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Let $G$ be soluble, so that $$1=G_0\lhd G_1\lhd \ldots \lhd G_n=G$$ where $G_{i+1}/G_i$ is abelian. Now let $N$ be a normal subgroup of $G$. The standard proof requires one to show that $$N/N=G_0N/N\lhd G_1N/N\lhd \ldots \lhd G_rN/N=G/N$$

In particular, that $G_iN\lhd G_{i+1}N$ and it is here that I'm stuck. I would appreciate some help.


Edit: I believe the following would be sufficient. Letting $g\in G_{i+1}$ and $n\in N$ we have

$$gnG_iN=gnG_iNn=gnNG_in=gNG_in=NgG_in=NG_ign=G_iNgn$$.

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Hint: $G_{i+1}N/G_iN \cong G_{i+1}/(G_iN \cap G_{i+1})$ and the latter is a quotient of the abelian $G_{i+1}/G_i$.

Let me make this somewhat more precise. I assume that you know that if $H \leq G$, $N \unlhd G$, then $HN$ is a subgroup of $G$ and in fact $HN=NH$. Further that you are familiar with the isomorphism theorems.

Lemma Let $H,K \leq G$ with $H \unlhd K$ and let $N \unlhd G$ then the following hold.
$(a)$ $HN \unlhd KN$
$(b)$ $KN/HN$ is a quotient of $K/H$
$(c)$ If $G$ is finite, $|KN:HN|$ divides $|K:H|$ with equality if and only if $K \cap N \subseteq H$.

Proof (a) let $h \in H, k \in K, m, n \in N$, we need to show that $(hm)^{kn} \in HN$. Now observe that $$(hm)^{kn}=n^{-1}k^{-1}hmkn=n^{-1}(k^{-1}hk)(k^{-1}mk)n=(k^{-1}hk)(k^{-1}h^{-1}k)n^{-1}(k^{-1}hk)(k^{-1}mk)n=h^k(n^{-1})^{k^{-1}hk}m^kn \in HN.$$
(b) $KN/HN \cong K/(K \cap HN)$. Note that $H \subseteq K \cap HN$. So $K/(K \cap HN) \cong (K/H)/((K \cap HN)/H)$ which is a quotient of $K/H$.
(c) From (b) it follows that $|KN:HN| \mid |K:H|$ and that we have equality if and only if $K \cap HN=H$. But by Dedekind's Modular Law, we have $K \cap HN=H(K \cap N)$. Hence (c) follows. Note that $K \cap N \subseteq H$ is equivalent to $K \cap N=H \cap N$.

Corollary Let $H,K \leq G$ with $H \unlhd K$ and let $N \unlhd G$ then if $K/H \in \{abelian, nilpotent, solvable, \pi-group\}$, then $KN/HN \in \{abelian, nilpotent, solvable, \pi-group\}.$

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    $\begingroup$ Yes, the bit that I don't understand is $G_iN\lhd G_{i+1}N$. $\endgroup$ – Leo Apr 18 '20 at 20:06
  • $\begingroup$ Ok, will write it out later, no time now. In general if $H \unlhd K$ and $N \unlhd G$, then $HN \unlhd KN$. $\endgroup$ – Nicky Hekster Apr 18 '20 at 20:46

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