Show that $x * y \in S$ for all $x,{\text{ }}y \in S$.

Show that the operation $*$ on the set $S$ is commutative.

Show that the operation $*$ on the set $S$ is associative.

Show that 2 is the identity element.

Show that each element $a \in S$ has an inverse.

$x,{\text{ }}y > 1 \Rightarrow (x - 1)(y - 1) > 0$     M1

$(x - 1)(y - 1) + 1 > 1$     A1

so $x * y \in S$ for all $x,{\text{ }}y \in S$     AG

[2 marks]

$x * y = (x - 1)(y - 1) + 1 = (y - 1)(x - 1) + 1 = y * x$     M1A1

so $*$ is commutative     AG

[2 marks]

$x * (y * z) = x * \left( {(y - 1)(z - 1) + 1} \right)$     M1

$= (x - 1)\left( {(y - 1)(z - 1) + 1 - 1} \right) + 1$     (A1)

$= (x - 1)(y - 1)(z - 1) + 1$     A1

$(x * y) * z = \left( {(x - 1)(y - 1) + 1} \right) * z$     M1

$= \left( {(x - 1)(y - 1) + 1 - 1} \right)(z - 1) + 1$

$= (x - 1)(y - 1)(z - 1) + 1$     A1

so $*$ is associative     AG

[5 marks]

$2 * x = (2 - 1)(x - 1) + 1 = x,{\text{ }}x * 2 = (x - 1)(2 - 1) + 1 = x$     M1

$2 * x = x * 2 = 2{\text{ }}(2 \in S)$     R1

Note:     Accept reference to commutativity instead of explicit expressions.

so 2 is the identity element     AG

[2 marks]

$a * {a^{ - 1}} = 2 \Rightarrow (a - 1)({a^{ - 1}} - 1) + 1 = 2$     M1

so ${a^{ - 1}} = 1 + \frac{1}{{a - 1}}$     A1

since $a - 1 > 0 \Rightarrow {a^{ - 1}} > 1{\text{ }}({a^{ - 1}} * a = a * {a^{ - 1}})$     R1

Note:     R1 dependent on M1.

so each element, $a \in S$, has an inverse     AG

[3 marks]