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.

In the following problem, the only parts I didn't understand were c) and e). The remaining I did. Please, help me!

Let $A,B$ be subgroups of $G$ that normalize each other. Assume that the set $$X=\{[a,b]:a\in A,b\in B\}$$ is finite. We will show that $[A,B]$ is finite. Note first that, without loss of generality, we may assume that $G=AB$. With this assumption $A$ and $B$ are normal subgroups of $G$. Let $U=[A,B]\subset $ $A\cap B$. Clearly $U\vartriangleleft $ $G$.

a) Show that $C_{G}(X)$ is a normal subgroup of finite index in G. Show that $C_{G}(X)$ centralizes $U$.

b) Deduce from part (a) that $C_{G}(X)\cap U$ is a central subgroup of $U$ and has finite index in $U$. Schur theorem implies that $U'$ is finite.

c) Show that, without loss of generality, we may assume that $U^{\prime }=1$.

d) Clearly the subset $\{[a,u]:a\in A,u\in U\}$ of $X$ is finite and these elements commute with each other. Show that $[a,u]^{2}=[a,u^{2}]$. Conclude that $[A,U]$ is finite.

e) Show that, without loss of generality, we may assume that $[A,U]=1$. Conclude that, without loss of generality $U$ is central in $G$.

f) Show that $X$ is closed under the squaring map $x\rightarrow $ $x^{2}$. Conclude that $[A,B]$ is finite.

proof of a)

$A,B\vartriangleleft G\Rightarrow X^{g}=X,\forall g\in G\Rightarrow N_{G}\left( X\right) =G$ e $C_{G}\left( X\right) \vartriangleleft G$ $ \Rightarrow G/C_{G}\left( X\right) \simeq Aut\left( X\right) $ is finite, because $X$ is.

As $U=\left[ A,B\right] =\left\langle X\right\rangle $ then $C_{G}\left( X\right) =C_{G}\left( U\right) .$ Therefore, $C_{G}\left( X\right) $ centralizes $U,$that is, $\left[ C_{G}\left( X\right) ,U\right] =1.$

\bigskip Proof of b)

As $C_{G}\left( X\right) \cap U=C_{U}\left( X\right) $, then $\left[ C_{U}\left( X\right) ,U\right] \leq \left[ C_{G}\left( X\right) ,U\right] =1. $ Therefore, $C_{U}\left( X\right) \leq Z\left( U\right) \leq U.$

As $\left( G:C_{G}\left( X\right) \right) $ is finite and $UC_{G}\left( X\right) /C_{G}\left( X\right) \simeq U/C_{U}\left( X\right) ,$ then $% U/C_{U}\left( X\right) $ is finite. Thereofore, by Schur Theorem $U^{\prime } $ is finite, as required.

I do not know how to prove that $U/U^{\prime }$ is finite$.$ I know that $ U/U^{\prime }$ is abelian and finitely generated because $X$ is finite (by hypothesis).

Excuse me, my English is not very good.

share|improve this question
1  
What have you tried ? –  Belgi May 14 '12 at 23:35
    
I don't know as doing the parts c) and e). –  User2040 May 14 '12 at 23:45
add comment

1 Answer

up vote 2 down vote accepted

You are trying to show that $U=[A,B]$ is finite. In part (b), you have proven that $U'$ is finite; if you can prove that $U/U'$ is finite, then this will suffice.

Since $U$ is normal in $G$ and $U'$ is characteristic in $U$, it follows that $U'$ is normal in $G$. Thus, instead of considering $U$ as a subgroup of $G$, you can reduce the problem (given the conclusion in (b) that $U'$ is finite) to considering $U/U'$ in $G/U'$. This is essentially the same as assuming that $U'=\{1\}$ (since $(U/U')'$ is trivial in $G/U'$).

Now, in (d) you've proven that $[A,U]$ is finite. Note that $[A,U]$ is a normal subgroup of $G$ that is contained in $U$: it is normal because both $A$ and $U$ are normal, so for every $g\in G$ we have $[a,u]^g = [a^g,u^g]\in [A,U]$. It is contained in $U$ because $U$ is normal so $[A,U]\subseteq [G,U]\subseteq U$.

Since $[A,U]$ is finite, if you can prove that $U/[A,U]$ is finite then you'll be done; so the same argument we used above to reduce to the case when $[U,U]=1$ can be used here: instead of considering $U$ in $G$, we consider $U/[A,U]$ in $G/[A,U]$ and proceed from there.

Now repeat the same argument, mutatis mutandis, to conclude that $[B,U]$ is finite and therefore that we may assume that $[B,U]=1$ by going to the quotient $G/[B,U]$. Hence, in this setting, $U$ will be a subgroup such that $[A,U]=1$ and $[B,U]=1$. Since $G=AB$, we easily conclude that $U$ is central: every element of $U$ commutes with each element of $a$ and of $b$, and every element of $G$ is a product of elements of $A$ and $B$.


