# 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 $Η$?

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You'd want to look at the normal closure of $A$ in $G$; since the normalizer contains in $H$, it is finite index, so $A^G$ is generated by finitely many copies of $A$. You would still have that $G/A^G$ is abelian-by-finite, but I'm not sure off-hand if $A^G$ is necessarily finite, even though $A$ has only finitely many conjugates in $G$. – Arturo Magidin Apr 6 '12 at 21:28
If $G$ is finitely generated, then the answer is yes. Let me know if you'd like details of that (I guess you might already know it and just be interested in the non-f.g. case). – Tara B Apr 7 '12 at 8:39
Ι am interested in general case. But can you give me details for f.g. case? – Dennis Apr 7 '12 at 9:08
Arturo: $A^G$ is generated by finitely many normal subgroups of $H$ and so is the product of those subgroups and is itself a finite normal subgroup of $H$, and hence also of $G$. – Derek Holt Apr 7 '12 at 11:01
It's also worth observing that a finitely generated finite-by-abelian group is abelian-by-finite, but that is not true in general. – Derek Holt Apr 7 '12 at 11:03

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).