Difference between revisions of "Free group"

m
 
(4 intermediate revisions by 3 users not shown)
Line 3: Line 3:
 
An example of an element of the free group on <math>I = \{1, 2\}</math> is <math>X_1X_2^{-1}X_1^{-1}X_2^3</math> (where by <math>X_2^3</math> we mean <math>X_2X_2X_2</math>).
 
An example of an element of the free group on <math>I = \{1, 2\}</math> is <math>X_1X_2^{-1}X_1^{-1}X_2^3</math> (where by <math>X_2^3</math> we mean <math>X_2X_2X_2</math>).
  
The [[inverse with respect to an operation | inverse]] of a given element of a free group can be found by reversing the string and the sign of all exponents.  Thus, the string given above has inverse <math>X_2^{-3}X_1X_2X_1^{-1}</math>.  This is easy to check: we just apply the group operation to the two strings to get <math>\displaystyle X_2^{-3}X_1X_2X_1^{-1}X_1X_2^{-1}X_1^{-1}X_2^3 = X_2^{-3}X_1X_2\cdot1\cdot X_2^{-1}X_1^{-1}X_2^3 = X_2^{-3}X_1X_2X_2^{-1}X_1^{-1}X_2^3 = X_2^{-3}X_1\cdot 1\cdotX_1^{-1}X_2^3 = X_2^{-3}X_1X_1^{-1}X_2^3 = X_2^{-3}\cdot 1\cdotX_2^3 = X_2^{-3}X_2^3 = 1</math>(Note that we implicitly used the [[associativity]] of the group operation repeatedly in this process.)  Showing that this string is a right inverse is equally straightforward.  The proof that this holds for every string is by [[induction]] using the same idea.
+
The [[inverse with respect to an operation | inverse]] of a given element of a free group can be found by reversing the string and the sign of all exponents.  Thus, the string given above has inverse <math>X_2^{-3}X_1X_2X_1^{-1}</math>.  This is easy to check: we just apply the group operation to the two strings to get <cmath>\begin{align*}X_2^{-3}X_1X_2X_1^{-1}X_1X_2^{-1}X_1^{-1}X_2^3 &= X_2^{-3}X_1X_2X_2^{-1}X_1^{-1}X_2^3\\ &= X_2^{-3}X_1X_1^{-1}X_2^3\\ &= X_2^{-3}X_2^3 = 1 \;.\end{align*}</cmath> (Note that we implicitly used the [[associativity]] of the group operation repeatedly in this process.)  Showing that this string is a right inverse is equally straightforward.  The proof that this holds for every string is by [[induction]] using the same idea.
  
 
More formally, free groups are defined by [[universal property|universal properties]]. A group <math>F</math> is called a free group on <math>I</math> if there is a function <math>\phi:I\to F</math> so that for any group <math>G</math> and a function <math>\theta:I\to G</math>, there is a unique group homomorphism <math>\psi:F\to G</math> so that <math>\theta=\psi\phi</math>, i.e. so that <math>\theta(i)=\psi\circ\phi(i)</math> for all <math>i\in I</math>. We often like to draw a diagram to represent this relationship; however WE DON'T HAVE THE xy PACKAGE INCLUDED, SO I CAN'T TeX IT.
 
More formally, free groups are defined by [[universal property|universal properties]]. A group <math>F</math> is called a free group on <math>I</math> if there is a function <math>\phi:I\to F</math> so that for any group <math>G</math> and a function <math>\theta:I\to G</math>, there is a unique group homomorphism <math>\psi:F\to G</math> so that <math>\theta=\psi\phi</math>, i.e. so that <math>\theta(i)=\psi\circ\phi(i)</math> for all <math>i\in I</math>. We often like to draw a diagram to represent this relationship; however WE DON'T HAVE THE xy PACKAGE INCLUDED, SO I CAN'T TeX IT.
 
  
 
===Commutivity in Free Groups===
 
===Commutivity in Free Groups===
 
The free group over <math>I</math> has "as little structure as possible."  Thus, we can show that two elements of the free group [[commute]] only when it's "necessary":
 
The free group over <math>I</math> has "as little structure as possible."  Thus, we can show that two elements of the free group [[commute]] only when it's "necessary":
If <math>u, v</math> are elements of the free group over a set <math>I</math> and <math>uv = vu</math>, then <math>u = v^n</math> or <math>v = u^n</math>.
+
If <math>u, v</math> are elements of the free group over a set <math>I</math> and <math>uv = vu</math>, then <math>u^m = v^n</math>, for some [[integer]]s <math>m</math> and <math>n</math>.
 +
 
 
====Proof====
 
====Proof====
  
  
 
{{stub}}
 
{{stub}}
 +
 +
[[Category:Group theory]]

Latest revision as of 16:54, 16 March 2012

A free group is a type of group that is of particular importance in combinatorics. Let $I$ be any nonempty index set. Informally, a free group on $I$ is the collection of finite strings of characters from the collection $\{X_i,X_i^{-1}:i\in I\}$ subject only to the criterion that $X_iX_i^{-1}=X_i^{-1}X_i=1$ where $1$ is the group identity and is equal to the empty string. The group operation is concatenation.

An example of an element of the free group on $I = \{1, 2\}$ is $X_1X_2^{-1}X_1^{-1}X_2^3$ (where by $X_2^3$ we mean $X_2X_2X_2$).

The inverse of a given element of a free group can be found by reversing the string and the sign of all exponents. Thus, the string given above has inverse $X_2^{-3}X_1X_2X_1^{-1}$. This is easy to check: we just apply the group operation to the two strings to get \begin{align*}X_2^{-3}X_1X_2X_1^{-1}X_1X_2^{-1}X_1^{-1}X_2^3 &= X_2^{-3}X_1X_2X_2^{-1}X_1^{-1}X_2^3\\ &=  X_2^{-3}X_1X_1^{-1}X_2^3\\ &= X_2^{-3}X_2^3 = 1 \;.\end{align*} (Note that we implicitly used the associativity of the group operation repeatedly in this process.) Showing that this string is a right inverse is equally straightforward. The proof that this holds for every string is by induction using the same idea.

More formally, free groups are defined by universal properties. A group $F$ is called a free group on $I$ if there is a function $\phi:I\to F$ so that for any group $G$ and a function $\theta:I\to G$, there is a unique group homomorphism $\psi:F\to G$ so that $\theta=\psi\phi$, i.e. so that $\theta(i)=\psi\circ\phi(i)$ for all $i\in I$. We often like to draw a diagram to represent this relationship; however WE DON'T HAVE THE xy PACKAGE INCLUDED, SO I CAN'T TeX IT.

Commutivity in Free Groups

The free group over $I$ has "as little structure as possible." Thus, we can show that two elements of the free group commute only when it's "necessary": If $u, v$ are elements of the free group over a set $I$ and $uv = vu$, then $u^m = v^n$, for some integers $m$ and $n$.

Proof

This article is a stub. Help us out by expanding it.