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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 $F$.

MANI
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    What does “free group of the free group” mean? You mean, the free group on the underlying set of $F$? – Arturo Magidin Nov 15 '20 at 05:41
  • yes sir, here $F$ is the underlying set. – MANI Nov 15 '20 at 05:42
  • @Shaun: so, you are interpreting the question “can $G$ be free”? If so, of course you can obtain any free group of rank less than or equal to that of $F$. – Arturo Magidin Nov 15 '20 at 05:42
  • There is a free group on any set, so of course you can make a free group on the underlying set of $F$. If $F$ were of rank $2$, say, generated by $x$ and $y$, then the free group on its underlying set would have free generators $x$, and $y$, and $x^2$, and $y^2$, and $xy$, and $e_F$, and so on. If $F$ is free on $X$, then it has cardinality $|X|\aleph_0$, so the free group on its underlying set has cardinality $|X|\aleph_0$. In particular, if $X$ is finite then this is not isomorphic to $F$, and if $X$ is infinite then it is isomorphic to $F$. But your phrasing is extremely confusing. – Arturo Magidin Nov 15 '20 at 05:44
  • Indeed, @ArturoMagidin. I didn't spot the glaring ambiguity. Sorry. – Shaun Nov 15 '20 at 05:45
  • In simple words my question is to obtain a free presentation of a free group $F$. – MANI Nov 15 '20 at 05:45
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    @MANI: That is neither what you wrote in the question, nor what you wrote in the comment. If you have a “simple way” of asking, why on Earth would you not ask it that way in the first place?! Yes, of course you can find a free presentation for a free group. The free group $F$ has free presentation $F/R$, where $F=F$ and $R={e}$. $R$ is the free group on the empty set. And you can also present it nontrivially by taking a free group of greater rank than $F$, and $R$ the normal subgroup generated by the “extra” generators. – Arturo Magidin Nov 15 '20 at 05:47
  • @ArturoMagidin sorry for the confusion sir and thanks for your help. Now i got my answer. – MANI Nov 15 '20 at 05:52

1 Answers1

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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 $\Bbb{F}$.

user847970
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    The free group on the underlying set of $F_{\omega}$ is free on a countable generating set, and hence isomorphic to $F_{\omega}$. It is false that you will always get $\mathbb{F}\not\cong F$ if that is the construction being used. It will be non-isomorphic if and only if $F$ is of finite rank. – Arturo Magidin Nov 15 '20 at 05:51