Let G be a group, and let |G|=51. Since 51=(17^1)*(3^1), by Sylow's First Theorem G has subgroups of order 17 and of order 3. By Sylow's Third Theorem, n_17 | 3 and n_17 is congruent to 1 mod 17, so n_17=1; and n_3 | 17 and n_3 is congruent to 1 mod 3, so n_3=1. Thus, there exists subgroups H and K of G such that H is the only subgroup of G of order 17 and K is the only subgroup of G of order 3. By Sylow's Second Theorem, since there is only one subgroup each of order 17 and 3, then H and K are both normal in G.
Now since H and K have prime order, they must both be both cyclic and Abelian. And since H intersect K is a subgroup of both H and K, then by Lagrange's Theorem the order of H intersect K divides both the order H and the order of K. So the order of H intersect K is 1, since 17 and 3 are prime. Then H intersect K is equal to the identity. Then by the Direct Product Theorem, HxK is isomorphic to HK. Since H and K are Abelian, HxK is Abelian, so HK is Abelian. And |HxK|=|H|x|K|=17*3=51, so |HK|=51, so HK=G, so G is Abelian. This proves that any group of order 51 is Abelian.
By Lagrange's Theorem, the order of any element of g divides the order of G, which is 51. Then o(g) is contained in {1,3,17,51}. Since H is the only subgroup of order 3, and K is the only subgroup of order 17, then since H intersect K is trivial, |H U K|=17+3-1=19. So there exists an element "g" in G such that "g" is not contained in H U K. Since the identity is in H U K, then o(g) is not equal to 1. Now if o(g)=3, then |
|=3, so =K. But this contradicts our assumption that g is not contained in H U K, so o(g) is not equal to 3 by contraposition. Similarly, if o(g)=17, then ||=17, so =H. But this contradicts our assumption that g is not contained in H U K, so o(g) is not equal to 17. Thus, o(g)=51, by process of elimination, so G is cyclic. This proves that any group of order 51 is cyclic.
So to answer your question, there aren't any non-cyclic abelian groups of order 51, but only because there aren't any non-cyclic groups of order 51 at all.