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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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    Can you do the case $N=G$ first? 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)2011-09-28
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    As for the hint from Robinson: it is what one uses to reduce to the case $N=G$ (for $H\cap N$ is playing $H$ in $N$).2011-09-28
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    @Steve D: Thank you very much for your help. I am sorry but I still can't get through. In case $N=G$, suppose $G=H\gamma_iG$, but $G\neq H\gamma_{i+1}G$. The quotient group $\overline{G}=G\gamma_{i+1}G$ is nilpotent, and $\overline{H}=H\(H\cap\gamma_{i+1}G)$ is a proper subgroup. Does "the nilpotent case" mean $G$ being nilpotent? Why can we reduce to this case according to these facts? Also, it is not quite clear to me how it is reduced to the case $N=G$.. I am sorry but I am not good at group theory. I'll be more than gratful if it will not cause you too much trouble to give some more hints.2011-09-29
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    Sorry for the mistake, $\overline{G} = G/ \gamma_{i+1}G$, $\overline{H} = H / (H \cap \gamma_{i+1}G)$. I cannot edit it now.2011-09-29
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    Sorry, when I said "reduce to the nilpotent case", I meant showing that if $G$ is nilpotent and $G=HG'$, then $H=G$. I was sketching a proof by contradiction. That is, suppose $G=H\gamma_i(G)$ and yet $G\neq H\gamma_{i+1}(G)$. Now quotient out by $\gamma_{i+1}(G)$, and we have $\overline{G}=\overline{H}\overline{G}'$.2011-09-29
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    Plus $\overline{G}$ is nilpotent, so now show this all implies $\overline{G}=\overline{H}$, which is the contradiction.2011-09-29
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    @Steve D: Thank you really a lot. Now this is clearer to me. But would you please allow me to ask just one more question? Suppose that $1 = \zeta_0(G) \leq \zeta_1(G) \leq \cdots \leq \zeta_c(G) =G$ be the upper central series of $G$ ($G$ is nilpotent), why does the fact $H\zeta_{j-1}(G)2011-09-30
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    Think about what happens when $G=HZ(G)$. Then every element can be written as $hz$; can you then simply a generic commutator $[hz,\hat{h}\hat{z}]$?2011-09-30
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    @Steve D: simply a generic commutator $[hz,\hat{h}\hat{z}]$? Would you please tell me what does this mean? I am sorry I am not an English speaker... Does $\hat{h}$ mean the image of $h$ in $G/ Z(G)$ under the canonical morphism? Sorry for the trouble I have brought to you.2011-10-08
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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)

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