1.6 Nilpotent and Solvable Group

Nilpotent Group

Lemma

Suppose $f:G\to H$ is a group homomorphism and $A⊲H$. Then $f^{-1}(A)⊲G$.

\begin{proof} Note that for any $g\in G$, $f(g^{-1}{f}^{-1}(A)g)=f(g{)}^{-1}Af(g)=A$ and so $f^{-1}(A{)}^g\subseteq f^{-1}(A)$. \end{proof}

Definition

Let $G$ be a finite group. Let $C_1(G)=C(G)$. Let $C_2(G)=\text{mbox}inverseimageofC(G/C_1(G))\text{mbox}oftheprojectionG\to G/C_1(G)$. Inductively, define $C_k(G)$ for $k\in {\mathbb{N}}_{+}$. This gives an increasing series $⟨e⟩⊲C_1(G)⊲C_2(G)⊲\cdots$. We say $G$ is nilpotent if $C_n(G)$ for some $n⩾0$.

Theorem

A finite $p$-group is nilpotent.

\begin{proof} Note that center of $p$-group is non-trivial. \end{proof}

Theorem

Direct product of nilpotent groups is also nilpotent.

\begin{proof} Note that $C_i(H\times K)=C_i(H)\times C_i(K)$. \end{proof}

Lemma

Let $G$ be a nilpotent group. If $H⪇G$, then $H⪇N_G(H)$.

\begin{proof} Write $C_0(G)=⟨e⟩$. Let $n$ be the largest number such that $C_n(G)<H$. Choose some $a\in C_{n+1}(G)\setminus H$. We claim that $a\in N_G(H)$. Note that $\overline{a}\in G/C_n(G)$ is in the center, that is, $(C_n(G)a)(C_n(G)h)=(C_n(G)h)(C_n(G)a)$ for all $h\in G$. Therefore, $[a,h]\in C_n(G)$. For all $h\in H$, we have that $aH{a}^{-1}\in H{C}_n(G)=H$ and so $a\in N_G(H)$. \end{proof}

Proposition

Let $G$ be a finite group. Then $G$ is nilpotent iff $G$ is the inner direct product of its Sylow subgroups.

\begin{proof} One direction is easy.

Now assume that $G$ is nilpotent. First, pick any Sylow subgroup $P$. If $P=G$, we have done. Otherwise $P<G$. If $P/⊲G$, then $N_G(P)<G$. However, by ^cb650b, $N_G(P)<N_G(N_G(P))$, which contradicts with ^b84lpf. Therefore, $P⊲G$.

If $p\ne q$ are distinct prime divisor of $|G|$, then for Sylow $p$-subgroup $P$ and Sylow $q$-subgroup $Q$ we have $P\cap Q=1$ and $PQ=P\times Q$ by this lemma. Therefore, $G$ is the inner direct product of its Sylow subgroups. \end{proof}

Corollary

Let $G$ be a nilpotent group. If $m\mid |G|$, then there exists subgroup of order $m$.

\begin{proof} By ^a9c9ed and Sylow theorems. \end{proof}

Solvable Group

Proposition

If $G$ is nilpotent, then $G$ is solvable.

\begin{proof} Since $G$ is nilpotent, we have $C_1(G)⪇C_2(G)⪇\cdots ⪇C_n(G)=G$. Note that $C(G/C_{n-1}(G))=G/C_{n-1}(G)$, then $G/C_{n-1}(G)$ is abelian and so $C_{n-1}(G)⩾G^{\prime }$. Consider $C_{n-1}$. Since $C_{n-1}/C_{n-2}=C(G/C_{n-2})$, we have that $C_{n-1}/C_{n-2}$ is abelian and so $C_{n-2}⩾C_{n-1}^{\prime }⩾G^{\prime \prime }$. Repeat this procedure and we finish the proof. \end{proof}

Example. $S_3$ is solvable, but it is not nilpotent.

Theorem

The followings hold:

• If $G$ is solvable, then its subgroups and quotient groups are solvable.
• Let $N⊲G$. Then $G$ is solvable iff $G/N$ and $N$ are solvable.

\begin{proof} i) Since $H<G$ yields that $H^{(n)}⩽G^{(n)}=1$ and $H$ is solvable. For any $K⊲G$, we have that $(G/K{)}^{(n)}=G^{(n)}K/K=K/K=1$ and $G/K$ is solvable.

ii) One direction is easy. Define $f:G\to G/N$. Then $(G/N{)}^{(n)}=G^{(n)}N/N=1$ and $G^{(n)}⩽N$. Thus, $G^{(m+k)}⩽N^{(k)}=\{e\}$. \end{proof}

Remark. If $G/N$ and $N$ are nilpotent, then $G$ may NOT nilpotent. For example, $S_3/A_3\simeq C_2$ and $A_3$ are nilpotent, but not for $S_3$.