Solvability of nilpotent groups I'm uncertain about my proof about this exercise regarding nilpotent groups. If someone could me help me out, that would be appreciated. There's a post about this problem, but it uses another definition.
A group $G$ is called nilpotent if $C^n (G)=\left\{1\right\}$ for some $n\geq 1$, where $C^1(G):=G$ and $C^{i+1}(G):=[G,C^{i}(G)]$.
Show that every nilpotent group $G$ is solvable.
Let $n\in\mathbb{N}$ be minimal with $C^n (G)=\left\{1\right\}$.
Proof by induction.
The case $n=1$ is clear.
For $n\geq2$ we have ($Z:=Z(G)$, the center of $G$)
\begin{equation*}
C^n (G)=\left\{1\right\} \iff [G,C^{n-1}(G)]=\left\{1\right\} \iff (\left\{1\right\}\neq) C^{n-1}(G) < Z 
\end{equation*}
(So $G$ has non-trivial center.)
Consider
\begin{equation*}
\pi:G \rightarrow G/Z, g\mapsto gZ.
\end{equation*}
We have $\pi(C^1(G))=\pi(G)=C^1(\pi(G))$. By sub-induction for $m=1,\dots,n-2$ and $x=\pi(g)\pi(c)\pi(g)^{-1}\pi(c)^{-1} \in C^{m+1}(\pi(G))$, $\pi(g)\in\pi(G)$, $\pi(c)\in C^{m}(\pi(G))$,
we have  $\pi(c)\in C^{m}(\pi(G))\subset\pi(C^{m}(G))$.
This means $\pi(c)=\pi(c')$, for some $c' \in C^{m}(G)$, thus $x=\pi(gc'g^{-1}c'^{-1})\in\pi(C^{m+1}(G))$. So $C^{m+1}(\pi(G)) \subset \pi(C^{m+1}(G))$.
This shows $\left\{1\right\}\subset C^{n-1}(\pi(G)) \subset \pi(C^{n-1}(G))\subset \pi(Z) = \left\{1\right\}$. By induction $\pi(G)=G/Z$ is solvable. Since Z is solvable this means G is solvable.
 A: Your proof is great! It is clear and correct.
In my opinion, I would just specify the part where you show  $C^m(π(G)) \subseteq π(C^m(G))$, for instance by saying :

We prove by induction on $m$ that $C^m(π(G)) \subseteq π(C^m(G))$. We have $\pi(C^1(G))=\pi(G)=C^1(\pi(G))$. Let $m ≥ 1$ and assume that 
  $C^{m}(\pi(G)) \subseteq \pi(C^{m}(G))$.
Let $x=\pi(g)\pi(c)\pi(g)^{-1}\pi(c)^{-1} \in C^{m+1}(\pi(G))$, that is : $x=\pi(g)\pi(c)\pi(g)^{-1}\pi(c)^{-1}$ with $\pi(g)\in\pi(G)$, $\pi(c)\in C^{m}(\pi(G))$.
  We have  $\pi(c)\in C^{m}(\pi(G))\subset \pi(C^{m}(G))$. […]

Otherwise everything seems to be fine!
A: Your proof is good. Here's a different possibility. Consider a nilpotent group, so we have
$$
G=C^1(G)\supseteq C^2(G)\supseteq \dots \supseteq C^{n}(G)
\supseteq C^{n+1}(G)=\{1\}
$$
Note that $C^{k+1}(G)$ is normal in $C^k(G)$ (actually in $G$), because for $g,h\in G$ and $c\in C^k(G)$ we have
$$
g[h,c]g^{-1}=[ghg^{-1},gcg^{-1}]
$$
which belongs to $[G,C^k(G)]$ (induction on $k$).
We want to prove that, for $1\le k\le n$, $C^k(G)/C^{k+1}(G)$ is abelian. This will prove $G$ is solvable.
Now, for $x,y\in C^k(G)$, we have, denoting by $\pi\colon C^k(G)\to C^k(G)/C^{k+1}(G)$ the canonical map,
$$
\pi(x)\pi(y)\pi(x^{-1})\pi(y^{-1})
=
\pi(xyx^{-1}y^{-1})=\pi(1)
$$
because $xyx^{-1}y^{-1}=[x,y]\in C^{k+1}(G)=[G,C^k(G)]$.
