Prove that \(f({e_G}) = {e_H}\), where \({e_G}\) is the identity element in \(G\) and \({e_H}\) is the identity

element in \(H\).

(i)     Prove that the kernel of \(f,{\text{ }}K = {\text{Ker}}(f)\), is closed under the group operation.

(ii)     Deduce that \(K\) is a subgroup of \(G\).

(i)     Prove that \(gk{g^{ - 1}} \in K\) for all \(g \in G,{\text{ }}k \in K\).

(ii)     Deduce that each left coset of K in G is also a right coset.

\(f(g) = f({e_G}g) = f({e_G})f(g)\) for \(g \in G\)     M1A1

\( \Rightarrow f({e_G}) = {e_H}\)     AG

[2 marks]

(i)     closure: let \({k_1}\) and \({k_2} \in K\), then \(f({k_1}{k_2}) = f({k_1})f({k_2})\)     M1A1

          \( = {e_H}{e_H} = {e_H}\)     A1

          hence \({k_1}{k_2} \in K\)     R1

(ii)     K is non-empty because \({e_G}\) belongs to K     R1

          a closed non-empty subset of a finite group is a subgroup     R1AG

[6 marks]

(i)     \(f(gk{g^{ - 1}}) = f(g)f(k)f({g^{ - 1}})\)     M1

\( = f(g){e_H}f({g^{ - 1}}) = f(g{g^{ - 1}})\)     A1

\( = f({e_G}) = {e_H}\)     A1

\( \Rightarrow gk{g^{ - 1}} \in K\)     AG

(ii)     clear definition of both left and right cosets, seen somewhere.     A1

use of part (i) to show \(gK \subseteq Kg\)     M1

similarly \(Kg \subseteq gK\)     A1

hence \(gK = Kg\)     AG

[6 marks]