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

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

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

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

\(f(a) = f(ae) = f(a)f(e)\)     M1A1

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

\(e' = f(e)\)

\( = f(x{x^{ - 1}})\)     M1

\( = f(x)f({x^{ - 1}})\)     A1

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

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

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

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.

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.

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.

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.