$\odot$ is commutative;

$*$ is associative;

$*$ is distributive over $\odot$ ;

$\odot$ has an identity element.

$a \odot b = \sqrt {ab}  = \sqrt {ba}  = b \odot a$     A1

since $a \odot b = b \odot a$ it follows that $\odot$ is commutative     R1

[2 marks]

$a * (b * c) = a * {b^2}{c^2} = {a^2}{b^4}{c^4}$     M1A1

$(a * b) * c = {a^2}{b^2} * c = {a^4}{b^4}{c^2}$     A1

these are different, therefore $*$ is not associative     R1

Note: Accept numerical counter-example.

[4 marks]

$a * (b \odot c) = a * \sqrt {bc}  = {a^2}bc$     M1A1

$(a * b) \odot (a * c) = {a^2}{b^2} \odot {a^2}{c^2} = {a^2}bc$     A1

these are equal so $*$ is distributive over $\odot$     R1

[4 marks]

the identity e would have to satisfy

$a \odot e = a$ for all a     M1

now $a \odot e = \sqrt {ae}  = a \Rightarrow e = a$     A1

therefore there is no identity element     A1

[3 marks]