https://artofproblemsolving.com/wiki/index.php?title=P-group&feed=atom&action=historyP-group - Revision history2024-03-29T05:15:22ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=P-group&diff=200713&oldid=prevRyanjwang at 02:50, 4 November 20232023-11-04T02:50:59Z<p></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>A '''<math>\boldsymbol{p}</math>-group''' is a [[finite]] [[group]] whose [[order (group theory) |order]] is a power of a [[prime]] <math>p</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>A '''<math>\boldsymbol{p}</math>-group''' is a [[finite]] [[group]] whose [[order (group theory) |order]] is a power of a [[prime]] <math>p</math>.</div></td></tr>
</table>Ryanjwanghttps://artofproblemsolving.com/wiki/index.php?title=P-group&diff=200712&oldid=prevRyanjwang at 02:50, 4 November 20232023-11-04T02:50:34Z<p></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 02:50, 4 November 2023</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>A '''<math>\boldsymbol{p}</math>-group''' is a [[finite]] [[group]] whose [[order (group theory) |order]] is a power of a [[prime]] <math>p</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>A '''<math>\boldsymbol{p}</math>-group''' is a [[finite]] [[group]] whose [[order (group theory) |order]] is a power of a [[prime]] <math>p</math>.</div></td></tr>
</table>Ryanjwanghttps://artofproblemsolving.com/wiki/index.php?title=P-group&diff=62624&oldid=prevMath154: /* Properties */ clarified how $G$ acts on itself2014-07-22T22:35:32Z<p><span dir="auto"><span class="autocomment">Properties: </span> clarified how $G$ acts on itself</span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 22:35, 22 July 2014</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>'''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.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>'''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.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>''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</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>''Proof.''  Let <math>G</math> act on itself <ins class="diffchange diffchange-inline">by conjugation</ins>.  Then the set of fixed points is the center <math>Z</math> of <math>G</math>; thus</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath> \lvert Z \rvert \equiv \lvert G \rvert \equiv 0 \not\equiv 1 \pmod{p}, </cmath></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><cmath> \lvert Z \rvert \equiv \lvert G \rvert \equiv 0 \not\equiv 1 \pmod{p}, </cmath></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>so <math>Z</math> is not trivial.  <math>\blacksquare</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>so <math>Z</math> is not trivial.  <math>\blacksquare</math></div></td></tr>
</table>Math154https://artofproblemsolving.com/wiki/index.php?title=P-group&diff=26251&oldid=prevBoy Soprano II: definition and results2008-06-02T19:41:21Z<p>definition and results</p>
<p><b>New page</b></p><div>{{title restriction|<math>p</math>-group}}<br />
<br />
A '''<math>\boldsymbol{p}</math>-group''' is a [[finite]] [[group]] whose [[order (group theory) |order]] is a power of a [[prime]] <math>p</math>.<br />
<br />
== Properties ==<br />
<br />
'''Lemma.''' Let <math>G</math> be a <math>p</math>-group acting on a finite set <math>S</math>; let <math>S^G</math> denote the set of fixed points of <math>S</math>. Then<br />
<cmath> \lvert S^G \rvert \equiv \lvert S \rvert \pmod{p} . </cmath><br />
<br />
''Proof.'' It is enough to show that <math>p</math> divides the cardinality of each [[orbit]] of <math>S</math> with more than one element. This follows directly from the [[orbit-stabilizer theorem]]. <math>\blacksquare</math><br />
<br />
'''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.<br />
