I want to solve the following exercise from Dummit & Foote's Abstract Algebra:
Prove that subgroups and quotient groups of solvable groups are solvable
My attempt:
Let $G$ be a solvable group with a chain $1=G_0 \triangleleft G_1 \triangleleft \dots \triangleleft G_s=G$ such that all factors $G_i/G_{i-1}$ are abelian.
Let $H \leq G$ be a subgroup. We claim that the chain $1=H_0 \leq H_1 \leq \dots \leq H_s=H$ where $H_i=G_i \cap H$ does the trick. Firstly we prove that $H_i \triangleleft H_{i+1}$ for all $i$. Let $x \in H_{i+1}=G_{i+1} \cap H$. Since $x \in H$ we have $x H x^{-1}=H$. Since $x \in G_{i+1} \triangleright G_i$ we have $xG_i x^{-1}=G_i$. Combining these gives $x(H \cap G_i) x^{-1}=H \cap G_i$, as required.
Next, we prove that the quotients $H_{i+1}/H_i=(G_{i+1} \cap H)/(G_i \cap H)$ are abelian. Let $x,y \in G_{i+1} \cap H$. We would like to prove that $xy (G_i \cap H)=yx (G_i \cap H)$. Observe that using the fact the cosets of $G_i$ in $G_{i+1}$ can be interchanged, we find
\begin{equation}
\begin{split}
&xy (G_i \cap H)=(xy G_i) \cap (xy H)=(x G_i) (y G_i) \cap H=(y G_i)(x G_i) \cap (yx H)\\
&=(yx G_i) \cap (yx H)=yx (G_i \cap H).
\end{split}
\end{equation}
Let $N \trianglelefteq G$ be a normal subgroup, and let $G/N$ be the corresponding quotient group. We suggest the following normal chain:
\begin{equation}
1=G_0/ N \triangleleft G_1/ N \triangleleft \dots \triangleleft G_s/ N =G/N.
\end{equation}
We first prove that $G_i/N \triangleleft G_{i+1}/N$:
Let $gN \in G_{i+1}/N,aN \in G_i/N$. We have $(gN)(aN)(gN)^{-1}=(gag^{-1})N$. Since $G_i \trianglelefteq G_{i+1}$ the element $gag^{-1} \in G_i$, which gives $(gN)(G_i/N)(gN)^{-1} \in G_i/N$ for all $gN \in G_{i+1}/N$.
The quotients (or factors) are
\begin{equation}
(G_{i+1}/N) \big/ (G_i/N),
\end{equation}
and we wish to prove they are abelian. Indeed, let $g_1N(G_iN),g_2 N(G_iN) \in (G_{i+1}/N) \big/ (G_i N)$ (i.e. $g_1,g_2 \in G_{i+1}$). These two arbitrary elements commute iff
\begin{equation}
[g_2Ng_1N]^{-1}[g_1Ng_2N] \in G_i/N
\end{equation}
which is equivalent to the condition $g_1^{-1}g_2^{-1}g_1g_2 \in G_i$. Since we have that the quotient $G_{i+1}/G_i$ is abelian we know that $(g_1G_i)(g_2G_i)=(g_2G_i)(g_1 G_i)$ which gives precisely this condition. We conclude that quotient groups of solvable groups are solvable.
Is my solution correct? If not, please help me fix it. Thanks!
Best Answer
As Eric Auld mentioned in the comment, the statement that $G_0/N=1/N$ is incorrect. To fix this, you may use the normal chain $$\begin{equation} 1=G_0N/ N \triangleleft G_1N/ N \triangleleft \dots \triangleleft G_sN/ N =G/N. \end{equation}$$