Difference between revisions of "2-Sylow subgroup is TI implies it is normal or there is exactly one conjugacy class of involutions"
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'''Proof''': Suppose <math>G</math> has two distinct conjugacy classes of involutions. Let <math>P,Q</math> be any two <math>2</math>-Sylow subgroups of <math>G</math>. First, note that every involution is conjugate to some element in any given 2-Sylow subgroup (fact (1), the domination part of Sylow's theorem), so every 2-Sylow subgroup intersects every conjugacy class of involutions. Thus, we may pick <math>x \in P, y \in Q</math> as representatives of the two (distinct) conjugacy classes of involutions. | '''Proof''': Suppose <math>G</math> has two distinct conjugacy classes of involutions. Let <math>P,Q</math> be any two <math>2</math>-Sylow subgroups of <math>G</math>. First, note that every involution is conjugate to some element in any given 2-Sylow subgroup (fact (1), the domination part of Sylow's theorem), so every 2-Sylow subgroup intersects every conjugacy class of involutions. Thus, we may pick <math>x \in P, y \in Q</math> as representatives of the two (distinct) conjugacy classes of involutions. | ||
− | By fact (2), there exists an involution <math>z</math> that centralizes both <math>x</math> and <math>y</math>. Since <math>z</math> is an involution, it is contained in a 2-Sylow subgroup <math>R</math>. Because the 2-Sylow subgroups are TI, <math>R</math> is the unique 2-Sylow subgroup containing <math>z</math>. Similarly, <math>P</math> is the unique 2-Sylow subgroup containing <math>x</math> and <math>Q</math> | + | By fact (2), there exists an involution <math>z</math> that centralizes both <math>x</math> and <math>y</math>. Since <math>z</math> is an involution, it is contained in a 2-Sylow subgroup <math>R</math>. Because the 2-Sylow subgroups are TI, <math>R</math> is the unique 2-Sylow subgroup containing <math>z</math>. Similarly, <math>P</math> is the unique 2-Sylow subgroup containing <math>x</math> and <math>Q</math> is the unique 2-Sylow subgroup containing <math>y</math>. |
Since <math>x</math> and <math>z</math> are commuting involutions, <math>\langle x,z \rangle</math> is a 2-group. By fact (1), there is a 2-Sylow subgroup containing this. This must equal both <math>P</math> and <math>R</math>, so <math>P = R</math>. Similarly, <math>\langle y,z \rangle</math> is a 2-group, so there is a 2-Sylow subgroup containing it, which must equal both <math>Q</math> and <math>R</math>, so <math>Q = R</math>. Thus, <math>P = Q</math>. Since <math>P,Q</math> were arbitrary, we conclude that any two 2-Sylow subgroups are equal, so there is a unique 2-Sylow subgroup which must therefore be normal. | Since <math>x</math> and <math>z</math> are commuting involutions, <math>\langle x,z \rangle</math> is a 2-group. By fact (1), there is a 2-Sylow subgroup containing this. This must equal both <math>P</math> and <math>R</math>, so <math>P = R</math>. Similarly, <math>\langle y,z \rangle</math> is a 2-group, so there is a 2-Sylow subgroup containing it, which must equal both <math>Q</math> and <math>R</math>, so <math>Q = R</math>. Thus, <math>P = Q</math>. Since <math>P,Q</math> were arbitrary, we conclude that any two 2-Sylow subgroups are equal, so there is a unique 2-Sylow subgroup which must therefore be normal. |
Latest revision as of 05:22, 12 August 2017
Statement
Suppose is a finite group of even order with the property that one (and hence all) of its 2-Sylow subgroup (?)s is a TI-subgroup (?). In other words, any two distinct 2-Sylow subgroups intersect trivially. Then, one of the following is true:
- has exactly one conjugacy class of involutions (i.e., any two elements of order two are conjugate).
- The 2-Sylow subgroup of is normal, or equivalently, has a unique 2-Sylow subgroup.
Facts used
- Sylow implies order-dominating
- Involutions are either conjugate or have an involution centralizing both of them
Proof
Given: A finite group of even order whose 2-Sylow subgroups are TI.
To prove: Either has a single conjugacy class of involutions or has a normal 2-Sylow subgroup.
Proof: Suppose has two distinct conjugacy classes of involutions. Let be any two -Sylow subgroups of . First, note that every involution is conjugate to some element in any given 2-Sylow subgroup (fact (1), the domination part of Sylow's theorem), so every 2-Sylow subgroup intersects every conjugacy class of involutions. Thus, we may pick as representatives of the two (distinct) conjugacy classes of involutions.
By fact (2), there exists an involution that centralizes both and . Since is an involution, it is contained in a 2-Sylow subgroup . Because the 2-Sylow subgroups are TI, is the unique 2-Sylow subgroup containing . Similarly, is the unique 2-Sylow subgroup containing and is the unique 2-Sylow subgroup containing .
Since and are commuting involutions, is a 2-group. By fact (1), there is a 2-Sylow subgroup containing this. This must equal both and , so . Similarly, is a 2-group, so there is a 2-Sylow subgroup containing it, which must equal both and , so . Thus, . Since were arbitrary, we conclude that any two 2-Sylow subgroups are equal, so there is a unique 2-Sylow subgroup which must therefore be normal.
References
Textbook references
- Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 302, Chapter 9 (Groups of even order), Theorem 1.4, ^{More info}