Free group

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 $\displaystyle X_2^{-3}X_1X_2X_1^{-1}X_1X_2^{-1}X_1^{-1}X_2^3 = X_2^{-3}X_1X_2\cdot1\cdot X_2^{-1}X_1^{-1}X_2^3 = X_2^{-3}X_1X_2X_2^{-1}X_1^{-1}X_2^3 = X_2^{-3}X_1\cdot 1\cdotX_1^{-1}X_2^3 = X_2^{-3}X_1X_1^{-1}X_2^3 = X_2^{-3}\cdot 1\cdotX_2^3 = X_2^{-3}X_2^3 = 1$ (Error compiling LaTeX. Unknown error_msg). (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 $u = v^n$ or $v = u^n$ .

Proof

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Free group

Commutivity in Free Groups

Proof