Final answer:
The given question can be answered by considering all possible group orders up to four and showing that, in each case, the group must be abelian due to the constraints on the group elements and the uniqueness of the identity and inverses in a group structure.
Step-by-step explanation:
The question asks to prove that if G is a group with at most four elements, then the group is abelian, which means that the group operation is commutative. This can be proved by considering the possible orders of G and using the definition of a group and the properties of group operations.
First, if |G|=1 or |G|=2, it is trivially abelian, since there is either one element or two elements. With one element, the group operation is automatically commutative. With two elements, say a and b, where one must be the identity, e, the operation must abide by ae=ea=a and be=eb=b, and since ab must be e (because every element must have an inverse), ba must also be e due to uniqueness of inverses, making the group abelian.
If |G|=3, let's denote the non-identity elements as a and b. Since ab cannot equal a or b (as this would contradict the cancellation property in groups), and it also cannot equal ab since then a would be the identity, it must equal e. By similar reasoning, ba=e. Hence, this group with three elements is abelian as well.
Lastly, if |G|=4, each element must have a unique inverse. Let the elements be e, a, b, and c such that ea=ae=a and similarly for b and c. Now consider ab, which cannot be a, b, or e (otherwise a or b would behave like an identity or an inverse of the other, contradicting the assumption of four distinct elements). Thus, ab=c and similarly, ba must be c, as there are no other possibilities. This argument can be applied similarly for the other products (ac, ca, bc, cb), all of which must be equal to the remaining element not in the product, hence the group is abelian.