What is a simple example of a free group?

Question

Can someone give me a simple example of a free group with a basis, given the definition below? I don't think I'm understanding the definition clearly.

For example if $F= (\Bbb Z, +)$, $X = \{0\}$, $\phi\colon\{0\} \rightarrow G$ is any function, then there should exist a unique homomorphism $\tilde \phi\colon \Bbb Z \rightarrow G$ such that $\tilde \phi(0) = \phi(0)$.

But if $\phi\colon\{0\} \mapsto \text{non identity element of $G$}$, then there's no way any homomorphism exists because identities of one group are mapped to the other group.

Answer 1

The simplest example is $(\mathbb{Z},+)$ with the basis $X=\{1\}$. Let $\varphi$ be any function from $X$ into a group $G$ and let $g=\varphi(1)$. Now, define $\tilde{\varphi}(m)=g^m$, for each integer $m$. Then $\tilde\varphi$ is a group homomorphism. Furthermore, it is the only group homomorphism from $(\mathbb{Z},+)$ into $G$ such that $\tilde\varphi(1)=\varphi(1)$.

Answer 2

You have a set $X$. The elements of $F_X$ the free group on $X$ are "reduced words" in $X$; finite sequences of symbols $x$ and $x^{-1}$ for $x\in X$ where no $x$ and $x^{-1}$ are adjacent and including the empty word, which I'll write as $1$ as it is the identity. So for $X=\{x,y,z\}$ reduced words include $1$, $x$, $y$, $z^{-1}$, $yx$, $xzx^{-1}$, $xxyx^{-1}z^{-1}xzy^{-1}z$ etc. To multiply, concatenate, and cancel any $xx^{-1}$s or $x^{-1}x$s.

If you have $X=\{x\}$ with one element, the reduced words are $1$, $xx\cdots x$ (with $n$ $x$s) and $x^{-1}x^{-1}\cdots x^{-1}$ (with $n$ $x^{-1}$s). If we write these instead as $x^0$, $x^n$ and $x^{-n}$ we see that $F_X\cong\Bbb Z$.

Rather perversely, in your notation you have chosen $0$ to be the element. This results in reduced words like $0000$ and $0^{-1}0^{-1}0^{-1}$. These can cause confusion...

Answer 3

The simplest non-abelian example is $F_2$, with generating set $S=\{a,b\}$. The elements are just reduced words of the form $a^{r_1}b^{s_1}\cdots a^{r_n}b^{s_n}$. In fact, $F_2$ can be shown to be isomorphic to a matrix subgroup of $SL_2(\mathbb{Z})$, namely the subgroup generated by $$ a = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}, \ \ \ \ \ b = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}, $$ see the Ping-Pong Lemma.

Answer 4

The identity in F is the empty word. This alleviates your problem because X consists of elements of F (not "0"). X has no identity. X isn't a group; it's just a set.

Answer 5

Whilst it is important to understand the formal definition of free groups, I shall give you a more geometric example heralding from algebraic topology, because intuitively it is more of use.

Imagine that you have a flower which has only two petals. Consider also that the meeting point of these two petals is of dimension 0 (namely it is a point). Now, the free group on two generators describes exactly the difficulty for a small insect to walk on the contour of the petals (no flying's allowed!). This "difficulty" is formally defined in algebraic topology as the fundamental group of that surface. Obviously, if you had a flower with k-petals (imagine a bouquet), then the free group on k generators would express the respective difficulty to walk on all these petals.