The groups \(\{ K,{\text{ }} * \} \) and \(\{ H,{\text{ }} \odot \} \) are defined by the following Cayley tables.

G    

H    

By considering a suitable function from G to H , show that a surjective homomorphism exists between these two groups. State the kernel of this homomorphism.

consider the function f given by

\(f(E) = e\)

\(f(A) = e\)

\(f(B) = a\)     M1A1

\(f(C) = a\)

then, it has to be shown that

\(f(X * Y) = f(X) \odot f(Y){\text{ for all }}X{\text{ , }}Y \in G\)     (M1)

consider

\(f\left( {(E{\text{ or }}A) * (E{\text{ or }}A)} \right) = f(E{\text{ or }}A) = e;{\text{ }}f(E{\text{ or }}A) \odot f(E{\text{ or }}A) = e \odot e = e\)     M1A1

\(f\left( {(E{\text{ or }}A) * (B{\text{ or }}C)} \right) = f(B{\text{ or }}C) = a;{\text{ }}f(E{\text{ or }}A) \odot f(B{\text{ or }}C) = e \odot a = a\)     A1

\(f\left( {(B{\text{ or }}C) * (B{\text{ or }}C)} \right) = f(E{\text{ or }}A) = e;{\text{ }}f(B{\text{ or }}C) \odot f(B{\text{ or }}C) = a \odot a = e\)     A1

since the groups are Abelian, there is no need to consider \(f\left( {(B{\text{ or }}C) * (E{\text{ or }}A)} \right)\)     R1

the required property is satisfied in all cases so the homomorphism exists

Note: A comprehensive proof using tables is acceptable.

the kernel is \(\{ E,{\text{ }}A\} \)     A1

[9 marks]