Difference between revisions of "Walther453Fall13 Topic2" - Rhea

Revision as of 14:40, 29 November 2013

Mark Rosinski, markrosi@purdue.edu Joseph Lam, lam5@purdue.edu Beichen Xiao, xiaob@purdue.edu

Outline:

Origin -Creator -History of the Sylow Theorems/ p-groups P-Groups -Definition -Regular p-groups

               -Relationship to Abelian Groups

-Application -Frattini Subgroup

       -Special p groups
    -Pro p-groups
    -Powerful p-groups

Sylow Theorems -Application

                -Theorem 1

-Theorem 2 -Theorem 3

       -Importance of Lagrange Theory

I've been using these websites:

and also the pdf emailed to you

Definitions:

Let p be a prime p $\in$ $\mathbb{Z}$ such that $\mathbb{Z}$ ≥0. A p-group is a group of order pⁿ.
A subgroup of order p^k for some k ≥ 1 is called a p-subgroup.
If |G| = p^$α$m where p does not divide m, then a subgroup of order p^$α$ is called a Sylow p-subgroup of G.

Definitons: Suppose p is a prime number. A p-group G is termed a regular p-group if it satisfies the following equivalent conditions:

For every a,b $\in G, there exists c \in [\langle a,b\rangle, \langle a,b \rangle] such that a^pb^p = (ab)^pc^p. *For every a,b \in G, there exist c_1,c_2, \dots, c_k \in [\langle a,b\rangle, \langle a,b \rangle] such that a^pb^p = (ab)^pc_1^pc_2^p \dots c_k^p. *For every a,b \in G and every natural number n, there exist c_1,c_2, \dots, c_k \in [\langle a,b\rangle, \langle a,b \rangle] such that a^qb^q = (ab)^qc_1^qc_2^q \dots c_k^q where q = p^n. == Sylow's Theorems == '''Definitions:''' Let G be a group of order pαm, where p is a prime, m≥1, and p does not divide m.  Then: #Sylp(G)  [[Category:MA453Fall2013Walther]]$

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 '''Definitons:'''
 Suppose p is a prime number. A p-group G is termed a regular p-group if it satisfies the following equivalent conditions:
-*For every a,b \in G, there exists c \in [\langle a,b\rangle, \langle a,b \rangle] such that a^pb^p = (ab)^pc^p.
+*For every a,b <math>\in G, there exists c \in [\langle a,b\rangle, \langle a,b \rangle] such that a^pb^p = (ab)^pc^p.
 *For every a,b \in G, there exist c_1,c_2, \dots, c_k \in [\langle a,b\rangle, \langle a,b \rangle] such that a^pb^p = (ab)^pc_1^pc_2^p \dots c_k^p.
 *For every a,b \in G and every natural number n, there exist c_1,c_2, \dots, c_k \in [\langle a,b\rangle, \langle a,b \rangle] such that a^qb^q = (ab)^qc_1^qc_2^q \dots c_k^q where q = p^n.