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?