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

 

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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.
Answer/Explanation

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.

Examiners report

[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.
Answer/Explanation

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 \(\{ G,{\text{ }} * \} \).

Answer/Explanation

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.
Answer/Explanation

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.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.

[N/A]

c.

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