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.

  • 5
    $\begingroup$ Nice fake. You had me going there. $\endgroup$
    – shalop
    May 18, 2019 at 3:41
  • $\begingroup$ You might ask the same question about the free group on one generator. $\endgroup$
    – Somos
    May 18, 2019 at 3:49
  • $\begingroup$ How do you get $a$ from your three generators? $\endgroup$
    – J.-E. Pin
    May 18, 2019 at 7:29
  • $\begingroup$ @J.-E.Pin $a = ab \cdot a^{-1} \cdot ab^{-1}$. A nicer triple would be $ab, a^{-1}, b^{-1}$. $\endgroup$
    – Adayah
    May 18, 2019 at 7:37
  • $\begingroup$ @Somos The free group on one generator is a bit easier to figure out :) $\endgroup$ May 18, 2019 at 14:08

2 Answers 2


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.

  • $\begingroup$ Nice proof and generalization! I had briefly considered the abelianization, but not carefully enough to see this reasoning. $\endgroup$ May 18, 2019 at 4:07

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.

  • 1
    $\begingroup$ This does not seem to be an $(n + 1)$-tuple, but rather a $(2n)$-tuple. $\endgroup$ May 18, 2019 at 10:11
  • 6
    $\begingroup$ @Yanior Weg: the $(n+1)$-st element of the tuple is the product $a_1\cdots a_n$. $\endgroup$
    – Pascal W
    May 18, 2019 at 11:34

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