finite nilpotent groupsFiniteNilpotentGroups2006-03-17 13:42:362006-05-12 15:49:27Topicnilpotent groupSylow subgroupsnilpotentcommutator\usepackage{latexsym}
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\newenvironment{example}{\textbf{Example.} }{$\Box$}The study of finite nilpotent groups mostly centers around the study of $p$-groups. This is because of the following two theorems.
\begin{thm}
\PMlinkname{Finite}{Finite} $p$-groups are nilpotent.
\end{thm}
\begin{proof}
From the class equation we know the center of a finite $p$-group is non-trivial.
Thus by induction the upper central series of a $p$-group $P$ terminates at $P$.
So $P$ is nilpotent.
\end{proof}
\begin{example}
Infinite $p$-groups may not always be nilpotent. In the extreme there are counterexamples like the Tarski monsters $T_p$. These are infinite $p$-groups in which every proper subgroup has order $p$. Therefore given any two non-trivial elements $x,y$ in which $y\notin \langle x\rangle$ generate $T_p$. In particular, the only central element is 1 so that the upper central series is trivial and therefore $T_p$ is not nilpotent.
Indeed, Tarski monsters are not in fact solvable groups which is a weaker property than nilpotent.
\end{example}
\begin{example}
Some infinite $p$-groups are nilpotent. Indeed, some infinite $p$-groups are even abelian such as $\mathbb{Z}_p^\infty$ -- the countable dimension vector space over the field $\mathbb{Z}_p$ -- and the Pr\"ufer group $\mathbb{Z}_{p^\infty}$ -- the inductive limit of $\mathbb{Z}_{p^n}$.
\end{example}
\begin{thm}\label{thm:nil}
Let $G$ be a finite group. Then all the following are equivalent.
\begin{enumerate}
\item\label{nil:1} $G$ is nilpotent.
\item\label{nil:2} Every Sylow subgroup of $G$ is normal.
\item\label{nil:3} For every prime $p\big{|}|G|$, there exists a unique Sylow $p$-subgroup of $G$.
\item\label{nil:4} $G$ is the direct product of its Sylow subgroups.
\end{enumerate}
\end{thm}
For the proof recal the following consequence of the Sylow theorems:
\begin{prop}
If $G$ is a finite group and $P$ a Sylow $p$-subgroup of $G$ then
\[ N_G(N_G(P)) = N_G(P).\]
\end{prop}
(See Subgroups Containing The Normalizers Of Sylow Subgroups Normalize Themselves)
Now we prove Theorem \ref{thm:nil}
\begin{proof}
(\ref{nil:1}) implies (\ref{nil:2}). Suppose that $G$ is nilpotent and that
$P$ is a Sylow $p$-subgroup of $G$. Then as $G$ is nilpotent, every subgroup of $G$ is subnormal in $G$, meaning, if $H$ is properly contained in $G$ then $N_G(H)$ properly contains $H$. Thus $N_G(N_G(P))$ is larger than $N_G(P)$ or $N_G(P)=G$. However because $P$ is a Sylow $p$-subgroup we know $N_G(P)=N_G(N_G(P))$ so we conclude $N_G(P)=G$. Therefore every Sylow $p$-subgroup of $G$ is normal in $G$.
(\ref{nil:2}) implies (\ref{nil:3}). Suppose every Sylow subgroup of $G$ is normal in $G$. Then by the Sylow theorems we know that for every prime $p$ dividing $|G|$ there is exactly one Sylow $p$-subgroup of $G$ -- as all Sylow $p$-subgroups are conjugate and here by assumption all are also normal.
(\ref{nil:3}) implies (\ref{nil:4}). Suppose that there is a unique Sylow $p$-subgroup of $G$ for every $p\big{|}|G|$. Then by the Sylow theorems every Sylow subgroup of $G$ is normal in $G$. Furthermore, if $P$ and $Q$ are two distinct Sylow subgroups then they they are Sylow subgroups for different primes so that by Lagrange's theorem their intersection is trivial. Let $P_1,\dots, P_k$ the Sylow subgroups of $G$. Then as each $P_i$ is normal in $G$ we have
$G=P_1\cdots P_k$ and we have also demonstrated $P_1\cdots P_i\intersect P_{i+1}=1$ for $2\leq i\leq k$ therefore $G$ is the direct product of $P_1,\dots, P_k$.
(\ref{nil:4}) implies (\ref{nil:1}). Suppose that $G$ is a product of its Sylow subgroups. Then as every Sylow subgroup is a $p$-group, $G$ is a product of nilpotent groups so $G$ itself is nilpotent.
\end{proof}