Finite-by-(abelian-by-finite) I want to prove that a (finite-by-abelian)-by-finite is also finite-by-(abelian-by-finite). If we have a (finite-by-abelian)-by-finite group $G$ that means that it has a normal subgroup of finite index $H$ which is finite- by-abelian. There is a finite normal subgroup $Α$ of $Η$ and the quotient $H/A$ is abelian. I cant continue this argument because i dont know if $A$ is normal in $G$. I think that we can choose $H$ fully invariant in $G$. But can we choose $A$ fully invariant in $Η$?
 A: I'm just writing out the details of Arturo's and Derek's comments, which is why I've made this community wiki.
Let $G/H = \{Hx_1, Hx_2, \ldots, Hx_n\}$.  Then the normal closure of $A$ in $G$ is
\begin{eqnarray*}
A^G &=& \langle A^g \mid g\in G\rangle\\
&=& \langle A^{hx_i} \mid h\in H, \; 1\leq i\leq n\rangle\\
&=& \langle A^{x_i} \mid 1\leq i\leq n\rangle.
\end{eqnarray*}
Also $A^G\leq H$, since $A\leq H$ and $H$ is normal in $G$.  
Each $A^{x_i}$ is normal in $H$, since for any $h\in H$ we have $x_i h = h' x_i$ for some $h'\in H$, so $(A^{x_i})^h = A^{x_i h} = A^{h' x_i} = A^{x_i}$.  Hence $A^{x_1},\ldots,A^{x_n}$ commute, and so $A^G = A^{x_1}A^{x_2}\cdots A^{x_n}$ and hence $A^G$ is finite, since it's a product of finitely many finite groups.
Now since $A^G$ contains $A$ and $G/A$ is abelian-by-finite, $G/A^G$ is also abelian-by-finite.  So $G$ is finite-by-(abelian-by-finite).
Suggested exercise: Show that a finitely generated (finite-by-abelian)-by-finite group is abelian-by-finite.  (Part of this is what Derek said about f.g. finite-by-abelian groups being abelian-by-finite and the rest is similar to the above.)
