Let G be a group of order 30. Since any subgroup of index 2 is normal, it suffices to show that G has a subgroup of order 15.

Let P be a Sylow 5-subgroup and Q a Sylow 3-subgroup of G. By the Sylow theorems, the number of Sylow 5-subgroups is n₅ = 1 or 6 and the number of Sylow 3-subgroups is n₃ = 1 or 10.

We claim that either P or Q is normal. For indirect proof, suppose that both P and Q are not normal. Then n₅ = 6 and n₃ = 10. This gives (5 - 1)(6) = 24 elements of order 5 and (3 - 1)(10) = 20 elements of order 3. So we have at least 44 elements in G, contradicting the fact that |G| = 30.

So either P or Q is normal. This means that PQ is a subgroup of G. Since |P| = 5 and |Q| = 3 are relatively prime, we have |PQ| = |P||Q| = 15. Hence, PQ is the desired subgroup.