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I was musing about this today and couldn't come up with an answer. Obviously it can be generated as a monoid by the four elements $a$, $b$, $a^{-1}$, and $b^{-1}$. After some playing around I was able to come up with three elements that generate it as a monoid: $ab$, $ab^{-1}$, and $a^{-1}$.
But I haven't been able to come up with two generators, nor an argument as to why that should be impossible.
The free group on two generators maps onto $\Bbb Z^2$. (This is its Abelianisation). If it were generated by two elements as a monoid, then so would $\Bbb Z^2$. But that's not so. If you have two elements $a$, $b$ of $\Bbb Z^2$ generating it as a monoid, they certainly generate it as an Abelian group, so they must be linearly independent as vectors. But in that case $-a-b$ is not in the submonoid of $\Bbb Z^2$ generated by $a$ and $b$.
Likewise, a free group on $n$ generators cannot be generated as a monoid by $n$ elements.
And, further generalizing the rank 2 case, the $(n+1)$-tuple $(a_1^{-1}, a_2^{-1},\ldots,a_n^{-1},a_1a_2\cdots a_n)$ generates all of the rank $n$ free group as a monoid.
