Aside in Dummit & Foote after the Second Isomorphism Theorem

Question

Let $G$ be a group, let $A \leq G$, and let $B \trianglelefteq G$. Then the Second Isomorphism Theorem says that $AB/B \cong A / A \cap B$. In a remark after the proof (page 98), D&F say that we also have in general that $|AB : A| = |B: A \cap B|$. Why is this?

Answer 1

The isomorphism $AB/B \cong A/(A \cap B)$ and the fact that the cosets of a subgroup form a partition of the whole group give $$\begin{aligned}|AB/B| = |A/(A \cap B)| &\implies \frac{|AB|}{|B|} = \frac{|A|}{|A \cap B|} \\ &\implies \frac{|AB|}{|A|} = \frac{|B|}{|A \cap B|} \\&\implies |AB : A| = |B : A \cap B|\end{aligned}$$

Edit: As mentioned by Arturo Magidin this argument works in the case where $A$ and $B$ are finite. For the general case, Ted Shifrin's suggested map gives a bijection.

Define $\varphi:B/(A \cap B) \to AB/A$ by $\varphi(b(A \cap B)) = bA$ (note that this is a function on the set of cosets and not necessarily on a quotient group). $\varphi$ is both well-defined and injective since for any $b_1, b_2 \in B$ we have $$b_1 A = b_2 A \iff b_1^{-1}b_2 \in A \iff b_1^{-1}b_2 \in A \cap B \iff b_1(A \cap B) = b_2(A \cap B).$$

Since $A$ normalizes $B$, any element of $AB/A$ is of the form $bA$ for some $b \in B$. We have that $\varphi(b(A \cap B)) = bA$ for any $b \in B$ so $\varphi$ is surjective.