P-group

Revision as of 17:35, 22 July 2014 by Math154 (talk | contribs) (Properties: clarified how $G$ acts on itself)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
The title of this article has been capitalized due to technical restrictions. The correct title should be $p$-group.

A $\boldsymbol{p}$-group is a finite group whose order is a power of a prime $p$.

Properties

Lemma. Let $G$ be a $p$-group acting on a finite set $S$; let $S^G$ denote the set of fixed points of $S$. Then \[\lvert S^G \rvert \equiv \lvert S \rvert \pmod{p} .\]

Proof. It is enough to show that $p$ divides the cardinality of each orbit of $S$ with more than one element. This follows directly from the orbit-stabilizer theorem. $\blacksquare$

Corollary. If $G$ is a non-trivial $p$-group, then the center of $G$ is non-trivial.

Proof. Let $G$ act on itself by conjugation. Then the set of fixed points is the center $Z$ of $G$; thus \[\lvert Z \rvert \equiv \lvert G \rvert \equiv 0 \not\equiv 1 \pmod{p},\] so $Z$ is not trivial. $\blacksquare$

Theorem. Let $G$ be a $p$-group of order $p^r$. Then there exists a series of subgroups \[G = G^1 \supseteq G^2 \supseteq \dotsb \supseteq G^{r+1} = \{e\}\] such that $G^k$ normalizes $G^{k+1}$, $(G,G^k) \subseteq G^{k+1}$, and $G^k/G^{k+1}$ is a cyclic group of order $p$, for all indices $k$.

Proof. We induct on the order of $G$. For $G= \{e\}$, the theorem is trivial. Let $Z$ be the center of $G$, and $x$ a non-identity element of $Z$. Let $p^s$ be the order of $x$. Then $x^{p^{s-1}}$ generates a cyclic group $A$ of order $p$; since $A$ is contained in $Z$, it is evidently a normal subgroup of $G$. Then $G/A$ is a $p$-group of order $p^{r-1}$. By inductive hypothesis, there is a sequence \[G/A = (G/A)^1 \supseteq (G/A)^2 \supseteq \dotsb \supseteq (G/A)^r = A\] satisfying the theorem's requirements.

Let $\phi$ be the canonical homomorphism from $G$ onto $G/A$, and for $1\le k \le r$, let $G^k$ be $\phi^{-1}((G/A)^k)$, and let $G^{r+1}=\{e\}$. Then for $k \in [1,r-1]$, $G^{k+1}$ is a normal subgroup of $G^k$, and $G^k/G^{k+1}$ is isomorphic to $(G/A)^k/(G/A)^{k+1}$; hence it is a cyclic group of order $p$. Also, \begin{align*} (G,G^k) & \subseteq \phi^{-1}(\phi(G),\phi(G^k)) = \phi^{-1}((G/A),(G/A)^k) \\ &\subseteq \phi^{-1}((G/A)^{k+1}) = G^{k+1} . \end{align*} Since $A$ is a cyclic group of order $p$ that lies in the center of $G$, the theorem's statements are true for $k=r$, as well. This completes the proof. $\blacksquare$

Corollary 1. Every $p$-group is nilpotent.

Corollary 2. If $G$ is a $p$-group, and $H$ is a proper subgroup of $G$, then the normalizer of $H$ is distinct from $H$.

This is a property of nilpotent groups in general.

Proposition. Let $H$ be a proper subgroup of a $p$-group $G$. Then there exists a normal subgroup $N$ of $G$ of index $p$ that contains $H$.

Proof. Since $G$ is nilpotent there exists a normal subgroup $A$ of $G$ such that $H \subseteq A$ and $G/A$ is abelian. Let $N$ be a maximal subgroup of $G$ containing $A$. Since $G$ is nilpotent, $N$ is normal in $G$. Since $N$ is evidently simple, it is cyclic, and hence of order $p$. $\blacksquare$

Corollary. Let $G$ be a $p$-group, and $H$ a subgroup of $G$ of index $p$. Then $H$ is a normal subgroup of $G$.

See also

Invalid username
Login to AoPS