# Quick question regarding the proof that the Quotient of a Soluble Groups is Soluble.

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

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

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 $$KN/HN \in \{abelian, nilpotent, solvable, \pi-group\}.$$

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