Introduction:
We are given a group G of order 57 which is non-cyclic. We are also given that G contains a unique subgroup H of order 19. We need to prove that for any element g not belonging to the subgroup H, the order of g must be 3.

Proof:
Step 1: Existence of an element of order 3:
Since the order of the group G is 57, by Lagrange's theorem, the order of each element of G must divide 57. Therefore, the possible orders of elements in G are 1, 3, 19, or 57.

Assume, for the sake of contradiction, that there exists an element g in G such that the order of g is not equal to 3. This means that the order of g is either 1, 19, or 57.

Step 2: Case 1 - Order of g is 1:
If the order of g is 1, then g is the identity element. However, since g does not belong to the subgroup H, g cannot be the identity element. Therefore, g cannot have an order of 1.

Step 3: Case 2 - Order of g is 19:
If the order of g is 19, then the subgroup generated by g must have order 19. However, since H is the unique subgroup of order 19 in G, this implies that g must belong to H. But we assumed that g does not belong to H. Hence, g cannot have an order of 19.

Step 4: Case 3 - Order of g is 57:
If the order of g is 57, then the subgroup generated by g must have order 57. However, since the subgroup H already has order 19, the subgroup generated by g must contain H. But this implies that the subgroup generated by g is equal to G. Therefore, g would be a generator of G, contradicting the given fact that G is non-cyclic.

Conclusion:
Since the order of g cannot be 1, 19, or 57, the only possibility left is that the order of g must be 3. Therefore, for any element g not belonging to the subgroup H, the order of g must be 3.