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Given a discrete group $\Gamma$ generated by $S$ let $K[S]$ denote the subring of the group-ring $K[\Gamma]$ generated by $S$ (over a commutative ring $K$). The ring $K[S]$ is thus a quotient of the free algebra generated by non-commutative variables indexed by $S$. Are there examples where the free monoid $S^*$ injects into $\Gamma$ but $K[S]$ is not free? (I guess the answer should be yes).

On the other hand, there should also be examples where $K[S]$ is the free algebra of non-commutative polynomials over $S$, although $\Gamma$ is not free (this implies of course that $S^*$ injects into $\Gamma$).

Remark: The subring $K[S]$ is of course isomorphic to $K[\Gamma]$ if $\Gamma$ is a torsion-group (or more generally, if all elements of $S$ represent torsion elements of $\Gamma$).

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    $\begingroup$ For your second question: certainly one can get free non-abelian subsemigroups $S$ inside certain solvable groups, and I suspect (though I don't have a proof) that in those setting $K[S]$ would be isomorphic to the free algebra generated by elements of $S$ $\endgroup$
    – Yemon Choi
    Commented Jan 7, 2015 at 19:04
  • $\begingroup$ Non-free groups generated by $S$ such that $S^*$ injects into them are easy to construct: most one-relator groups should have this property if the relation is a bit complicated and involves at least one inverse of a generator in $S$. $\endgroup$
    – Roland Bacher
    Commented Jan 7, 2015 at 19:10
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    $\begingroup$ Maybe I'm missing something, but it seems to me the answer to your first question is an obvious no. Isn't $K[S]$ just the monoid ring of the submonoid generated by $S$? $\endgroup$
    – Eric Wofsey
    Commented Jan 7, 2015 at 19:14
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    $\begingroup$ As Eric says the elements of S are linearly independent so you get the monoid ring. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 7, 2015 at 20:32
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    $\begingroup$ It's not obvious from the perspective of the universal property (or equivalently, generators and relations) of a monoid ring, but it is obvious from its explicit description as the ring of formal linear combinations (the subring of the set of formal linear combinations of $\Gamma$ generated by $S$ is just the set of formal linear combination of products of elements of $S$). $\endgroup$
    – Eric Wofsey
    Commented Jan 7, 2015 at 22:37

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