Problems in group theory

<p>Using combinatorial and character-theoretic methods the following theorems are proved.</p> <p><em>Theorem I</em> Let G be a finite simple group containing two non-conjugate subgroups A, B such that:</p> <p><ol type="i"&...

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Main Author: Stewart, W
Format: Thesis
Published: 1967
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author Stewart, W
author_facet Stewart, W
author_sort Stewart, W
collection OXFORD
description <p>Using combinatorial and character-theoretic methods the following theorems are proved.</p> <p><em>Theorem I</em> Let G be a finite simple group containing two non-conjugate subgroups A, B such that:</p> <p><ol type="i"><li><p>if a ε A, a &amp;notequals; 1, then C<sub>G</sub> (a) = A; if b ε B, b &amp;notequals; 1, then</p> <p>C<sub>G</sub>(b) = B;</p></li> <li>| N<sub>G</sub>(A)/A | = 2, | N<sub>G</sub>(B)/B | = 2.</li></ol></p> <p>Then G is isomorphic to some PSL (2,2<sup>a</sup>).</p> <p><em>Theorem II</em> Let G be a finite simple group containing a subgroup A such that:</p> <p><ol type="i"><li>if a ε A, a &amp;notequals; 1, then C<sub>G</sub>(a) = A;</li> <li>| N<sub>G</sub>(A)/A | = 2;</li> <li>3 divides | A | .</li></ol></p> <p>Then G is isomorphic to some PSL (2,q).</p>
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spelling oxford-uuid:6841b5f6-8bc0-407c-bad1-051381b564032022-03-26T18:43:39ZProblems in group theoryThesishttp://purl.org/coar/resource_type/c_db06uuid:6841b5f6-8bc0-407c-bad1-051381b56403Polonsky Theses Digitisation Project1967Stewart, W<p>Using combinatorial and character-theoretic methods the following theorems are proved.</p> <p><em>Theorem I</em> Let G be a finite simple group containing two non-conjugate subgroups A, B such that:</p> <p><ol type="i"><li><p>if a ε A, a &amp;notequals; 1, then C<sub>G</sub> (a) = A; if b ε B, b &amp;notequals; 1, then</p> <p>C<sub>G</sub>(b) = B;</p></li> <li>| N<sub>G</sub>(A)/A | = 2, | N<sub>G</sub>(B)/B | = 2.</li></ol></p> <p>Then G is isomorphic to some PSL (2,2<sup>a</sup>).</p> <p><em>Theorem II</em> Let G be a finite simple group containing a subgroup A such that:</p> <p><ol type="i"><li>if a ε A, a &amp;notequals; 1, then C<sub>G</sub>(a) = A;</li> <li>| N<sub>G</sub>(A)/A | = 2;</li> <li>3 divides | A | .</li></ol></p> <p>Then G is isomorphic to some PSL (2,q).</p>
spellingShingle Stewart, W
Problems in group theory
title Problems in group theory
title_full Problems in group theory
title_fullStr Problems in group theory
title_full_unstemmed Problems in group theory
title_short Problems in group theory
title_sort problems in group theory
work_keys_str_mv AT stewartw problemsingrouptheory