Fréchet-Urysohn spaces in free topological groups

Let $F(X)$ and $A(X)$ be respectively the free topological group and the free Abelian topological group on a Tychonoff space $X$. For every natural number $n$ we denote by $F_n(X)$ ($A_n(X)$) the subset of $F(X)$($A(X)$) consisting of all words of reduced length $\leq n$. It is well known that if a space $X$ is not discrete, then neither $F(X)$ nor $A(X)$ is Fréchet-Urysohn, and hence first countable. On the other hand, it is seen that both $F_2(X)$ and $A_2(X)$ are Fréchet-Urysohn for a paracompact Fréchet-Urysohn space $X$. In this paper, we prove first that for a metrizable space $X$, $F_3(X)$ ($A_3(X)$) is Fréchet-Urysohn if and only if the set of all non-isolated points of $X$ is compact and $F_5(X)$ is Fréchet-Urysohn if and only if $X$ is compact or discrete. As applications, we characterize the metrizable space $X$ such that $A_n(X)$ is Fréchet-Urysohn for each $n\geq3$ and $F_n(X)$ is Fréchet-Urysohn for each $n\geq3$ except for $n=4$. In addition, however, there is a first countable, and hence Fréchet-Urysohn subspace $Y$ of $F(X)$ ($A(X)$) which is not contained in any $F_n(X)$ ($A_n(X)$). We shall show that if such a space $Y$is first countable, then it has a special form in $F(X)$ ($A(X)$). On the other hand, we give an example showing that if the space $Y$ is Fréchet-Urysohn, then it need not have the form.