Example for $[G : G\cap H] \neq [H:G\cap H]$ with isomorphic subgroups $G \cong H$?

Question

For isomorphic finite subgroups $G,H$ of a group $A$ it holds $[G : G\cap H] = [H: G\cap H]$, since $[G : G\cap H] = \frac{|G|}{|G\cap H|} = \frac{|H|}{|G\cap H|} =[H: G\cap H]$.

Does this also hold for general isomorphic subgroups which may not be finite?

Does it hold if one of the indices is finite?

I think it does not hold in general, unless the isomorphism $G \to H$ takes $G\cap H$ to $G\cap H$.

Concretely, has someone an example for a group $A$ with subgroup $G$ such that $[G : G \cap (aGa^{-1})] \neq [aGa^{-1} : G \cap (aGa^{-1})]$ for some $a \in A$?

Answer 1

This is an answer to your "concrete" question at the end of your post. Your first question is much easier and has been answered by user1729.

A standard example for this sort of question is the solvable Baumslag-Solitar group $$A = \langle x,y \mid y^{-1}xy=x^{2} \rangle.$$ Let $G = \langle x \rangle$ and $H = y^{-1}Gy = \langle x^2 \rangle$. Then $H < G$ so $G \cap H = H$ with $|G:G\cap H| = 2$, $|H: G \cap H| = 1$.

Answer 2

Take $A=G\cong\mathbb{Z}$, and let $H$ be any proper subgroup of finite index in $G$. Then $G\cong H$, and moreover $$ |G:G\cap H|=|G:H|\neq1=|H:H|=|H:G\cap H|. $$ So no, it does not hold infinite groups, even if both indices are finite.