\begin{definition}
Let $G$ be a group, $C_1(G) = C(G)$, and define $C_i(G)$ to be the inverse image of $C(G/C_{i-1}(G))$ under the canonical projection $G \to G/C_{i-1}(G)$. The ascending central series of $G$ is
$$ \langle e \rangle \leq C_1(G) \leq C_2(G) \leq \cdots $$
$G$ is said to be nilpotent iff $\exists n \in \mathbb{N}, C_n(G) = G$.
\end{definition}\begin{theorem}
Every finite $p$-group is nilpotent.
\end{theorem}

Proof.$G$ and all its
nontrivial quotients are
$p$-groups and thus have
nontrivial centers. Thus if
$G \neq C_i(G)$,
$C_i(G) < C_{i+1}(G)$.
Since
$G$ is finite,
$\exists n \in \mathbb{N}, C_n(G) = G$.Â â—»

\begin{theorem}
The direct product of a finite number of nilpotent groups is nilpotent.
\end{theorem}

Proof. Suppose
$G = H \times K$. Assume
that
$C_i(G) = C_i(H) \times C_i(K)$.
Let
$\pi_H: H \to H/C_i(H)$
be the canonical epimorphism. Similarly for
$\pi_K$. Verify that the
canonical epimorphism
$\phi: G \to G/C_i(G)$ is
the composition
$$G = H \times K \xrightarrow{\pi} H/C_i(H) \times K/C_i(K) \xrightarrow{\psi} (H \times K)/(C_i(H) \times C_i(K)) = G/C_i(G)$$
where
$\pi = \pi_H \times \pi_K$
and
$\psi$ is an isomorphism.
Consequently,
\begin{align*}C_{i+1}(G) &= \phi^{-1}[C(G/C_i(G))] = \pi^{-1} \circ \psi^{-1} [C(G/C_i(G))] \\
&= \pi^{-1} [C(H/C_i(H)) \times C(K/C_i(K))] \\
&= \pi_H^{-1}[C(H/C_i(H))] \times \pi_K^{-1} [C(K/C_i(K))] \\
&= C_{i+1}(H) \times C_{i+1}(K)\end{align*} The inductive
step is proved so
$C_i(G) = C_i(H) \times C_i(K)$
for all
$i$. Since
$H, K$ are nilpotent,
$\exists n \in \mathbb{N}$,
$C_n(H) = H, C_n(K) = K$.
Thus
$C_n(G) = G$.Â â—»

\begin{lemma}
If $H$ is a proper subgroup of a nilpotent group $G$, then $H$ is a proper subgroup of its normalizer $N_G(H)$.
\end{lemma}

Proof. Let
$C_0(G) = \langle e \rangle$
and let
$n$ be the largest index
such that
$C_n(G) \leq H$. Choose
$a \in C_{n+1}(G) \setminus H$.
$\forall h \in H, ah C_n(G) = haC_n(G)$.
Thus
$\exists h' \in C_n(G), ah=hah'$.
This implies
$a \in N_G(H)$. Thus
$a \in N_G(H) \setminus H$.Â â—»

\begin{proposition}
A finite group is nilpotent iff it is the direct product of its Sylow subgroups.
\end{proposition}

Proof. If
$G$ is the direct product
of its Sylow
$p$-subgroups, then it is
nilpotent by our previous results. If
$G$ is nilpotent and
$P$ is a Sylow
$p$-subgroup of
$G$ for some prime
$p$, then either
$G=P$ or
$P$ is a proper subgroup.
In the latter case,
$P$ is a proper subgroup
of $N_G(P)$. Since
$N_G(N_G(P)) = N_G(P)$,
$N_G(P) = G$. Thus
$P \triangleleft G$ and
hence the unique Sylow
$p$-subgroup of
$G$. Let
$|G| = p_1^{n_1} p_2^{n_2} \cdots p_k^{n_k}$
where
$p_i$ are distinct primes
and let
$P_1, P_2, \cdots, P_k$
be the corresponding proper normal Sylow subgroups of
$G$. For
$i \neq j$,
$P_i \cap P_j = \langle e \rangle$.
Thus for any
$x \in P_i, y \in P_j$,
$xy = yx$. It is easy to
see that
$G = P_1 P_2 \cdots P_k$
is an inner direct product.Â â—»

