Prove that the function \(f\) is a homomorphism from the group \(\{ \mathbb{Z},{\text{ }} + \} {\text{ to }}\{ D,{\text{ }}{ \times _3}\} \).

Find the kernel of \(f\).

Prove that \(\{ {\text{Ker}}(f),{\text{ }} + \} \) is a subgroup of \(\{ \mathbb{Z},{\text{ }} + \} \).

consider the cases, \(a\) and \(b\) both even, one is even and one is odd and \(a\) and \(b\) are both odd     (M1)

calculating \(f(a + b)\) and \(f(a){ \times _3}f(b)\) in at least one case     M1

if \(a\) is even and \(b\)  is even, then \(a + b\) is even

so\(\;\;\;f(a + b) = 1.\;\;\;f(a){ \times _3}f(b) = 1{ \times _3}1 = 1\)     A1

so\(\;\;\;f(a + b) = f(a){ \times _3}f(b)\)

if one is even and the other is odd, then \(a + b\) is odd

so\(\;\;\;f(a + b) = 2.\;\;\;f(a){ \times _3}f(b) = 1{ \times _3}2 = 2\)     A1

so\(\;\;\;f(a + b) = f(a){ \times _3}f(b)\)

if \(a\) is odd and \(b\) is odd, then \(a + b\) is even

so\(\;\;\;f(a + b) = 1.\;\;\;f(a){ \times _3}f(b) = 2{ \times _3}2 = 1\)     A1

so\(\;\;\;f(a + b) = f(a){ \times _3}f(b)\)

as\(\;\;\;f(a + b) = f(a){ \times _3}f(b)\;\;\;\)in all cases, so\(\;\;\;f:\mathbb{Z} \to D\) is a homomorphism     R1AG

[6 marks]

\(1\) is the identity of \(\{ D,{\text{ }}{ \times _3}\} \)     (M1)(A1)

so\(\;\;\;{\text{Ker}}(f)\) is all even numbers     A1

[3 marks]

METHOD 1

sum of any two even numbers is even so closure applies     A1

associative as it is a subset of \(\{ \mathbb{Z},{\text{ }} + \} \)     A1

identity is \(0\), which is in the kernel     A1

the inverse of any even number is also even     A1

METHOD 2

\({\text{ker}}(f) \ne \emptyset \)

\({b^{ - 1}} \in {\text{ker}}(f)\) for any \(b\)

\(a{b^{ - 1}} \in {\text{ker}}(f)\) for any \(a\) and \(b\)

Note:     Allow a general proof that the Kernel is always a subgroup.

[4 marks]

Total [13 marks]