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]