It's impossible. Let's prove by contradiction. Suppose that such a subgroup exists. Denote the entire group as G and the subgroup as G1.
G1 is nontrivial, so there is g1ϵG1, g1≠0. G1 is also proper, so there is g2ϵG\G1.
g1 and 13 are pairwise coprime because 13 is prime. So we can apply the Chinese remainder theorem with k=2, n1=g1, n2=13, a1=0 and a2=g2.
So by the theorem there is integer x such that
x ≡ 0 mod g1 AND x ≡ g2 mod 13.
x ≡ 0 mod g1 means that there is integer m such that x = m*g1. Then x = g1 + g1 + ... + g1 (m times) so (x mod 13)ϵG1. But also (x mod 13) = (g2 mod 13) which is NOT in G1.
So we have a contradiction.
Answer: no such subgroup.
(of course 13 could be replaced by any prime number)
The order of a subgroup must be a divisor of the order of the group. The only positive divisors of 13 are 1 (the order of only the trivial group) and 13 (the order of only the group itself).