To prove that a set with a given binary operation is a "commutative group" you need to show
1) a(bc)= (ab)c (associative law)
2) there exist an "identity", e, such that ae= a for all a (existence of identity)
3) every element, a, has an inverse, \(\displaystyle a^{-1}\) such that \(\displaystyle a(a^{-1})= e\) (existence of inverses)
4) for any elements, a, b, ab = ba (commutative law)
(A general "group", not necessarily commutative, only has to satisfy the first three.)
Here the members of the group are all non zero rational numbers. Any such number can be written as "\(\displaystyle \frac{m}{n}\)" where m and n are integers and m is not 0.
The associative law requires that you show that \(\displaystyle \left[\left(\frac{m}{n}\right)\left(\frac{p}{q}\right)\right]\left(\frac{r}{s}\right)= \left(\frac{m}{n}\right)\left[\left(\frac{p}{q}\right)\left(\frac{r}{s}\right)\right]\). That is, \(\displaystyle \frac{(mp)r}{(nq)s}= \frac{m(pr)}{n(qs)}\). I presume you are allowed to use the fact that multiplication of integers is associative.
The identity element and the inverse of \(\displaystyle \frac{a}{b}\) should be obvious (It is important that 0 iss not included- 0 does not have a multiplicative inverse).
To prove the commutative law, you need to show that \(\displaystyle \left(\frac{a}{b}\right)\left(\frac{p}{q}\right)= \left(\frac{p}{q}\right)\left(\frac{a}{b}\right)\). Again use the fact that multiplication of integers is commutative.