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Let $G_i \triangleleft G_{i+1}$ both subgroups of $G$. Let $N$ be a normal group.

Does $G_iN \triangleleft G_{i+1}N$?

Does $(G_iN/N) \triangleleft (G_{i+1}N/N)$?

I know that $q:G\longrightarrow G/N$ preserves normality. Hence if $G_iN \triangleleft G_{i+1}N$ then their quotients would be normal.

I have tried by considering an element in $g\in G_iN$ and $h\in G_{i+1}N$ and trying to find out if $ghg^{-1}\in G_iN$. But I am a bit confused of those multilpication groups. And that is why I can not play with this expression.

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  • $\begingroup$ What do you know? What have you tried? $\endgroup$ – verret Jan 22 at 20:55
  • $\begingroup$ Edited. @verret $\endgroup$ – idriskameni Jan 22 at 20:58
  • $\begingroup$ Have you heard of the second/third isomorphism theorem? $\endgroup$ – DonAntonio Jan 22 at 21:00
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    $\begingroup$ Yes I do. I have been playing with that expression too. $G_i/(G_i\cap H)\triangleleft G_{i+1}/(G_{i+1}\cap H)$. But still nothing. $\endgroup$ – idriskameni Jan 22 at 21:02
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    $\begingroup$ I have tried a lot of things. Is not that I haven't worked on it. Don't you think so. I mean, it is easy to come and coment these things. $\endgroup$ – idriskameni Jan 22 at 21:02
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I want to change the notation to make stuff clearer: Suppose $H\lhd K< G$, and let $N\lhd G$ (so really $H=G_i$ and $K=G_{i+1}$). Then a proof that $HN\lhd KN$ by considering a conjugator is:

Suppose $k\in K$, $h\in H$, and $n, n'\in N$. Firstly, note that $$n^{-1}k^{-1}(hn')kn=(n^{-1}k^{-1}hkn)\cdot (n^{-1}k^{-1}n'kn)$$ Clearly $(n^{-1}k^{-1}n'kn)\in N$, while $(n^{-1}k^{-1}hkn)\in HN$ by the following: $$\begin{align*} k^{-1}hk&\in H\\ \Rightarrow k^{-1}hk\cdot (k^{-1}hk)^{-1}n^{-1}(k^{-1}hk)&\in HN\\ n^{-1}(k^{-1}hk)&\in HN\\ n^{-1}(k^{-1}hk)n&\in HN \end{align*} $$ The result follows.

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Hint:

The first one should be almost immediate, and for the second one:

$$G_iN/N\cong G_i/(G_i\cap N)\;,\;\;\; G_{i+1}/(G_{i+1}\cap N)\cong G_{i+1}N/N$$

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    $\begingroup$ I am afraid that I cannot see how this helps solve the problem. The second one follows directly from the first which, as you say. is straightforward. $\endgroup$ – Derek Holt Jan 23 at 8:47
  • $\begingroup$ @DerekHolt That isomorphism helps, imo. to see that (1) those quotient groups are well-defined and (2) one is normal in the other. I also think both are more or less straightforward (though this usually depends on the beholder...). Did I miss anything? $\endgroup$ – DonAntonio Jan 23 at 8:55
  • $\begingroup$ I don't really see how the isomorphisms help prove that one is normal in the other. (The downvote is not from me.) $\endgroup$ – Derek Holt Jan 23 at 9:11

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