A free group contains no notrivial elements of finite order
This statement seems obvious and trivial, but I cannot think of a nice proof besides going and getting my hands dirty with elements of the group. Is there a nice proof of this fact?
A free group contains no notrivial elements of finite order
This statement seems obvious and trivial, but I cannot think of a nice proof besides going and getting my hands dirty with elements of the group. Is there a nice proof of this fact?
Assume $F$ is the free group on a set $X$ and $a \in F$ has finite order $e$. Since $F(X)$ is the directed colimit of the $F(Y)$ with $Y \subseteq X$ with $Y$ finite, we may assume as well that $X$ is finite. Since every element is conjugated to a cyclically reduced word, we may assume that $a=x_{k_1}^{e_1} \dotsc x_{k_n}^{e_n}$ is cyclically reduced with $e_i \neq 0$; in particular $x_{k_1} \neq x_{k_n}$. Then $a^e=x_{k_1}^{e_1} \dotsc x_{k_n}^{e_n} x_{k_1}^{e_1} \dotsc x_{k_n}^{e_n} \dotsc $ is again (cyclically) reduced. On the other hand it is trivial, thus it's length must be $0$. This shows $a=1$.