Prove that for all \(a \in G,{\text{ }}f({a^{ - 1}}) = {\left( {f(a)} \right)^{ - 1}}\).

Let \(\{ H,{\text{ }} \circ \} \) be the cyclic group of order seven, and let \(p\) be a generator.

Let \(x \in G\) such that \(f(x) = {p^{\text{2}}}\).

Find \(f({x^{ - 1}})\).

Given that \(f(x * y) = p\), find \(f(y)\).

\(f({e_G}) = {e_H} \Rightarrow f(a * {a^{ - 1}}) = {e_H}\)     M1

\(f\) is a homomorphism so \(f(a * {a^{ - 1}}) = f(a) \circ f({a^{ - 1}}) = {e_H}\)     M1A1

by definition \(f(a) \circ {\left( {f(a)} \right)^{ - 1}} = {e_H}\) so \(f({a^{ - 1}}) = {\left( {f(a)} \right)^{ - 1}}\) (by the left-cancellation law)     R1

[4 marks]

from (a) \(f({x^{ - 1}}) = {\left( {f(x)} \right)^{ - 1}}\)

hence \(f({x^{ - 1}}) = {({p^2})^{ - 1}} = {p^5}\)     M1A1

[2 marks]

\(f(x * y) = f(x) \circ f(y)\;\;\;\)(homomorphism)     (M1)

\({p^2} \circ f(y) = p\)     A1

\(f(y) = {p^5} \circ p\)     (M1)

\( = {p^6}\)     A1

[4 marks]

Total [10 marks]

Part (a) was well answered by those who understood what a homomorphism is. However many candidates simply did not have this knowledge and consequently could not get into the question.

Part (b) was well answered, even by those who could not do (a). However, there were many who having not understood what a homomorphism is, made no attempt on this easy question part. Understandably many lost a mark through not simplifying \({p^{ - 2}}\) to \({p^5}\).

Those who knew what a homomorphism is generally obtained good marks in part (c).