A question about the proof of Ito's theorem (Metabelian) 
Let $G = AB$, where $A$ and $B$ are abelian subgroups of $G$. Then
$G'$ is abelian.

That is the first part of Ito's theorem, of which the proof has always been well-known for its simplicity.
Ito's basic idea :
$(i)$ $[\,A, B~]=G'$.
$(ii)$ All elements of the form $a^{-1}b^{-1}ab\,(a\in A,\,b\in B\,)$ are commutative. That is, $[\,A, B~]$ is abelian.
My questions :
It is $(i)$ that confuses me. In the paper, Ito said that, for $(i)$, it suffices to show $[\,A, B~]$ is invariant under similarity transformations with all $a\in A$, $b\in B$, in the third paragraph of the first page.
Why? What does that mean? I can see something, though, the reason is still very much not clear to me.
Once in Jack Schmidt's answer it was mentioned that equation $(1)(2)$ help to show $$[\,A, B~]=[\,AB, AB~]~(=[\,G, G~]=G').$$
I believe so, but how can I manage to do it?
Any help will be welcome.
PS:
[1] An alternative link to the proof is here.
 A: Write $[x,y] = x^{-1} y^{-1} xy$ and $x^y = y^{-1} xy$. Then we have the identities $[x, yz] = [x,z] [x,y]^z$ and $[xy,z] = [x,z]^y [y,z]$. 
Now if $G = AB$, you see from the identities that each commutator in $G$ is a product of conjugates of some commutators $[a,b]$ or $[b,a]$ with $a \in A$ and $b \in B$.
Thus if you are able to show that the set $\{[a,b] : a \in A, b \in B\}$ is invariant under conjugation by elements of $A$ and $B$, it follows that $[G,G] = [A,B]$.
A: Mikko Korhonen's answer and Derek Holt's comments are perfect for my questions, and I'd like to sort out my thoughts by adding a fully-detailed one.

Define $[x,y] = x^{-1} y^{-1} xy$ and $x^y = y^{-1} xy$. It's easy to show the identities  $[x, yz] = [x,z] [x,y]^z$ and $[xy,z] = [x,z]^y [y,z]$. 
Since $G =AB$, for all $g_1,\,g_2\in G$, there exist $a_1,\,a_2\in A $, $b_1,\,b_2\in B $ such that $g_1= a_1 b_1 $, $g_2= a_2b_2. $ Therefore, ( the result can vary according to the convention of writing $g=ab[\text{or } ba]$ and how we use the identities )
$$[g_1, g_2]= [a_1b_1, a_2b_2]=[a_1 b_1, b_2] [a_1 b_1, a_2]^{b_2}=\left([a_1,b_2]^{b_1}\cdot [b_1,b_2]  \right)\cdot\left([a_1,a_2]^{b_1}\cdot [b_1,a_2]\right) ^{b_2}=[a_1,b_2]^{b_1}\cdot[b_1,a_2]^{b_2}.$$
If we can show $[A,B]$ is invariant under conjugation by elements of $A$ and $B$, we will be able to get $G'=[A,B]$, since we have  $G'=[G,G]\subseteq [A,B]$.
That is "it suffices to show $[A, B]$ are invariant..." as Ito said.

That's for my first question. Next, let me add more details to the second one following Derek Holt's nice comment.

For all $a'\in A$, $b'\in B$, we now compute $[a,b]^{a'}$ and $[a,b]^{b'}$ with $A$, $B$ abelian:
\begin{align*}
[a,b]^{a'}=a'^{-1}(a^{-1}b^{-1}ab)\,a'&=a^{-1}a'^{-1}b^{-1}aba'\\&=a^{-1}a'^{-1}b^{-1}(a'a'^{-1})aba'=a^{-1}(a'^{-1}b^{-1}a')(a'^{-1}a)ba'\\&=a^{-1}(a'^{-1}b^{-1}a')\,a\,(a'^{-1}ba')=a^{-1}(a'^{-1}ba')^{-1}a\,(a'^{-1}b^{-1}a')=[a,b^{a'}]
\\&~\\
[a,b]^{b'}=b'^{-1}(a^{-1}b^{-1}ab)\,b'&=b'^{-1}a^{-1}b^{-1}ab'b\\&=b'^{-1}a^{-1}b^{-1}(b'b'^{-1})ab'b=b'^{-1}a^{-1}(b^{-1}b')(b'^{-1}ab')b\\&=(b'^{-1}a^{-1}b')b^{-1}(b'^{-1}ab')b=(b'^{-1}ab')^{-1}b^{-1}(b'^{-1}ab')b=[a^{b'},b]
\end{align*}
$G=AB=BA$, so there exist $\widetilde{a},\,\widehat{a}\in A $, $\widetilde{b},\,\widehat{b}\in B $ such that $b^{a’} =\widetilde{a}\widetilde{b}$, $a^{b’} =\widehat{b} \widehat{a} $. However, due to one of communicator identities, $$[a,b]^{a’}=[a,b^{a'} ]=[a,\widetilde{a} \widetilde{b}]=[a, \widetilde{b}][a, \widetilde{a}]^{\widetilde{b}}= [a, \widetilde{b}]\in [A,B]; $$ $$
[a,b]^{b’}=[a^{b’},b ]=[\widehat{b} \widehat{a},b]= [\widehat{b}, b]^{\widehat{a}}[\widehat{a},b] =[\widehat{a}, b]\in [A,B]. $$
That is, $[A,B]$ is invariant under similarity transformations.
QED.
