Statement a) A group of order 289 is abelian.
To determine whether this statement is true or false, we need to consider the possible groups of order 289. By the fundamental theorem of finite abelian groups, any group of order 289 must be isomorphic to Z289 or Z17 × Z17.

- Z289 is cyclic, so it is abelian.
- Z17 × Z17 is isomorphic to the direct product of two cyclic groups of prime order, so it is also abelian.

Therefore, every group of order 289 is abelian, and statement a) is true.

Statement b) In S3, there are four elements satisfying x^2 = e and six elements satisfying y^3 = e.
S3 is the symmetric group on three elements, which consists of all permutations of three objects. We need to count the number of elements that satisfy the given equations.

- For x^2 = e, the only elements that satisfy this equation are the identity element e and the three 2-cycles (transpositions) such as (12), (23), and (13). So, there are four elements satisfying x^2 = e.

- For y^3 = e, the only elements that satisfy this equation are the identity element e and the three 3-cycles such as (123), (132), and (321). So, there are six elements satisfying y^3 = e.

Therefore, statement b) is true.

Statement c) Every proper subgroup of S3 is cyclic.
A proper subgroup of S3 is a subgroup that is not equal to the whole group S3. We can list all the proper subgroups of S3 and check if they are cyclic.

- The proper subgroups of S3 are {e}, {(12), (23), (13)}, {(123), (132), (321)}, and {(e), (123), (132)}.

- The first two subgroups {e} and {(12), (23), (13)} are cyclic because they consist of a single element and a 2-cycle respectively.

- The subgroup {(123), (132), (321)} is also cyclic because it is generated by the element (123), which is a 3-cycle.

- However, the subgroup {(e), (123), (132)} is not cyclic. It is isomorphic to the Klein four-group, which is non-cyclic.

Therefore, not every proper subgroup of S3 is cyclic, and statement c) is false.

Statement d) (Z30, t30) has four subgroups of order 15.
To determine the number of subgroups of order 15 in (Z30, t30), we need to find the elements of order 15 in Z30.

- The elements of order 15 in Z30 are the integers relatively prime to 30 and congruent to 1 modulo 2 and modulo 3. These elements are 1, 7, 11, 13, 17, 19, 23, and 29.

- Each of these elements generates a cyclic subgroup of order 15.

Therefore, (Z30, t30) has eight subgroups of order 15, not four. Therefore, statement d) is false.

In summary, the correct statements are a) A group of order 289 is abelian