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.

$H$, $N$ are subnormal subgroups in the finite group $G$ and $G = H*N$. Show: $(H*N)^{\infty} = H^{\infty}*N^{\infty}$. (And $G^{\infty} := \bigcap\limits_{i\geq 0}G^{i}$, and $G^{i+1} = [G, G^{i}]$ from the lower central series.)

We got the hint to start with $H,N$ normal in $G$; prove this with induction and later tranfer this to the case where $H$ and $N$ are subnormal.

So I started: The case i=1 holds apperently.

For the induction step I'd need some help. I have so far:

Since $G$ is finite, we know that the central series stagnates at some point, let's say:
$\bigcap\limits_{i\geq 0}G^{i} = G^{m}$ for some $m$.

The same holds then for $H^{\infty}$ and $N^{\infty}$. I named them:
$H^{\infty} = H^{k}$ for some $k$ and $N^{\infty} = N^{l}$ for some $l$.

So I get $G^{\infty} = G^{m} = (H*N)^{m} \overset{?}{=} H^{\infty}*N^{\infty} = H^k*N^l$.
So I have to show that $m=k=l$ right?

Is this so far OK and a good way to approach the problem? I got at this point already stuck.. So I'd be very happy if someone had a little hint how to go on :)

All the best, Sara!

share|improve this question
    
It might help to take a look at the various formulas for commutators that involve a product in one of the entries, since the elements will look like that. –  Tobias Kildetoft Dec 6 '12 at 15:31
    
Thanks! :) I tried now the induction step: Induct. assumption: $G^{n} = (H*N)^{n} = H^{n}N^{n}$ Now for $n\mapsto (n+1):$ $G^{n+1} = (G^{n})' \overset{assump.}{=} (H^{n}*N^{n})'$ right? And $(H^{n}*N^{n})' = [H^{n}*N^{n}, H^{n}*N^{n}]$ which is generated by $[hn, h'n']$, $h,h' \in H^{n}, n,n' \in N^{n}$ (is that right?) Now I have don't know how to show that this is the same as $H^{n+1}*N^{n+1}$.... –  Sara Dec 7 '12 at 13:53

1 Answer 1

Here are some identities that will prove useful for this:

Let $G$ be any group and let $x,y,z,w\in G$. Write $x^y = y^{-1}xy$. Then we have the following identities (that follow directly from the definitions):

$$[xy,z] = [x,z]^y[y,z]$$ $$[x,zw] = [x,w][x,z]^w$$

And you can then use these to see what $[xy,zw]$ is in terms of products of commutators and their conjugates.

This should at least allow you to finish the part where the subgroups are normal, and in fact you just need that they normalize each other to use this directly.

Here are some more details (remember that we are still dealing with the case where $H$ and $N$ are normal): Since the subgroups are normal, rather than dealing with $H^nN^n$ we can deal with the subgroup generated by all elements from $H^n$ and $N^n$ (since this is the same group). So in order to show that $[nh,n'h']$ is in $H^nN^n$ we just need to show that it can be written as a product of elements in $H^n$ and $N^n$. Now, we get terms of various forms when we expand that commutator. We get terms like $[h,h']$ which are certainly in $H^n$ (since $h$ and $h'$ are). The same is true for those terms we get of the form $[n,n']$. We also get terms of the form, $[h,n']$ (or with the roles reversed), but since $H$ and $N$ normalize each other, these terms are in fact in both $H$ and $N$, so they too are ok. Finally, we get terms of the form $x^y$ with $x$ being some commutator and $y$ being some element from $H$ or $N$. But again, as the two subgroups normalize each other, such an element belongs to the same subgroup that $x$ does. As the original commutator is a product of terms of these forms, this shows that the given commutator belong to the correct subgroup.

share|improve this answer
    
Hey! Thanks for the answer so far! I tried some thinks, but to be honest I am still stuck. The thing I want to show is, that $[hn,h'n'] = [h,h']*[n,n']$ right? with $h,h'\in H^{n}$ and $n,n' \in N^{n}$, because $[hn,h'n']$ is the generator of $[H^{n}*N^{n},H^{n}*N^{n}]$ and $[h,h']*[n,n']$ of $[H^{n},H^{n}]*[N^{n},N^{n}]$? Is that right so far? But I just don't get the commutators to do so... I calculated $[hn,h'n'] = [h,n']^{n}*[h,h']^{nn'}*[n,n']*[n,h']^{n'}$ Could you maybe give me some more hint? Thanks and best, Sara :) –  Sara Dec 9 '12 at 12:06
    
you just need to show that the commutator of the products is some product of elements in the commutator subgroups. And if the two subgroups normalize each other, then they also each normalize any commutator subgroup of the other, so conjugation is also "allowed". –  Tobias Kildetoft Dec 9 '12 at 15:50
    
ok... sorry, but I still don't know what I can do then with my expression $[h,n']^{n}*[h,h']^{nn'}*[n,n']*[n,h']^{n'}$ I'm sorry.. I feel kind of blocked. Or don't I need this thing? If the conjugation 'does no matter' you mean I can write $[h,n']*[h,h']*[n,n']*[n,h']$? Sorry, I hope you can give me some more help! :) –  Sara Dec 9 '12 at 16:58
    
If $x$ is in some commutator subgroup of $H$, $y$ is in some commutator subgroup of $N$ and $N$ normalizes $H$, then $x^y$ is in the same commutator subgroup of $H$ that $x$ is. This is what I mean by conjugation not mattering. –  Tobias Kildetoft Dec 10 '12 at 2:01
    
hm OK. But can you maybe help me with the induction step? I am a little lost how to manage the commutators.. But anyway, thanks for the help so far!! :) –  Sara Dec 10 '12 at 3:32

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.