\begin{corollary}
If $G$ is a finite nilpotent group and $m \mid |G|$, $G$ has a subgroup of order $m$.
\end{corollary}\begin{definition}
Let $G$ be a group. The subgroup generated by $\{aba^{-1}b^{-1} \mid a, b \in G\}$ is called the commutator subgroup of $G$ and is denoted $G'$. The elements $[a,b] = aba^{-1}b^{-1}$ are called commutators.
\end{definition}

$G$ is abelian iff
$G' = \langle e \rangle$.

\begin{theorem}
If $G$ is a group, then $G' \trianglelefteq G$ and $G/G'$ is abelian. If $N \trianglelefteq G$, then $G/N$ is abelian iff $G' \leq N$.
\end{theorem}

Proof. Note that for every endomorphism
$f: G \to G$,
$f(G') \leq G'$. Hence,
$G' \trianglelefteq G$.
Since
$(ab)(ba)^{-1} \in G', abG=baG$
and hence
$G/G'$ is abelian. If
$G/N$ is abelian,
$abN=baN$ for all
$a, b \in G$, whence
$ab(ba)^{-1} \in N$.
Therefore
$G' \leq N$. If
$G' \leq N, ab(ba)^{-1} \in N$
so $abN = baN$.Â â—»

\begin{definition}
Let $G$ be a group and $G^{(1)} = G'$. Define $G^{(i)} = (G^{(i-1)})'$. $G^{(i)}$ is called the $i$th derived subgroup of $G$. The derived series of $G$ is $G \geq G^{(1)} \geq G^{(2)} \geq \cdots$. A group $G$ is said to be solvable iff $G^{(n)} = \langle e \rangle$ for some $n$.
\end{definition}\begin{proposition}
Every nilpotent group is solvable.
\end{proposition}

Proof. By definition,
$C_i(G)/C_{i-1}(G) = C(G/C_{i-1}(G))$
is abelian.
$C_i(G)' \leq C_{i-1}(G)$
and
$C(G)' = \langle e \rangle$.
For some
$n$,
$G = C_n(G)$. Therefore
$C(G/C_{n-1}(G)) = G/C_{n-1}(G)$
is abelian and hence
$G' \leq C_{n-1}(G)$.
$G^{(2)} \leq C_{n-1}(G)' \leq C_{n-2}(G)$.
Continuing,
$G^{(n)} \leq C(G)' = \langle e \rangle$.
Hence
$G$ is solvable.Â â—»

\begin{theorem}
\begin{enumerate}[label=(\roman*)]
\item Every subgroup and every homomorphic image of a solvable group is solvable.
\item If $N$ is a normal subgroup of a group $G$ such that $N$ and $G/N$ are solvable, then $G$ is solvable.
\end{enumerate}
\end{theorem}

Proof. (i) If
$f: G \to H$ is a
homomorphism [resp. epimorphism] then
$f(G^{(i)}) \leq H^{(i)}$
[resp.
$f(G^{(i)}) = H^{(i)}$]
for all
$i$. Suppose
$f$ is an epimorphism and
$G$ is solvable. For some
$n$,
$\langle e \rangle = f(G^{(n)}) = H^{(n)}$,
whence
$H$ is solvable.
Similarly for the subgroup.

(ii) Let
$f: G \to G/N$ be the
canonical epimorphism.
$\exists n \in \mathbb{N}, f(G^{(n)}) = (G/N)^{(n)} = \langle e \rangle$.
Hence
$G^{(n)} \leq \operatorname{ker} f = N$.
By (i),
$G^{(n)}$ is solvable.
Hence