Question

Is `(R,**)` a commutative group if `**` is defined in `R` by `a**b= 3ab`?

Answer 1

Not if by `R` you mean `RR`, but yes if `R = RR^"*" = RR "\" { 0 }`

Explanation:

Writing `RR^"* "` for the set `RR "\" { 0 }` ...

Let us check the axioms for a commutative group:

Closure:

If `\ a, b in RR^"* "` then `\ a * b = 3ab in RR^"* "`

Associativity:

`(a * b) * c = 3ab * c = 3(3ab)c = 3a(3bc) = a * (3bc) = a * (b * c)`

Identity, namely `1/3`:

`1/3 * a = 3(1/3)a = a`

`a * 1/3 = 3(a)(1/3) = a`

Inverse (if `a != 0`):

`1/(9a) * a = 3(1/(9a))a = 1/3`

`a * 1/(9a) = 3a(1/(9a)) = 1/3`

Commutativity:

`a * b = 3ab = 3ba = b * a`

So the only problem is that `0` has no multiplicative inverse.