Does existence of free submonoid implies free subgroup Suppose that we have group $G$ that, regarded as a monoid, has a submonoid $M$, which is free with $r$ generators. Does it imply that is has a subgroup $H$, free and with $r$ generators?
For $r=1$ it is clear, and the case $r=2$ will be, I think, the only interesting one.
The first approach might be that $H=\langle M \rangle$ works, but that's not true: for
$$ A=\begin{pmatrix}1&0\\1&1 \end{pmatrix}, \textrm{ and }B=\begin{pmatrix}1&1\\0&1 \end{pmatrix} $$
it's not hard to check that they generate a free submonoid, but we have $(A B^{-1} A)^4 = I$. However, $\langle A^2,B^2 \rangle$ is free, so it isn't a counterexample.

No. In fact, if $G$ is a finitely generated solvable group then either $G$ is virtually nilpotent (contains a nilpotent subgroup of finite index) or $G$ contains a free submonoid on two generators. See
Rosenblatt, Joseph Max, Invariant measures and growth conditions, Trans. Am. Math. Soc. 193, 33-53 (1974). ZBL0246.28017.
On the other hand, a solvable group cannot contain nonabelian free subgroups.
One of the simplest examples of a solvable group which is not virtually nilpotent is the Baumslag-Solitar group $G=BS(2,1)$ with the presentation
$$
<a,b| aba^{-1}=b^2>.
$$
To see that this group is not virtually nilpotent one constructs a (faithful) representation of $G$ to the group of affine transformations of the real line:
$$
b(t)=t+1, a(t)=2t. 
$$
Specifically, in the case of $G$ one can take $x=a^{-2}$ and $y=bxb^{-1}$ and use ping-pong on the real line to check that $x, y$ generate a free submonoid in $G$.
