Central Series - Normal subgroups $H$, $N$ 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!
 A: 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.
