Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $Η$?

share|improve this question
2  
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
1  
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
1  
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
show 1 more comment

1 Answer

up vote 1 down vote accepted

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.)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.