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$H$, $N$ 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!

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    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.2012-12-06
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    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 $H^{n+1}*N^{n+1}$....2012-12-07

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