I am reading Stewart book on Galois theory, one of the theorem proves that if $G$ is solvable and $N$ is a normal subgroup of $G$, then the quotient of $G$, $G/N$ is solvable. I can't fully understand one of the step in the proof.
Since G is solvable we have a series such that $1=G_0\lhd G_1 \lhd … \lhd G_r=G$, and $G_{i+1}/G_i$ is abelian.
Without any comment or explanation the book states that
$1=N/N=G_0N/N \lhd G_1N/N \lhd … \lhd G_rN/N=G/N$.
I guess I can use the correspondence (or 3rd isomorphism) theorem to work on the quotients and prove that $G_iN/N \lhd G_{i+1}N/N$, but my understanding is that to use it I first have to prove that $G_iN \lhd G_{i+1}N$.
Is it correct?
In that case, it seems to me that from $G_i \lhd G_{i+1}$ it's not so obvious that $G_iN \lhd G_{i+1}N$.
I managed to prove it using the reasoning below, but I am wondering if it's correct and if there's an easier/more elegant way to prove it:
$G_iN \le G $ since $N \lhd G$ and $G_i \le G$ (Dummit & Foote, Corollary 3.15), then $G_iN=NG_i$ (Dummit & Foote, Proposition 3.14)
$G_iN \le G_{i+1}N$ since $N \lhd G_{i+1}N$ and $G_i \le G_{i+1}N$ (Dummit & Foote, Corollary 3.15)
So far I have proved that $G_iN$ is a subgroup of $G_{i+1}N$, now I have to prove that is normal. To achieve it I use the fact that a subgroup K of a group H is normal if $hK=Kh$ for all $h \in H$ (Dummit & Foote, Theorem 3.6), and the results just above:
$hG_i=G_ih$ for all $h \in G_{i+1}$ since $G_i \lhd G_{i+1}$
$gN=Ng$ for all $g \in G$ since $N \lhd G$
$hn \in G_{i+1}N$, $hnG_iN=hnNG_i=hNG_i=hG_iN=G_ihN=G_ihNn=G_iNhn$
Best Answer
Proposition. Let $G$ be a group, $A,B,N\leq G$, and assume that $N\triangleleft G$ and $A\triangleleft B$. Then $AN\triangleleft BN$.
Proof. Because $N$ is normal, $AN$ and $BN$ are both subgroups of $G$. Since $A\leq B$, we also have $AN\leq BN$.
In addition,, for every $g\in G$, we have $gN=Ng$; so for each $x\in N$ there exist $y$ and $z$ in $N$ such that $gx=zg$ and $xg=gy$.
Now let $a\in A$, $b\in B$, $n\in N$, and $m\in N$. We want to show that $(bm)(an)(bm)^{-1}\in AN$. We have for suitable choices of elements of $N$, then $$\begin{align*} (bm)(an)(bm)^{-1} &= b(ma)nm^{-1}b^{-1}\\ &= b(am')nm^{-1}b^{-1} &&\text{for some }m'\in N\\ &= bam''b^{-1} &&\text{with }m''=m'nm^{-1}\in N\\ &= bab^{-1}m''' &&\text{for some }m'''\in N. \end{align*}$$ Now, $bab^{-1}\in A$, since $A\triangleleft B$; so $(bm)(an)(bm)^{-1}\in AN$, as desired. $\Box$
To finish the proof we need to show that $(G_{i+1}N)/(G_iN)$ is abelian for each $i$. We have: $$\frac{G_{i+1}N}{G_iN} = \frac{G_{i+1}(G_iN)}{G_iN} \cong \frac{G_{i+1}}{G_{i+1}\cap(G_iN)}.$$ Now, $G_i\leq G_iN$, so $G_i\leq G_{i+1}\cap G_iN$. So this final quotient is itself a quotient of $G_{i+1}/G_i$, which is abelian. So $(G_{i+1}N)/(G_iN)$ is abelian, as desired.