Equation in Free Group

Question

Let $F$ be a free group and $a \in F$. Assume that for any natural $n>1$ the equation $x^n=a$ has solutions (that is, $a$ is infinitely divisible). Show that $a=1$.

Answer 1

Without loss of generality, $a$ may be supposed cyclically reduced. If $x^n=a$, then $x$ is cyclically reduced, hence $\ell g(x^n)=n \cdot \ell g(x)$, where $\ell g( \cdot)$ is the length function on a free basis. So $n \ell g(x)= \ell g (a)$ for all $n \geq 1$, hence $\ell g(a)=0$ ie. $a=1$.

Answer 2

Here is a proof that uses a basic property of free groups: any element $\ne e$ maps to by a morphism to a $\it{finite}$ group onto an element $\ne e$. The reason for it is this: there do not exists non-trivial universal equalities valid in any finite group.

Take an element $a\ne e$ and map it to a finite group of cardinality $n$ to an element $\ne e$. It's clear that the equation $x^n = a$ cannot have any solution in $F$.