Skip to main content Contents Index Prev Up Next \(\DeclareMathOperator{\ann}{ann}
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Section 4.1 Free Groups
Definition 4.1 . Free Group.
Let \(A\) be a set. Consider a new set of symbols \(A^{-1}=\{a^{-1} \mid a \in A\}\text{.}\) The free group on \(A\text{,}\) denoted \(F(A)\text{,}\) is the set of all finite words written using symbols in \(A\cup A^{-1}\text{,}\) including the empty word, where two words are equal if one is obtained from the other by erasing a pair of consecutive symbols \(aa^{-1}\) or \(a^{-1}a\text{.}\) In symbols,
\begin{equation*}
F(A)=\{a_1^{i_1}a_2^{i_2}\cdots a_m^{i_m} \mid m\geq 0, a_j\in A, i_j\in\{-1,1\}\}.
\end{equation*}
The set \(F(A)\) is a group in which any two words are multiplied by concatenation.
Example 4.2 .
The free group on a singleton set \(A={x}\) is the infinite cyclic group \(C_\infty\text{.}\) It is easy to visualize this since \(C_\infty\cong \mathbb{Z}\text{.}\)
It is already somewhat challenging to visualize the free group on two generators, \(F({x,y})\text{.}\) The best one can do to represent it is an infinite graph called the Cayley graph of this group. It is obtained by starting at the origin (the center of the picture below), then branching out in four directions by a length of 1, then branching out similarly by a length of 1/2, then by 1/4, then by 1/8, etc (I stopped there, to avoid cluttering the picture too much but to get the complete Cayley graph one continues infinitely obtaining a fractal kind of picture). Then every element of \(F ({x, y})\) corresponds in a rather natural way to exactly one dot in this diagram. Indeed, we can place the empty word at the center, and we can agree that every \(x\) in a word takes us one step to the right, every \(x^{-1}\) to the left, every \(y\) up, and every \(y^{-1}\) down. For example, the word \(yx^{-1}yx\) takes us here:
Theorem 4.3 . Universal Mapping Property for Free Groups.
Let \(A\) be a set, let \(F(A)\) be the free group on \(A\text{,}\) let \(H\) be a group, and let \(g:A \to H\) be a function. Then there is a unique homomorphism \(f:F(A) \to H\) satisfying \(f(a) = g(a)\) for all \(a \in A\text{.}\)
Proof.
Set \(f:F(A) \to H\) to be given by \(f(a_1^{i_1}a_2^{i_2}\cdots a_m^{i_m})=g(a_1)^{i_1}g(a_2)^{i_2}\cdots g(a_m)^{a_m}\) for any \(m\geq 0, a_j\in A, i_j\in\{-1,1\}\text{.}\) One checks that \(f\) is well defined by noting that
\begin{align*}
f(a_1^{i_1}a_2^{i_2}\cdots aa^{-1}\cdots a_m^{i_m}) & =g(a_1)^{i_1}g(a_2)^{i_2}\cdots g(a)g(a)^{-1}\cdots g(a_m)^{a_m}\\
&=f(a_1^{i_1}a_2^{i_2}\cdots a_m^{i_m})
\end{align*}
for any \(a\in G\) and similarly for inserting \(a^{-1}a\text{.}\) The fact that \(f\) is a group homomorphism and its uniqueness are left as an exercise.
