Explanation:

To find the order of an element g that does not belong to the unique subgroup H of order 19 in the group G, we need to use some basic group theory concepts.

1. Lagrange's theorem:
Lagrange's theorem states that the order of any subgroup of a finite group divides the order of the group. In this case, since H is a subgroup of G and has order 19, the order of G must be a multiple of 19.

2. Order of an element:
The order of an element g in a group G is the smallest positive integer n such that g^n = e, where e is the identity element of the group. The order of an element also represents the order of the subgroup generated by that element.

3. Unique subgroup of order 19:
Given that G contains a unique subgroup H of order 19, we know that the order of G is a multiple of 19. Since G is not cyclic of order 57, the order of G must be greater than 57.

4. Order of g:
Now, let's consider an element g that does not belong to H. Since G is not cyclic of order 57, g cannot have order 57. The order of g should be a divisor of the order of G.

5. Possible orders of g:
Since the order of G is a multiple of 19, the possible orders of g can be multiples of 19. However, g cannot have order 19 because it does not belong to H. Therefore, the possible orders of g can be multiples of 19 greater than 19.

6. Order of g = 57:
If the order of g is 57, then the subgroup generated by g would have order 57. But since G is not cyclic of order 57, we know that there cannot be a subgroup of order 57. Therefore, the order of g cannot be 57.

7. Order of g = 3:
Since the order of g cannot be 57, we need to find the next possible order. The next possible order is 3, which is a multiple of 19 greater than 19. We can verify that the order of g is indeed 3 by checking if g^3 = e and if there are no positive integers k < 3="" such="" that="" g^k="e." if="" g^3="e" and="" there="" are="" no="" smaller="" positive="" integers="" k="" such="" that="" g^k="e," then="" the="" order="" of="" g="" is="" />

Therefore, the order of g is 3.