Consider $N \unlhd G$ and write $\overline{A}:= A/N$ for a subgroup $A \leq G$ containing $N$.
I will write an element of the quotient group $G/N$ as coset $gN$.
I want to prove that for $A \leq B$ we have
$$[B:A]= [\overline{B}: \overline{A}].$$
If everything is finite, this is an easy corollary of Lagrange, but I'm proving the general case. Here is my attempt, but possibly I'm overcomplicating things.
Define $$\phi: B/A \to \overline{B}/\overline{A}: bA \mapsto (bN)\overline{A}$$
We show that this map is a bijection.
- Well defined: $bA = b'A \implies b^{-1}b' \in A$ $$\implies (bN)\overline{A}= \{bN aN\mid a \in A\} = \{b(b^{-1}b'a)N \mid a \in A\} = \{b'aN \mid a \in A\} = (b'N)\overline{A}$$
Here, we used the fact that $A \to A: a \mapsto b^{-1}b'a$ is a bijection.
Injective: $(bN)\overline{A}= (b'N)\overline{A}\implies (bN)^{-1}b'N\in \overline{A}$ $\implies \exists a \in A: b^{-1}b'N = aN \implies a^{-1}b^{-1}b' \in N \subseteq A\implies b^{-1}b' \in A \implies b' A = bA$
Surjective: Clear.
Since $\phi$ is bijective, domain and codomain have equal cardinalities and the result follows.
Is my attempt correct?