Proving isomorphic groups have the same number of subgroups.

Question

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!

Answer 1

If $A$ is a subgroup of $G$, then $\theta(A) $ is a subgroup of $H$. You can prove this directly.

Next, if $A_1$, $A_2$ are subgroups with $A_1\neq A_2$, then $\theta(A_1)\neq\theta(A_2)$.

Lastly, if $B$ is a subgroup of $H$, then $A=\theta^{-1}(B)$ is a subgroup of $G$ with $\theta(A) =B$.

This gives you a bijection between the subgroups of $G$ and the subgroups of $H$, hence in particular they have the same number.

Answer 2

Consider an isomorphism $f: G\to H$. Then the map

$$K \mapsto f(K)$$

defines a bijection (why?) from the subgroups of $G$ to the subgroups of $H$ (why?). Since bijections preserve cardinalities, you are done.

Answer 3

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 $H$?