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If $H \leq G$, $N \lhd G$, G=HN', then $G = H( \gamma _i N)$ for all $i$.

Here, $\gamma_iN$ are the terms in the lower central series of $N$, i.e., $\gamma_1 = N$ and $\gamma_{i+1}N = [\gamma_i N, N] $.

There is a hint:

N =(H \cap N) N'.

I have no idea what to do with the hint, and how to get a proof. I am wondering if I can get some help.

Thank you very much.

[This is on page 128 of A Course in the Theory of Groups (GTM 80) written by Derek J.S. Robinson.]

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    Right, so we are now in the case where $G$ is nilpotent and $G=HZ(G)$. What you want to show is $[hz,\hat{h}\hat{z}]$ is in $H$. I meant "simplify" - it was my mistake :)2011-10-08

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I am sorry if it is not good here to answer my own question. I just want to collect all the ingredients @Steve D has given to me in order to make it more convinient to read.

First, reduce to the case $N = G$. As N = (H \cap N) N', $H\cap N$ is playing $H$ in $N$.

Second, reduce to the case $G$ is nilpotent. Suppose $G=H\gamma_{i}(G)$ but $G\neq H\gamma_{i+1}(G)$; then quotienting out by $\gamma_{i+1}(G)$ makes $G$ nilpotent, and the image of $H$ in the quotient is proper. Thus we can reduce to the nilpotent case.

Now choose a $j$ so that $H\zeta_{j-1}(G). Then it is reduced to the case $G = HZ(G)$. Every element can be written as $hz$; Then simplify a generic commutator $[hz,\hat{h}\hat{z}]$. Show that $[hz,\hat{h}\hat{z}]$ is in $H$. Show this implies HG'\le H\zeta_{j-1}(G), a contradiction to the hypothesis.

Sincerlest thanks to Steve. I hope I am not mistaken.