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]