Let $k$ be a field, $G = \langle S \rangle/N(R)$ be a group specified by a set of generators $S$ and a set of relations $R$ (the brackets denote the free group, and $N(R)$ means the conjugate closure of $R$). Let $kG$ be the group algebra of $G$. It is an associative, unital, non commutative $k$-algebra; as a vector space it is the free vector space on $G.$

It seems natural to me that one should be able to identify $kG$ with a suitable quotient of the free algebra on $S$, namely $A := \langle S \rangle_k/I(1-R),$ where $\langle \cdot \rangle_k$ means the free (associative, unital, non commutative) algebra on the set $S$ and $I(1-R)$ is the ideal generated by $1-R.$

Hence I thought it would make sense for $A$ to satisfy the universal property of the group algebra of $G$, namely that for any other algebra $B$ and monoid homomorphism $f:G \to B,$ there should exist an algebra homomorphism $\tilde{f}:A \to B$ such that $\tilde{f} \circ \iota = f.$ Here $\iota$ should be a natural monoid homomorphism of $G$ into $A.$

However, the "obvious" map $\iota(gN(R)) = g + I(1-R)$ isn't even well-defined. What am I doing wrong here? I am obviously quite confused because I have a hard time even formulating a proper question. Is the result (i.e. that $kG \cong A$ as algebras) even true?

Thanks for your time, and sorry for being all over the place.

EDIT: As Rob pointed out in the comments, this fails trivially when inverse elements of $S$ appear in $G$ but not in $A.$ What can we say if we add the requirement that $G$ be finite, or more generally that any element of $G$ can be represented as a word on the alphabet $S$?

One could also try to work on the algebra side and add formal inverses to the generators of $A.$ Thoughts?

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    $\begingroup$ Take $S = \{x\}$ and $R$ empty, then $A$ is the polynomial ring $k[x]$, but $kG$ is the ring $k[x, x^{-1}]$ of Laurent polynomials, which isn't a quotient of $k[x]$. $\endgroup$ – Rob Arthan Feb 4 '15 at 17:57
  • $\begingroup$ @RobArthan Thank you. Would you mind taking a look at my updated question? $\endgroup$ – anon Feb 6 '15 at 15:34
  • $\begingroup$ The requirements that you have added now imply that $kG$ will be a homomorphic image of $A$ under the natural homomorphism. Now you have to prove that that homomorphism is one-to-one. $\endgroup$ – Rob Arthan Feb 9 '15 at 23:53

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