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]