# Is the free group on two generators generated by two elements as a monoid?

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.

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

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