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Date May 2015 Marks available 5 Reference code 15M.2.hl.TZ0.9
Level HL only Paper 2 Time zone TZ0
Command term Show that Question number 9 Adapted from N/A

Question

Let \(f\) be a homomorphism of a group \(G\) onto a group \(H\).

Show that if \(e\) is the identity in \(G\), then \(f(e)\) is the identity in \(H\).

[2]
a.

Show that if \(x\) is an element of \(G\), then \(f({x^{ - 1}}) = {\left( {f(x)} \right)^{ - 1}}\).

[2]
b.

Show that if \(G\) is Abelian, then \(H\) must also be Abelian.

[5]
c.

Show that if \(S\) is a subgroup of \(G\), then \(f(S)\) is a subgroup of \(H\).

[4]
d.

Markscheme

\(f(a) = f(ae) = f(a)f(e)\)     M1A1

hence \(f(e)\) is the identity in \(H\)     AG

a.

\(e' = f(e)\)

\( = f(x{x^{ - 1}})\)     M1

\( = f(x)f({x^{ - 1}})\)     A1

hence \(f({x^{ - 1}}) = {\left( {f(x)} \right)^{ - 1}}\)     AG

b.

let \(a',{\text{ }}b' \in H\), we need to show that \(a'b' = b'a'\)     (M1)

since \(f\) is onto \(H\) there exists \(a,{\text{ }}b \in G\) such that \(f(a) = a'\)

and \(f(b) = b'\)     (M1)

now \(a'b' = f(a)f(b) = f(ab)\)     A1

since \(f(ab) = f(ba)\)     M1

\(f(ba) = f(b)f(a) = b'a'\)     A1

hence Abelian     AG

c.

METHOD 1

\(e' = f(e)\) and \(f({x^{ - 1}}) = {\left( {f(x)} \right)^{ - 1}}\) from above     A1A1

let \(f(a)\) and \(f(b)\) be two elements in \(f(S)\)

then \(f(a)f(b) = f(ab)\)     M1

\( \Rightarrow f(a)f(b) \in f(S)\)     A1

hence closed under the operation of \(H\)

\(f(S)\) is a subgroup of \(H\)     AG

 

METHOD 2

\(f(S)\) contains the identity, so is non empty     A1

Suppose \(f(a),{\text{ }}f(b) \in f(S)\)

Consider \(f(a)f{(b)^{ - 1}}\)     M1

\( = f(a)f({b^{ - 1}})\)     (from (b))     A1

\( = f(a{b^{ - 1}})\)     (homomorphism)     A1

\( \in f(S)\) as \(a{b^{ - 1}} \in H\)

So \(f(S)\) is a subgroup of \(H\) (by a subgroup theorem)     AG

d.

Examiners report

It was pleasing to see a small number of wholly correct responses on this final question. Although the majority of candidates gained some marks, the majority failed to gain full marks because they failed to show full formal understanding of the situation.

a.

It was pleasing to see a small number of wholly correct responses on this final question. Although the majority of candidates gained some marks, the majority failed to gain full marks because they failed to show full formal understanding of the situation.

b.

It was pleasing to see a small number of wholly correct responses on this final question. Although the majority of candidates gained some marks, the majority failed to gain full marks because they failed to show full formal understanding of the situation.

c.

It was pleasing to see a small number of wholly correct responses on this final question. Although the majority of candidates gained some marks, the majority failed to gain full marks because they failed to show full formal understanding of the situation.

d.

Syllabus sections

Topic 4 - Sets, relations and groups » 4.12 » Definition of a group homomorphism.

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