The problem statement is:
Suppose $G_1, G_2$ are groups and $H_1 < G_1, H_2 < G_2$ and $\phi: G_1 \to G_2$ an isomorphism with $\phi(H_1) = H_2$ Prove that $[G_1 : H_1] = [G_2 : H_2]$
So, for this problem I can do it in the case that $G_1, G_2$ finite (since it's almost trivial)
Since $\phi$ an isomorphism then we know $\lvert G_1 \rvert = \lvert G_2 \rvert$ and since the image of $H_1$ is all of $H_2$ then it must be the case that $\lvert H_1 \rvert = \lvert H_2 \rvert$. Thus, by Lagrange's Theorem $$[G_1:H_1] = \frac{\lvert G_1 \rvert}{\lvert H_1 \rvert} = \frac{\lvert G_2 \rvert}{\lvert H_2 \rvert} = [G_2 : H_2]$$
However the general case where either group is infinite is where I'm having trouble. Since we can't appeal to Lagranges Theorem, as well the fact that $\phi$ is an isomorphism does not necessarily mean that $\lvert G_1 \rvert = \lvert G_2 \rvert$. Does the fact that $\phi(H_1) = H_2$ imply that $\lvert H_1 \rvert = \lvert H_2 \rvert$ in this case though? What exactly should I be thinking about to get moving in the right direction? I know this question is asking if there's an isomorphism between $2$ general groups each with subgroup such that the image of one subgroup under the isomorphism is exactly the other subgroup then the number of left cosets of $H_i$ in $G_i$ for $i = 1,2$ must be equal.
The only other piece of information that could be useful that comes to mind is the fact that $G_i/H_i$ is a partition of $G_i$. Hence, for example for $G_1$ we know that
- $\forall g_i, g_k \in G_1\ \text{either}\ g_iH_1 = g_kH_1\ \text{or}\ g_iH_1 \cap g_kH_1 = \emptyset$
- $\bigcup_{g_iH_1 \in G_1/H_1} g_iH_1 = G_1$
But not sure how to proceed.