G is isomorphic to H.
Prove that if G has a subgroup of order n, H has a subgroup of order n.
G is isomorphic to H means there is an operation preserving bijection from G to H.
The Attempt at a Solution
I don't know if this is the right approach, but
G has a subgroup of order n implies e is in the subgroup, and since e in G maps to e in H, e is in the potential new subgroup of H.
Suppose g is an element in the subgroup of G. Then, g maps to some h in the new subgroup of H.
Man, I think I'm proving the wrong thing here. I'm heading toward proving a subgroup in H, not that the order is the same. I need a new direction. Any one have a direction idea?