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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.



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

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

  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.