Let $G=\frac{F}{R}$ be the presentation of a group $G$. Clearly here $F$ is free group. Can we obtain a free group $\mathbb{F}$ of the group $F$. Is $\mathbb{F}\cong F?$ What will be the presentation of $\mathbb{F}.$ actually I am studying the presentation of direct product of groups Group Presentation of the Direct Product.. Here free product of two groups is used before, that's why I am thinking about this.
In simple words my question is to obtain a free presentation of a free group $F$.
Yes, you can make a free group $\mathbb{F}$ from any set, including the underlying set of a (possibly free) group $F$.
That said, $\mathbb{F} \not\cong F$. For example, suppose $F$ was the free group on a set $\{a, b\}$, then $a$ and $a^{-1}$ are distinct elements of $F$, which happen to multiply to give the identity. But, when generating $\Bbb{F}$, each distinct element of $F$ becomes an unrelated generator of $\Bbb{F}$. So, in $\Bbb{F}$, $a$ and $a^{-1}$ are interpreted as strings of length $1$, with the two different symbols $a$ and $a^{-1}$. Making the free group $\Bbb{F}$ ignores the former relationship between these elements of $F$, and so $aa^{-1}$ becomes an irreducible word in $\Bbb{F}$.
In a similar manner, $a^2$ and $a$ are also distinct elements of $F$, and as such, $a^2a$, $aa^2$, and $a^3$ are three distinct elements of $\Bbb{F}$.