Difference between revisions of "Notes:Free group"
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| − | ==[[Abstract Algebra - Pierre Antoine Grillet|Grillet - Abstract Algebra]]== | + | * This is about the [[Free group generated by]] but may include the [[free product of groups]]! |
| + | __TOC__ | ||
| + | ==[[Books:Abstract Algebra - Pierre Antoine Grillet|Grillet - Abstract Algebra]]== | ||
This is taken from '''section 6 of chapter 1''' starting on page 27. | This is taken from '''section 6 of chapter 1''' starting on page 27. | ||
===Reduction=== | ===Reduction=== | ||
| Line 26: | Line 28: | ||
===Sequences of reductions=== | ===Sequences of reductions=== | ||
# We write {{M|a\overset{1}{\rightarrow} b}} if | # We write {{M|a\overset{1}{\rightarrow} b}} if | ||
| + | ==[[Books:Introduction to Topological Manifolds - John M. Lee|Lee - Topological Manifolds]]== | ||
| + | ===Free group generated by=== | ||
| + | Let {{M|S:\eq\{\sigma\} }} - a set containing a single thing. Then: | ||
| + | * {{M|F(S)}} - the free group generated by {{M|S}} (we may write {{M|F(\sigma)}} instead, for short) is defined as follows: | ||
| + | *# The set of the [[group]] is {{M|F(\sigma):\eq\{\sigma\}\times\mathbb{Z} }} - the set of all [[tuples]] of the form {{M|(\sigma,m)}} for {{M|m\in\mathbb{Z} }} | ||
| + | *# The operation is: {{M|(\sigma,a)\cdot(\sigma,b):\eq(\sigma,a+b)}} | ||
| + | * We identify {{M|\sigma}} with {{M|(\sigma,1)}}, thus: | ||
| + | ** {{M|\sigma^m\eq(\sigma,m\cdot(1))\eq(\sigma,m)}} | ||
| + | |||
| + | |||
| + | Now suppose {{M|S}} is some arbitrary set, then: | ||
| + | * {{M|F(S):\eq\underset{\sigma\in S}{\Huge \ast}F(\sigma)}} - the {{link|free product|group}} of the groups {{M|F(\sigma)}} for each {{M|\sigma\in S}} | ||
| + | ===Free product=== | ||
| + | Quite simple: | ||
| + | * {{M|\underset{\alpha\in I}{\huge\ast}G_\alpha}} is a [[quotient by an equivalence relation]] on the [[free monoid generated by]] the set that is the [[disjoint union]]: {{M|\coprod_{\alpha\in I}G_\alpha}} where: | ||
| + | ** Two words in the monoid are considered equivalent if one can be reduced to the other. | ||
| + | ** The rules for reduction are: | ||
| + | **# (two elements in the word from the same group are combined into one that is their product) | ||
| + | **# (any identity elements are discarded) | ||
| + | |||
| + | A bit of [[factor (function)|factorisation]] later and you've got an associative operation on the quotient with identity, just need to show inverse then. | ||
==Notes== | ==Notes== | ||
<references group="Note"/> | <references group="Note"/> | ||
Latest revision as of 20:36, 10 December 2016
- This is about the Free group generated by but may include the free product of groups!
Contents
Grillet - Abstract Algebra
This is taken from section 6 of chapter 1 starting on page 27.
Reduction
- Let [ilmath]X[/ilmath] be a set.
- Let [ilmath]X'[/ilmath] be a disjoint set
- Let [ilmath]A:X\rightarrow X'[/ilmath] be a bijection, and let [ilmath]A':\eq A^{-1}:X'\rightarrow X[/ilmath] be the inverse bijection
- Let [ilmath]Y:\eq X\cup X'[/ilmath]
Caveat:Apparently we denote [ilmath]A[/ilmath] by [ilmath]x\mapsto x'[/ilmath] and [ilmath]A'[/ilmath] by [ilmath]y\mapsto y'[/ilmath] such that [ilmath](x')'\eq x[/ilmath] and [ilmath](y')'\eq y[/ilmath] - I am unsure of this.
Words in the "alphabet" [ilmath]Y[/ilmath] are finite, but possibly empty, sequences of elements of [ilmath]Y[/ilmath].
Next:
- Let [ilmath]W[/ilmath] be the free monoid generated by [ilmath]Y[/ilmath], where, as usual, multiplication is concatenation[Note 1]
Reduced word
A word, [ilmath]a\in W[/ilmath] with [ilmath]a\eq(a_1,\ldots,a_n)[/ilmath] is reduced when:
- [ilmath]\forall i\in\{1,\ldots,n-1\}[a_{i+1}\neq a_i'][/ilmath]
For example:
- [ilmath](x,y,z)[/ilmath] - reduced
- [ilmath](x,x,x)[/ilmath] - reduced
- [ilmath](x,y,y',z)[/ilmath] - NOT reduced
Reduction deletes subsequences of the form [ilmath](a_i,a'_i)[/ilmath] until a reduced word is reached.
Sequences of reductions
- We write [ilmath]a\overset{1}{\rightarrow} b[/ilmath] if
Lee - Topological Manifolds
Free group generated by
Let [ilmath]S:\eq\{\sigma\} [/ilmath] - a set containing a single thing. Then:
- [ilmath]F(S)[/ilmath] - the free group generated by [ilmath]S[/ilmath] (we may write [ilmath]F(\sigma)[/ilmath] instead, for short) is defined as follows:
- We identify [ilmath]\sigma[/ilmath] with [ilmath](\sigma,1)[/ilmath], thus:
- [ilmath]\sigma^m\eq(\sigma,m\cdot(1))\eq(\sigma,m)[/ilmath]
Now suppose [ilmath]S[/ilmath] is some arbitrary set, then:
- [ilmath]F(S):\eq\underset{\sigma\in S}{\Huge \ast}F(\sigma)[/ilmath] - the free product of the groups [ilmath]F(\sigma)[/ilmath] for each [ilmath]\sigma\in S[/ilmath]
Free product
Quite simple:
- [ilmath]\underset{\alpha\in I}{\huge\ast}G_\alpha[/ilmath] is a quotient by an equivalence relation on the free monoid generated by the set that is the disjoint union: [ilmath]\coprod_{\alpha\in I}G_\alpha[/ilmath] where:
- Two words in the monoid are considered equivalent if one can be reduced to the other.
- The rules for reduction are:
- (two elements in the word from the same group are combined into one that is their product)
- (any identity elements are discarded)
A bit of factorisation later and you've got an associative operation on the quotient with identity, just need to show inverse then.
Notes
- ↑ Obviously, concatenation of finite sequences [ilmath]a:\eq(a_1,\ldots,a_\ell)[/ilmath] and [ilmath]b:\eq(b_1,\ldots,b_m)[/ilmath] is:
- [ilmath]a\cdot b:\eq(a_1,\ldots,a_\ell,b_1,\ldots,b_m)[/ilmath]
