Difference between revisions of "Notes:Free group"

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!

Grillet - Abstract Algebra

This is taken from section 6 of chapter 1 starting on page 27.

Reduction

• Let $X$ be a set.
• Let $X'$ be a disjoint set
• Let $A:X\rightarrow X'$ be a bijection, and let $A':\eq A^{-1}:X'\rightarrow X$ be the inverse bijection
• Let $Y:\eq X\cup X'$

Caveat:Apparently we denote $A$ by $x\mapsto x'$ and $A'$ by $y\mapsto y'$ such that $(x')'\eq x$ and $(y')'\eq y$ - I am unsure of this.

Words in the "alphabet" $Y$ are finite, but possibly empty, sequences of elements of $Y$.

Next:

• Let $W$ be the free monoid generated by $Y$, where, as usual, multiplication is concatenation^{[Note 1]}

Reduced word

A word, $a\in W$ with $a\eq(a_1,\ldots,a_n)$ is reduced when:

• $\forall i\in\{1,\ldots,n-1\}[a_{i+1}\neq a_i']$

For example:

1. $(x,y,z)$ - reduced
2. $(x,x,x)$ - reduced
3. $(x,y,y',z)$ - NOT reduced

Reduction deletes subsequences of the form $(a_i,a'_i)$ until a reduced word is reached.

Sequences of reductions

1. We write $a\overset{1}{\rightarrow} b$ if

Lee - Topological Manifolds

Free group generated by

Let $S:\eq\{\sigma\}$ - a set containing a single thing. Then:

• $F(S)$ - the free group generated by $S$ (we may write $F(\sigma)$ instead, for short) is defined as follows:
  1. The set of the group is $F(\sigma):\eq\{\sigma\}\times\mathbb{Z}$ - the set of all tuples of the form $(\sigma,m)$ for $m\in\mathbb{Z}$
  2. The operation is: $(\sigma,a)\cdot(\sigma,b):\eq(\sigma,a+b)$
• We identify $\sigma$ with $(\sigma,1)$, thus:
  • $\sigma^m\eq(\sigma,m\cdot(1))\eq(\sigma,m)$

Now suppose $S$ is some arbitrary set, then:

• $F(S):\eq\underset{\sigma\in S}{\Huge \ast}F(\sigma)$ - the free product of the groups $F(\sigma)$ for each $\sigma\in S$

Free product

Quite simple:

• $\underset{\alpha\in I}{\huge\ast}G_\alpha$ is a quotient by an equivalence relation on the free monoid generated by the set that is the disjoint union: $\coprod_{\alpha\in I}G_\alpha$ where:
  • Two words in the monoid are considered equivalent if one can be reduced to the other.
  • The rules for reduction are:
    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.

Notes

1. Jump up ↑ Obviously, concatenation of finite sequences $a:\eq(a_1,\ldots,a_\ell)$ and $b:\eq(b_1,\ldots,b_m)$ is:
   • $a\cdot b:\eq(a_1,\ldots,a_\ell,b_1,\ldots,b_m)$