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Magnus showed that the multiplicative group generated by the $n$ formal power series $1+X_1, \dots,1+X_n$ in $n$ non-commuting variables $X_1,\dots,X_n$ (over an arbitrary field) is a free subgroup of rank $n$ in the multiplicative group of invertible formal power series.

Consider the multiplicative group generated by $1+X_1,1+X_2,1+X_1+X_2$, say, over the field of $2$ elements. Is this group the free group on $3$ generators?

More generally, given $k$ homogenous linear polynomials $l_1,\dots,l_k$ (over an arbitrary field) in $n$ non-commuting variables with $l_1,\dots,l_k$ pairwise linearly independent, is the multiplicative group generated by $1+l_1,\dots,1+l_k$ the free group on $k$ generators?

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1 Answer 1

Over $\mathbb Q$ the three elements $1+X_1, 1+X_2, 1+X_1+X_2$ generate a free subgroup of rank 3. Indeed, let $G$ be the group generated by these three elements. Consider the $2\times 2$-integer matrices $A=1+2E_{1,2}$, $B=1+2E_{2,1}$, where $E_{i,j}$ is the matrix with 1 in the $i,j$-place and 0 everywhere else. It is well known (Poincare and Sanov) that $A,B$ generate a free group of rank 2. Now consider three matrices $X=A^2B^2, Y=A^3B^{-3}, Z=A^5B^{-1}$. It is easy to check that $X+Y-1=Z$. Hence there exists a homomorphism from $G$ to the group generated by $X,Y,Z$ (it follows from the fact that there exists a ring homomorphism from the free non-commutative ring Ring$\langle X_1, X_2\rangle$ to the ring of matrices generated by $X-1, Y-1$ and under this homomorphism $1+X_1$ is mapped to $X$, $1+X_2$ is mapped to $Y$, $1+X_1+X_2$ is mapped to $Z$). But the subgroup of the free group Group$\langle A,B\rangle$ generated by $A^2B^2, A^3B^{-3}, A^5B^{-1}$ is free of rank 3 (apply Stallings foldings). Hence the group $G$ is also free of rank 3. In order to do it over the field ${\mathbb F}_2={\mathbb Z}/2{\mathbb Z}$, you need to consider matrices over the ring of polynomials ${\mathbb F}_2[t]$. The problem about arbitrary linear forms is more complicated but I think it can be treated the same way.

Edit: As Andreas correctly pointed out, there is no homomorphism from the ring of the formal power series to the ring of matrices over $\mathbb Q$. The argument above only shows that the semigroup generated by these elements is free.

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The argument is not complete, since one has to define a homomorphism from the ring of formal power series. However, one can show that there is no $k$-algebra homomorphism $\phi \colon k \langle X_1,X_2\rangle \to M_2 k$. –  Andreas Thom Oct 5 '10 at 5:38
    
@Andreas: OK, it was too late when I was writing it. I will fix that. It should not be a real problem. –  Mark Sapir Oct 5 '10 at 11:37
    
Nice idea! I knew however already that the corresponding semi-group is free (my proof uses a metric on the group of rational formal power series of the form $1+\dots$). –  Roland Bacher Oct 5 '10 at 17:14

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