Nilpotent and solvable groups

\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, $\exists n \in \mathbb{N}, C_n(G) = G$. ◻

\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 $C_n(G) = G$. ◻

\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 $a \in N_G(H) \setminus H$. ◻

\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. ◻

\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 $abN = baN$. ◻

\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. ◻

\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