Proof of $G$ is solvable implies $G/N$ is solvable.

I want to show that if $$N$$ is normal in $$G$$ then $$G$$ is solvable implies $$G/N$$ is solvable.

Now, $$G$$ is solvable implies there exists a chain
$$\{e\}=G_0 \trianglelefteq G_1 \trianglelefteq G_2 \trianglelefteq G_3 \cdots \trianglelefteq G_s=G$$, such that $$G_i\trianglelefteq G_{i+1}$$ and $$G_{i+1}/G_i$$ is abelian.

We can consider the chain $$\overline{N} =G_0N/N \trianglelefteq G_1N/N \trianglelefteq G_2/N \trianglelefteq G_3N/N \cdots \trianglelefteq G_sN/N=G/N$$ I want to show that

1. $$G_iN/N \trianglelefteq G_{i+1}N/N$$ which is equivalent to showing $$G_iN\trianglelefteq G_{i+1}N$$

2. and $$\frac{G_{i+1}N/N} { G_{i}N/N }$$ which is isomorphic to $$\frac{G_{i+1}N}{G_iN}$$ is abelian.

With a lot of brut force somehow I can prove the first part. But I am unable to prove the second part. Can someone suggest me an elegant proof of (1) and any proof of (2)?

I am including a proof of (1) which I have done

1. follows immediately from the third isomorphism theorem. Consider the map $$\phi: G_{i+1}N/N \to G_{i+1}N/G_iN$$. It is easy to check that this is a homomorphism with kernel $$G_iN/N$$.

Having obtained $$\frac{G_{i+1}N/N}{G_iN/N} \simeq G_{i+1}N/G_iN$$ we need to show that the right-hand side is an abelian group. Elaborating on the same argument used here Quotients of Solvable Groups are Solvable:

consider the commutators $$[xn, ym]$$ for $$x, y \in G_{i+1}$$, $$n,m \in N$$. We want $$[xn,ym] \in G_iN$$, given that $$[x,y] \in G_i$$. Now, $$[xn,ym]G_iN = ([xn,ym]N)G_i$$, so it suffices to show that $$[xn,ym]N = [x,y]N$$. This is again brute force

$$xnymn^{-1}x^{-1}m^{-1}y^{-1}N = xyx^{-1}y^{-1}N$$

$$\iff nymn^{-1}x^{-1}mN = yx^{-1}N$$

$$\iff (y^{-1}ny)(mn^{-1})(x^{-1}mx)N = N$$, and this is true because $$N$$ is normal.

• Why is $G_i$ normal in $G_{i+1}N$? – Babai Mar 10 '19 at 21:15
• Sorry! I made a mistake and I removed it. I will have to think of another proof - the only ones I know of use brute force. – vxnture Mar 10 '19 at 21:34
• I can already show the isomorphism you have just shown above. What I am struggling to show is enough of those isomorphic quotients are abelian. – Babai Mar 10 '19 at 21:54
• math.stackexchange.com/questions/1339609/… there is a proof here (with commutators) – vxnture Mar 10 '19 at 21:59
• There is a gap in that proof too, I am not able to verify it – Babai Mar 11 '19 at 5:24