Let $k$ be a field, $G$ a group and $R$ a $k$-algebra (i.e. a ring $R$ with a homomorphism $i : k \rightarrow Z(R)$). The claim is that there is a natural bijection between the set of $k$-algebra homomorphisms $k[G] \rightarrow R$ and the set of group homomorphisms $G \rightarrow R^*$. I suppose this statement could be denoted

Hom$_{i[k]}(k[G], R) \approx$ Hom$(G,R^*)$

(as $k$-algebra homomorphisms are, I believe, simply those ring homomorphisms which are $i[k]-$linear). (Note also that $k$ itself is a $k$-algebra: as $Z(k) = k$, we can simply take $i$ to be the identity.)

Am I getting this right so far?

So given a $k$-algebra homomorphism $\phi : k[G] \rightarrow R$, I should be able to map it onto a group homomorphism $f : G \rightarrow R^*$, and back.

I would say we want to do something like "$f = \phi|_G$", as it seems clear that then $f[G] \subseteq R^*$.


First we define $\varphi: \text{Hom}_{i(k)}(k[G], R) \to \text{Hom}(G,R^*)$. Let $f:k[G] \to R$ be a $k$-algebra homomorphism. Define $\varphi(f)$ by $\varphi(f)(g) = f(g)$ for each $g \in G$.

I claim $\varphi(f)$ is a group homomorphism from $G$ to $R^*$. First note that, for any $g \in G$, $$\varphi(f)(g)\varphi(f)(g^{-1}) = f(g)f(g^{-1}) = f(gg^{-1}) = f(1) = 1.$$ Thus, for any $g \in G$, $\varphi(f)(g) \in R^*$. It follows quickly from the definition that $\varphi(f)$ is a group homomorphism.

To see that $\varphi$ is bijective, define $\psi:\text{Hom}(G,R^*) \to \text{Hom}_{i(k)}(k[G], R)$ as follows. Given $f:G \to R^*$ define $\psi(f)\left(\sum_{g \in G} c_g g\right) = \sum_{g \in G}i(c_g)f(g)$. The fact that $\psi(f)$ is a $k$-algebra homomorphism will follow from the fact that $f$ is a group homomorphism. You can also quickly check that $\varphi = \psi^{-1}$.


For simplicity I will treat $G$ as a subset of $k[G]$ and $k$ as a subset of $R$ (note that your $i$ homomorphism is always injective).

The multiplication in $k[G]$ is inherited from $G$. And so if $\phi:k[G]\to R$ is a $k$-algebra homomorphism then for any $g\in G$ we have

$$1=\phi(1)=\phi(gg^{-1})=\phi(g)\phi(g^{-1})$$ $$1=\phi(1)=\phi(g^{-1}g)=\phi(g^{-1})\phi(g)$$

Note that $\phi$ has to be unital. So $\phi(g)$ is invertible with the inverse $\phi(g^{-1})$. Meaning $\phi(G)\subseteq R^*$. And therefore $\phi$ induces a group homomorphism $G\to R^*$, simply the restriction of $\phi$.

On the other hand every element in $k[G]$ can be uniquely written as a finite sum $\sum_{g\in G}\lambda_g g$. So if $f:G\to R^*$ is a group homomorphism then $f$ extends to $F:k[G]\to R$ via

$$F\big(\sum \lambda g\big):=\sum\lambda f(g)$$

All is left is to prove all the missing axioms.


I can't say I like the notation $i[k]$ but never mind.

If you have a group homomorphism $f:G\to R^*$, define $\phi:k[G]\to R$ by $$\phi:\sum_{g\in G}a_g g\mapsto\sum_{g\in G}i(a_g)f(g).$$

  • $\begingroup$ Ah you prefer im $i$? That's fine too but I like to see what I'm putting in ;). $\endgroup$ – Jos van Nieuwman Jun 10 at 20:30
  • 1
    $\begingroup$ @JosvanNieuwman I'd prefer $\text{Hom}_{k-\text{alg}}(k[G],R)$. $\endgroup$ – Lord Shark the Unknown Jun 11 at 1:43

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