Since you seem to be having trouble with the way the argument is structures, let me restructure the argument to present the proof as a proof in three cases instead. I will not establish the claims in the parts of the exercise, only show you a different way in which they fit together to reach the desired conclusion. I do this because you say that you've understood and proven all the parts except (c) and (e). If this claim is not correct, that is, if you in fact do not understand or have failed to prove (d) and (f), then say so! It's not a problem, but if you are misrepresenting the state of your knowledge, then you invite further confusion when you receive answers that take you at your word, and you waste the time of those who will have to explain things several times because you were actually confused about other stuff.

What we want to prove is the following:

Suppose $G$ is a group, and $A$ and $B$ are subgroups of $G$ that normalize each other. If the set $X=\{[a,b]\mid a\in A,b\in B\}$ is finite, then $[A,B]$, the subgroup generated by $X$, is also finite.

We will do this by examining several simpler cases before establishing the claim in full generality. Note that $G$ does not play any role except of context, so we may assume that $G=\langle A,B\rangle = AB$ (the last equality because $A$ and $B$ normalize each other), and therefore that $A$ and $B$ are each normal in $G$.

Let $U=[A,B] = \langle X \rangle$.

Case 1. $[A,U]=[B,U]=1$; that is, elements of $U$ commute with elements of $A$ and of $B$.

Note that since $G=AB$, this implies that $U$ is central. Using (f), we show that $X$ is closed under the squaring map; that is, that for every $a\in A$ and $b\in B$ there exist $a'\in A$ and $b'\in B$ such that $[a,b]^2 = [a',b']$; and from there that $U$ is finite. So Case 1 follows from (f).

Case 2. $[U,U]=1$. That is, we assume that $U$ is abelian. Using (d), we show that $[A,U]$ is finite; repeating the argument with $B$ instead of $A$ we again use (d) to show that $[B,U]$ is finite. Since $U$ is abelian, and $[A,U],[B,U]$ are subgroups of $U$, it follows that $N=\langle [A,U],[B,U]\rangle$ is finite. We now consider $U/N$ in $G/N$. Note that $A/N$, $B/N$ and $XN=\{[aN,bN]\mid a\in A,b\in B\}$ satisfy the hypothesis of the desired result, and that in this case we are in Case 1, since $[A/N,U/N]=[B/N,U/N]=1$ in $G/N$. So by Case 1 we can conclude that $U/N$ is finite. Since $N$ is finite and $U/N$ is finite, it follows that $U$ is finite in this case and we are done.

Case 3. General case. Using (a) and (b), we prove that $[U,U]$ is finite. Now we consider $U/U'$ in $G/U'$. Again, $A/U'$, $B/U'$, and $XU' = \{[aU',bU']\mid a\in A,b\in B\}$ satisfy the hypothesis of this case, and $[U/U',U/U']=1$ in $G/U'$. By Case 2, we can conclude that $U/U'$ is finite. Since by (b) we know that $U'$ is finite, it follows that $U$ is finite, which proves the general case.

That's the structure of the argument, except that instead of doing the special cases first, it proceeds by reducing to simpler cases (using (c) to reduce from Case 1 to Case 2; and (e) to reduce from Case 2 to Case 1) before establishing Case 1 at the end.

share|improve this answer
    
Only one more doubt: Will $U/U^{\prime }$ be finite why $U^{\prime }$ it is finite and $\varphi :G\rightarrow G/U^{\prime }$ is surjective? –  User2040 May 15 '12 at 12:37
1  
That's three questions, not one. Of course $\varphi\colon G\to G/U'$ is surjective: it's the canonical projection. As to why $U'$ is finite, you claimed that you had proven all except (c) and (e), and the finiteness of $U'$ is in part (b). And "will $U/U'$ be finite?" That's what you are trying to prove. The point is that since $U'$ has been proven finite, in order for $U$ to be finite it suffices for $U/U'$ to be finite, since $|U| = |U/U'|\times|U'|$. –  Arturo Magidin May 15 '12 at 16:42
    
I do not know how to prove that $U/U^{\prime }$ is finite$.$ I know that $ U/U^{\prime }$ is abelian and finitely generated because $X$ is finite (by hypothesis). Excuse me, my English is not very good. –  User2040 May 15 '12 at 23:35
1  
@Lima: The rest of the problem is about proving $U/U'$ is finite. In (f) you end up concluding that $[A,B]=U$ is finite; in fact, you are showing that $U/\langle[U,A],[U,B]\rangle$ is finite at the end of (f). But you have already shown, in (b), that $[U,U]$ is finite, and in (d) that $[A,U]/[U,U]$ and $[B,U]/[U,U]$ are both finite, hence that $\langle[A,U],[B,U]\rangle/[U,U]$ is finite. Since (f) gives you that $[U,U]/\langle [A,U],[B,U]\rangle$ is finite, (d) gives you that $\langle [A,U],[B,U]\rangle/[U,U]$ is finite, and (b) that $[U,U]$ is finite, it follows that $U$ is finite. –  Arturo Magidin May 15 '12 at 23:42
    
@Lima: Ok! thank you! –  User2040 May 16 '12 at 0:34
show 1 more 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.