Problem understanding Itô's proof about product of abelian groups I am trying to understand Itô's Theorem that states the following: Let the group $G=AB$ be the product of two abelian subgroups $A$ and $B$. Then $G$ is metabelian.
I am following the book 'Product of groups' by Amberg, Franciosi and De Giovanni which is as follows:
Let $a$, $a_{1}$ be elements of $A$ and $b$, $b_{1}$ elements of $B$. Write $b^{a_{1}}=a_{2}b_{2}$ and $a^{b_{1}}=a_{3}b_{3}$ with $a_{2}$, $a_{3}$ in $A$ and $b_{2}$, $b_{3}$ in $B$. Then, $$[a,b]^{a_{1}b_{1}}=[a,b^{a_{1}}]^{b_{1}}=[a,b_{2}]^{b_{1}}=[a^{b_{1}},b_{2}]=[a_{3},b_{2}]$$ and $$[a,b]^{b_{1}a_{1}}=[a^{b_{1}},b]^{a_{1}}=[a_{3},b]^{a_{1}}=[a_{3},b^{a_{1}}]=[a_{3},b_{2}].$$
This proves that the commutators $[a,b]$ and $[a_{1},b_{1}]$ commute. Since the factor group $G/[A,B]$ is abelian, it follows that $G^\prime=[A,B]$, and hence $G^\prime$ is abelian.
I don't understand the second and third equalities: Why is $[a,b^{a_{1}}]^{b_{1}}=[a,b_{2}]^{b_{1}}?$ I have tried to do it by using that $a_{2}b_{2}=b_{2}^\prime a_{2} ^\prime$ for some $b_{2}^\prime \in B$ and $a_{2}^\prime \in A$, but I obtain that $$[a,a_{2}b_{2}]^{b_{1}}={b_{1}}^{-1}a{{b_{2}^\prime}}a^{-1}{b_{2}^\prime}^{-1}b_{1},$$ but how do I know that $b_{2}=b_{2}^\prime?$

Sorry, my comment was wrong. These equations are derived using the definitions $[x,y] = xyx^{-1}y^{-1}$. You can check that we then have the commutator identities:
$$[x,zy] = [x,z]z[x,y]z^{-1}\ \ \ \mathrm{and}\ \ \ [xz,y] = x[z,y]x^{-1}[x,y].$$
So in your first equation we have:$$[a,b^{a_1}] = [a,a_2b_2] = [a,b_2]a_2[a,a_2]a_2^{-1} = [a,b_2],$$ and in the second one:$$[a^{b_1},b] =[a_3b_3,b]= a_3[b_3,b]a_3^{-1}[a_3,b]=[a_3,b],$$ and also, as in the first equation, $[a_3,b^{a_1}] = [a_3,b_2]$.
