# Why is it true that $|AB:A|=|B:A\cap B|$ even if $A$ is not normal in $AB$? (Second Isomorphism Theorem)

I just read about the First and Second Isomorphism Theorems in the book Abstract Algebra by Dummit and Foote. After proving the Second Isomorphism Theorem, they said:

Proposition 13 isn't really important for my question (I guess) but anyway it is the one that states that if $H$ and $K$ are two finite subgroups of $G$ then $|HK|=\dfrac{|H||K|}{|H\cap K|}$.

I didn't understand why $|AB:A|=|B:A\cap B|$. It is obvious if $AB$ is finite because using the Second Isomorphism theorem we have $|AB:B|=|A:A\cap B|$ because $AB/ B \cong A/ A\cap B$. Then $$|AB:B|=\dfrac{|AB|}{|B|}\quad\textsf{and}\quad|A:A\cap B|=\dfrac{|A|}{|A\cap B|}\\[0.3in] \implies\dfrac{|AB|}{|B|}=\dfrac{|A|}{|A\cap B|}\\[0.3in] \implies\dfrac{|AB|}{|A|}=\dfrac{|B|}{|A\cap B|}$$ which proves that $|AB:A|=|B:A\cap B|$, but this doesn't prove it if $A$ or $B$ is infinite and I believe there's a simple general proof that under the asssumptions in Theorem 18 we get $|AB:A|=|B:A\cap B|$. Could anyone please help me? If the proof is easy (and I guess so because they stated the relation in the book like if it was obvious) then could you give me hints?

• Another related post can be found here. Commented Aug 31, 2016 at 13:18

How do we usually prove the second isomorphism theorem? We compose two homomorphisms $$A\xrightarrow{\large\;\;\text{inclusion}\;\;}AB\xrightarrow{\large\;\;\text{quotient}\;\;}AB/B$$ observe that the composite homomorphism is surjective with kernel $A\cap B\trianglelefteq A$, and then apply the first isomorphism theorem for groups to conclude that $A/A\cap B\cong AB/B$.
What can we do when our quotients are not groups (i.e., the relevant subgroups aren't normal)? We can still apply the first isomorphism theorem for sets to the function gotten by composing $$B\xrightarrow{\large\;\;\text{inclusion}\;\;}AB\xrightarrow{\large\;\;\text{send }x\,\mapsto\, xA\;\;}AB/A$$ and observe that two elements of $B$ are sent to the same place in $AB/A$ if and only if they differ by an element of $A\cap B$. Since the composite map is also surjective, the first isomorphism theorem for sets tells us there is a bijection $$B/A\cap B\cong AB/A$$ and hence $|AB:A|=|B:A\cap B|$.
For any function of sets $f:A\to B$, define an equivalence relation on $A$ by $$a_1\sim a_2\;\text{ when }\;f(a_1)=f(a_2)$$ Then $f$ factors through a surjection, a bijection, and an injection as follows: $$\large A\; \xrightarrow[\text{surjective}]{\;\large a\;\mapsto\;[a]\;}\;A/{\sim}\; \xrightarrow[\text{bijective}]{\;\large [a]\;\mapsto\;f(a)\;} \;f(A)\;\xrightarrow[\text{injective}]{\;\large f(a)\;\mapsto\;f(a)\;}\;B$$
• That's a really nice proof. In fact I learned by your post inclusion homomorphism and first isomorphism theorem for sets (easy to prove but seems really important). One point is that I couldn't prove that the composite map $B\xrightarrow{\large\;\;\text{inclusion}\;\;}AB\xrightarrow{\large\;\;\text{send }x\,\mapsto\, xA\;\;}AB/A$ without using the assumption $A\le N_G(B)$. So thank you very much for your post. You put effort on that and even on editing my question (sorry for writing $AB\backslash B$ instead of $AB/B$ don't know why did I do such thing). Commented Jul 14, 2015 at 1:15
• @Scientifica: It may help to observe that given a group $G$ and subgroups $H,K\subseteq G$, neither of which are assumed to be normal, we have $$HK=KH\iff HK\text{ is a subgroup of }G$$ Therefore in our situation, we have $AB=BA$, and it should be clear why for any $x\in BA$, the coset $xA\in BA/A$ is equal to $yA\in BA/A$ for some $y\in B$. (This is why the composite map is surjective.) Commented Jul 14, 2015 at 8:16
• Yes you're right! Forgot about that (Proposition 14 in the book). In fact we prove that $AB\le G$ in the theorem using $A\le N_G(B)$ but still I shouldn't forget the equivalence you stated. My bad! +1 for your comment :D Commented Jul 14, 2015 at 10:14