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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
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]