Difference between revisions of "P-group"

Revision as of 17:35, 22 July 2014

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 $G$ .

@@ Line 12: / Line 12: @@
 '''Corollary.'''  If <math>G</math> is a non-[[trivial group |trivial]] <math>p</math>-group, then the [[center (algebra) |center]] of <math>G</math> is non-trivial.
-''Proof.''  Let <math>G</math> act on itself.  Then the set of fixed points is the center <math>Z</math> of <math>G</math>; thus
+''Proof.''  Let <math>G</math> act on itself by conjugation.  Then the set of fixed points is the center <math>Z</math> of <math>G</math>; thus
 <cmath> \lvert Z \rvert \equiv \lvert G \rvert \equiv 0 \not\equiv 1 \pmod{p}, </cmath>
 so <math>Z</math> is not trivial.  <math>\blacksquare</math>

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Difference between revisions of "P-group"

Revision as of 17:35, 22 July 2014

Properties

See also