proof that if G is a solvable group then quotient of G is solvable

Question

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 $N \lhd G$

$hn \in G_{i+1}N$, $hnG_iN=hnNG_i=hNG_i=hG_iN=G_ihN=G_ihNn=G_iNhn$

Answer 1

Here is a more general observation, and probably why Herstein doesn't feel the need to do the step-by-step:

Lemma. Let $G$ be a group, and let $\phi\colon G\to K$ be a group homomorphism. If $N\triangleleft G$, then $\phi(N)\triangleleft \phi(G)$.

Proof. Let $x\in\phi(N)$ and $y\in \phi(G)$. Then there exists $n\in N$ and $g\in G$ such that $x=\phi(n)$ and $y=\phi(g)$. Then $$yxy^{-1} = \phi(g)\phi(n)\phi(g)^{-1} = \phi(gng^{-1})\in \phi(N).$$ Therefore, $\phi(N)\triangleleft \phi(G)$, as claimed. $\Box$

From which we get:

Theorem. Let $G$ be a group, $N\triangleleft G$, and $A\leq B$ subgroups of $G$. If $A\triangleleft B$, then $\frac{AN}{N}\triangleleft \frac{BN}{N}$.

Proof. Let $\pi\colon G\to G/N$ be the canonical projection. Then from Lemma 1 we have that $\pi(A) = \frac{AN}{N}\triangleleft \frac{BN}{N}$, as claimed. $\Box$

And in fact, since we also have the following:

Lemma 2. Let $N\triangleleft G$. If $A$ and $B$ are subgroups of $G$ such that $N\leq A\leq B$, then $A\triangleleft B$ if and only if $\frac{A}{N}\triangleleft \frac{B}{N}$. In other words, the correspondence in the Lattice Isomorphism Theorem is normality-preserving.

Proof. Let $\pi\colon G\to G/N$ be the canonical projection. Restricting to $B$, we have that that if $A\triangleleft B$, then $\frac{A}{N} = \phi(A)\triangleleft \phi(B)=\frac{B}{N}$, by Lemma 1. This proves the "only if" clause.

Conversely, assume that $\frac{A}{N}\triangleleft \frac{B}{N}$, and let $a\in A$, $b\in B$. Then $(bab^{-1})N = (bN)(aN)(bN)^{-1} \in AN = A$, since $\frac{A}{N}\triangleleft \frac{B}{N}$. In particular, $bab^{-1}\in A$, as desired. $\Box$

We conclude that

Theorem. Let $N\triangleleft G$, and let $A\leq B$ be subgroups of $G$. If $A\triangleleft B$, then $AN\triangleleft BN$ and $\frac{AN}{N}\triangleleft \frac{BN}{N}$.

Answer 2

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. $\Box$

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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.