Notes:Free group

• This is about the Free group generated by but may include the free product of groups!

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

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:

1. [ilmath](x,y,z)[/ilmath] - reduced
2. [ilmath](x,x,x)[/ilmath] - reduced
3. [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

1. 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:
  1. The set of the group is [ilmath]F(\sigma):\eq\{\sigma\}\times\mathbb{Z} [/ilmath] - the set of all tuples of the form [ilmath](\sigma,m)[/ilmath] for [ilmath]m\in\mathbb{Z} [/ilmath]
  2. The operation is: [ilmath](\sigma,a)\cdot(\sigma,b):\eq(\sigma,a+b)[/ilmath]
• 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:
    1. (two elements in the word from the same group are combined into one that is their product)
    2. (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.