Proof that commutative groupoid $(G,*)$ is a semigroup

Question

Prove that the commutative groupoid $(G,*)$ is a semigroup if the following equation is true for every $x,y,z \in G$ : $$(x*y)*z=(z*x)*y$$

So what I know is that for a groupoid to be a semigroup we need associativity, which means that $(x*y)*z=x*(y*z)$

So what I thought is to transform the equation we know is true for this groupoid, but every single of the transformations I wanted to do required either removing the brackets which is impossible if we don't still know that associativity is true for this groupoid.

My other thought is to assume that it is correct and check, but I don't know what $*$ does in this equation, so I can't just assume that from

$(x*y)*z = x*(y*z)$

$z=\large{x*(y*z)\over (x*y)}$

I hope that I'm incorrect and the way of doing this is actually this simple.

Answer 1

Let $x,y,z\in G$ be given. Applying the condition twice and the commutativity of $G$ once, we get:

$$(x*y)*z=(z*x)*y=(y*z)*x=x*(y*z)$$

Therefore, $G$ is associative.