TITLE: Proving isomorphic groups have the same number of subgroups.
QUESTION [5 upvotes]: Suppose $G$ and $H$ are groups with identities $e_G$ and $e_H$, respectively, and $\theta :G\to H$ is an isomorphism. Prove that if $G$ has exactly $n$ subgroups (where $n$ is some positive integer), then $H$ has exactly $n$ subgroups. 
I've been looking at this problem for my Abstract Algebra course and have been really stuck. I know that isomorphic groups have the same algebraic structure but my professor told me not to use that fact in the proof. I would greatly appreciate any help with this proof!

REPLY [3 votes]: You'll find that $\mathcal{G}\le G$ is isomorphic to $\theta(\mathcal{G})=\{\theta(g)\mid g\in\mathcal{G}\}$, which is easy enough to see is a subgroup of $H$. But $\theta$ is an isomorphism; in particular, it has an inverse and it is surjective. What does this tell you about the subgroups of $H$?