2 * x * 4 * x * 4 = 2$$.

[4]
e.

## Markscheme

A3

Note:     Award A3 for correct table, A2 for one or two errors, A1 for three or four errors and A0 otherwise.

[3 marks]

a.

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]

b.

A3

Note:     Award A3 for all correct values, A2 for 5 correct, A1 for 4 correct and A0 otherwise.

[3 marks]

c.

(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]

d.

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]

e.

## Examiners report

The majority of candidates were able to complete the Cayley table correctly.

a.

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.

b.

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

c.

[N/A]

d.

The majority found only one solution, usually the obvious $$x = 2$$, but sometimes only the less obvious $$x = 7$$.

e.

## Question

The group $$\{ G,{\text{ }} * \}$$ is defined on the set $$G$$ with binary operation $$*$$. $$H$$ is a subset of $$G$$ defined by $$H = \{ x:{\text{ }}x \in G,{\text{ }}a * x * {a^{ – 1}} = x{\text{ for all }}a \in G\}$$. Prove that $$\{ H,{\text{ }} * \}$$ is a subgroup of $$\{ G,{\text{ }} * \}$$.

## Markscheme

associativity: This follows from associativity in $$\{ G,{\text{ }} * \}$$     R1

the identity $$e \in H$$ since $$a * e * {a^{ – 1}} = a * {a^{ – 1}} = e$$ (for all $$a \in G$$)     R1

Note:     Condone the use of the commutativity of e if that is involved in an alternative simplification of the LHS.

closure: Let $$x,{\text{ }}y \in H$$ so that $$a * x * {a^{ – 1}} = x$$ and $$a * y * {a^{ – 1}} = y$$ for all $$a \in G$$     (M1)

multiplying, $$x * y = a * x * {a^{ – 1}} * a * y * {a^{ – 1}}$$ (for all $$a \in G$$)     A1

$$= a * x * y * {a^{ – 1}}$$    A1

therefore $$x * y \in H$$ (proving closure)     R1

inverse: Let $$x \in H$$ so that $$a * x * {a^{ – 1}} = x$$ (for all $$a \in G$$)

$${x^{ – 1}} = {(a * x * {a^{ – 1}})^{ – 1}}$$    M1

$$= a * {x^{ – 1}} * {a^{ – 1}}$$    A1

therefore $${x^{ – 1}} \in H$$     R1

hence $$\{ H,{\text{ }} * \}$$ is a subgroup of $$\{ G,{\text{ }} * \}$$     AG

Note:     Accuracy marks cannot be awarded if commutativity is assumed for general elements of $$G$$.

[9 marks]

## Examiners report

This is an abstract question, clearly defined on a subset. Far too many candidates almost immediately deduced, erroneously, that the full group was Abelian. Almost no marks were then available.

## 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 $$\{ G,{\text{ }}{ \times _{18}}\}$$.

[4]
a.i.

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

[2]
a.ii.

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

[1]
b.

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

[4]
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.