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]