It is perhaps around 1950, proved by Higman and Neumann that the Frattini subgroup of a Free group is trivial.
I want to know, now, is there elementary proof of this?
If yes, please write short sketch or give directions for the proof; I would like to complete the proof.
I will give the idea of proof,
Let $<a_1,...,a_n>=F$ where $F$ is a free group with generaters $n$ elements.
Assume that $\Phi(G)\neq 1.$
I will show the very basic case, If the word $a_1a_2\in \Phi(G)$, then notice that
$$<a_1,a_1a_2,a_3,a_4...,a_n>=F$$
As $a_1a_2$ is a nongenerotor,
$$<a_1,a_3,a_4..,a_n>=F$$
which is a contradiction.
Hence, the idea is that it does not contain any non- trivial words with this methods.