<br />
''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<br />
<cmath> \lvert Z \rvert \equiv \lvert G \rvert \equiv 0 \not\equiv 1 \pmod{p}, </cmath><br />
so <math>Z</math> is not trivial. <math>\blacksquare</math><br />
<br />
'''Theorem.''' Let <math>G</math> be a <math>p</math>-group of order <math>p^r</math>. Then there exists a series of subgroups<br />
<cmath> G = G^1 \supseteq G^2 \supseteq \dotsb \supseteq G^{r+1} = \{e\} </cmath><br />
such that <math>G^k</math> [[normalizer |normalizes]] <math>G^{k+1}</math>, <math>(G,G^k) \subseteq G^{k+1}</math>, and <math>G^k/G^{k+1}</math> is a [[cyclic group]] of order <math>p</math>, for all indices <math>k</math>.<br />
<br />
''Proof.'' We induct on the order of <math>G</math>. For <math>G= \{e\}</math>, the theorem is trivial. Let <math>Z</math> be the center of <math>G</math>, and <math>x</math> a non-identity element of <math>Z</math>. Let <math>p^s</math> be the order of <math>x</math>. Then <math>x^{p^{s-1}}</math> generates a cyclic group <math>A</math> of order <math>p</math>; since <math>A</math> is contained in <math>Z</math>, it is evidently a [[normal subgroup]] of <math>G</math>. Then <math>G/A</math> is a <math>p</math>-group of order <math>p^{r-1}</math>. By inductive hypothesis, there is a sequence<br />
<cmath> G/A = (G/A)^1 \supseteq (G/A)^2 \supseteq \dotsb \supseteq (G/A)^r = A </cmath><br />
satisfying the theorem's requirements.<br />
<br />
Let <math>\phi</math> be the canonical [[homomorphism]] from <math>G</math> onto <math>G/A</math>, and for <math>1\le k \le r</math>, let <math>G^k</math> be <math>\phi^{-1}((G/A)^k)</math>, and let <math>G^{r+1}=\{e\}</math>. Then for <math>k \in [1,r-1]</math>, <math>G^{k+1}</math> is a normal subgroup of <math>G^k</math>, and <math>G^k/G^{k+1}</math> is [[isomorphic]] to <math>(G/A)^k/(G/A)^{k+1}</math>; hence it is a cyclic group of order <math>p</math>. Also,<br />
<cmath> \begin{align*}<br />
(G,G^k) & \subseteq \phi^{-1}(\phi(G),\phi(G^k)) = \phi^{-1}((G/A),(G/A)^k) \\<br />
&\subseteq \phi^{-1}((G/A)^{k+1}) = G^{k+1} .<br />
\end{align*} </cmath><br />
Since <math>A</math> is a cyclic group of order <math>p</math> that lies in the center of <math>G</math>, the theorem's statements are true for <math>k=r</math>, as well. This completes the proof. <math>\blacksquare</math><br />
<br />
'''Corollary 1.''' Every <math>p</math>-group is [[nilpotent group |nilpotent]].<br />
<br />
'''Corollary 2.''' If <math>G</math> is a <math>p</math>-group, and <math>H</math> is a proper [[subgroup]] of <math>G</math>, then the normalizer of <math>H</math> is distinct from <math>H</math>.<br />
<br />
This is a property of nilpotent groups in general.<br />
<br />
'''Proposition.''' Let <math>H</math> be a proper subgroup of a <math>p</math>-group <math>G</math>. Then there exists a normal subgroup <math>N</math> of <math>G</math> of index <math>p</math> that contains <math>H</math>.<br />
<br />
''Proof.'' Since <math>G</math> is nilpotent there exists a normal subgroup <math>A</math> of <math>G</math> such that <math>H \subseteq A</math> and <math>G/A</math> is abelian. Let <math>N</math> be a maximal subgroup of <math>G</math> containing <math>A</math>. Since <math>G</math> is nilpotent, <math>N</math> is normal in <math>G</math>. Since <math>N</math> is evidently [[simple group |simple]], it is cyclic, and hence of order <math>p</math>. <math>\blacksquare</math><br />
<br />
'''Corollary.''' Let <math>G</math> be a <math>p</math>-group, and <math>H</math> a subgroup of <math>G</math> of index <math>p</math>. Then <math>H</math> is a normal subgroup of <math>G</math>.<br />
<br />
== See also ==<br />
<br />
* [[Sylow p-subgroup]]<br />
* [[Sylow Theorems]]<br />
<br />
[[Category:Group theory]]</div>Boy Soprano II