Prove that \(f({e_G}) = {e_H}\).

Prove that \({\text{Ker}}(f)\) is a subgroup of \(\{ G,{\text{ }} * \} \).

let \(a \in G\) and \(f(a) \in H\)

\(f\) is a homomorphism so \(f(a * {e_G}) = f(a) \circ f({e_G})\)     (M1)

\(f(a) = f(a) \circ f({e_G})\)     A1

\({e_H} = f({e_G})\)     AG

[2 marks]

from part (a) \({e_G} \in {\text{Ker}}(f)\) and associativity follows from G    R1

let \(a,{\text{ }}b \in {\text{Ker}}(f)\)

\(f(a * b) = f(a) \circ f(b) = {e_H} \circ {e_H} = {e_H}\)     A1

hence closed since \(a * b \in {\text{Ker}}(f)\)

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

hence \({a^{ - 1}} \in {\text{Ker}}(f)\)     R1

hence \({\text{Ker}}(f)\) is subgroup of \(G\)     AG

[6 marks]