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]