Show that $\{ \mathbb{Z},{\text{ }} * \}$ is an Abelian group.

Show that there is no element of order 2.

Find a proper subgroup of $\{ \mathbb{Z},{\text{ }} * \}$.

Show that the groups $\{ \mathbb{Z},{\text{ }} * \}$ and $\{ \mathbb{Z},{\text{ }} \circ \}$ are isomorphic.

closure: $\{ \mathbb{Z},{\text{ }} * \}$ is closed because $a + b - 3 \in \mathbb{Z}$     R1

identity: $a * e = a + e - 3 = a$     (M1)

$e = 3$     A1

inverse: $a * {a^{ - 1}} = a + {a^{ - 1}} - 3 = 3$     (M1)

${a^{ - 1}} = 6 - a$     A1

associative: $a * (b * c) = a * (b + c - 3) = a + b + c - 6$     A1

$\left( {a{\text{ }}*{\text{ }}b} \right){\text{ }}*{\text{ }}c{\text{ }} = \left( {a{\text{ }} + {\text{ }}b{\text{ }} - {\text{ }}3} \right)*{\text{ }}c{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c{\text{ }} - {\text{ }}6$    A1

associative because $a * (b * c) = (a * b) * c$     R1

$b * a = b + a - 3 = a + b - 3 = a * b$ therefore commutative hence Abelian     R1

hence $\{ \mathbb{Z},{\text{ }} * \}$ is an Abelian group     AG

[9 marks]

if $a$ is of order 2 then $a * a = 2a - 3 = 3$ therefore $a = 3$     A1

which is a contradiction

since $e = 3$ and has order 1     R1

Note:     R1 for recognising that the identity has order 1.

[2 marks]

for example $S = \{ - 6,{\text{ }} - 3,{\text{ }}0,{\text{ }}3,{\text{ }}6 \ldots \}$ or $S = \{ \ldots ,{\text{ }} - 1,{\text{ }}1,{\text{ }}3,{\text{ }}5,{\text{ }}7 \ldots \}$     A1R1

Note:     R1 for deducing, justifying or verifying that $\left\{ {S, * } \right\}$ is indeed a proper subgroup.

[2 marks]

we need to show that $f(a * b) = f(a) \circ f(b)$     R1

$f(a * b) = f(a + b - 3) = a + b - 9$     A1

$f(a) \circ f(b) = (a - 6) \circ (b - 6) = a + b - 9$     A1

hence isomorphic     AG

Note:     R1 for recognising that $f$ preserves the operation; award R1A0A0 for an attempt to show that $f(a \circ b) = f(a) * f(b)$.

[3 marks]