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I cannot figure out how to prove that the submonoid of $F_n$ generated by elements $\alpha_1,\dots , \alpha_n$ will never be the whole group.

It clearly is possible to generate the free group on $a_1, \dots , a_n$ as a monoid with $n+1$ generators, e.g. with $a_1, \dots, a_n , a_1^{-1}a_2^{-1} \dots a_n^{-1}$ and it is intuitively obvious that this should be minimal, but I can't prove it.

This is easy to prove in free Abelian groups as the generators must be linearly independent for even the group they generate to be of rank $n$, but being linearly independent means they generate the free commutative monoid, which is not isomorphic to the free Abelian group.

It seems like the problem requires some analogue of linear independence and rank for non-Abelian groups.

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  • $\begingroup$ What definition of free group do you use? Is it a set of strings with concatenation as group operation, or some set with a universal property, or something else? $\endgroup$
    – student91
    Commented Jun 6 at 7:39
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    $\begingroup$ Can't you just abelianize? $\endgroup$
    – Trebor
    Commented Jun 6 at 7:44
  • $\begingroup$ @Trebor Oh. I guess that solves it. $\endgroup$
    – Zoe Allen
    Commented Jun 6 at 7:46

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