Basic Theorems Regarding Free Groups on a Set

On the The Free Group on a Set X, F(X) page we defined the free group $F(X)$ generated by the set $X$. We will now look at some basic results regarding free groups.

Proposition 1: The free group generated $1$ element, $F_1$ is isomorphic to $(\mathbb{Z}, +)$.

• Proof: Let $X = \{ x \}$, so that $F_1 = F(X)$. Then every word on $X$ is of the form $x^n$ where $x^0$ is defined to be the identity word. Let $\varphi : \mathbb{Z} \to F_1$ be defined for all $n \in \mathbb{Z}$ by $\varphi(n) = x^n$.

• Clearly $\varphi$ is bijection. It is also a homomorphism since for all $m, n \in \mathbb{Z}$ we have that:

(1)

\begin{align} \quad \varphi(m + n) = x^{m+n} = x^mx^n = \varphi(m)\varphi(n) \end{align}

• Where we reduce $x^{m}x^{n}$ if $m$ and $n$ have different signs. $\blacksquare$

Proposition 2: Let $X$ be a nonempty set. Then $F(X)$ is abelian if and only if $|X| = 1$.

• Proof: $\Rightarrow$ Suppose that $|X| \neq 1$. We will show that $F(X)$ is nonabelian. Since $|X| \geq 2$, let $x_1, x_2 \in X$ with $x_1 \neq x_2$. Then $x_1x_2x_1^{-1}x_2^{-1}$ is a reduced word on $X$ that is not the identity word which we will denote by $1$, i.e., $x_1x_2x_1^{-1}x_2^{-1} \neq 1$.

• Concatenate both words by $x_2$ to get $x_1x_2x_1 \neq x_2$. Then concatenate both words by $x_1$ to get that $x_1x_2 \neq x_2x_1$. So there exists $x_1, x_2 \in X$ with $x_1x_2 \neq x_2x_1$. So $F(X)$ is nonabelian.

• $\Leftarrow$ Suppose that $|X| = 1$. Then $F(X)$ is isomorphic to $F_1$. By proposition 1, $F_1$ is isomorphic to $(\mathbb{Z}, +)$, which is an abelian group. So $F(X) = F_1$ is abelian. $\blacksquare$

Proposition 3: Let $X$ be a set. If $H$ is a subgroup of $F(X)$ then there exists a set $Y$ such that $H$ is isomorphic to $F(Y)$. That is, a subgroup of a free group is a free group.