Introduction:
To solve this problem, we need to determine the number of non-isomorphic subgroups of a group G with order 17.

Explanation:
1. By Lagrange's Theorem, the order of any subgroup of G must be a divisor of 17. Therefore, the possible orders of subgroups are 1, 17, and the divisors of 17.
2. The subgroup of order 1 is the trivial subgroup, which consists only of the identity element.
3. The subgroup of order 17 is the entire group G itself.
4. Let's consider the subgroups of order less than 17.
5. If G has a subgroup of order p, where p is a prime divisor of 17, then by Lagrange's Theorem, G must have an element of order p. Since 17 is a prime number, the only possible prime divisors of 17 are 1 and 17 itself.
6. If G has a subgroup of order 1, then it is the trivial subgroup.
7. If G has a subgroup of order 17, then it is the entire group G itself.
8. Therefore, the only possible subgroups of G with order less than 17 are the trivial subgroup and the entire group itself.
9. As a result, there are only two non-isomorphic subgroups of G, namely the trivial subgroup and the entire group G itself.
10. Hence, the correct option is b) 2.

Summary:
The group G of order 17 has only two non-isomorphic subgroups, namely the trivial subgroup and the entire group itself. Therefore, the correct option is b) 2.