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

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 $$\phi:\sum_{g\in G}a_g g\mapsto\sum_{g\in G}i(a_g)f(g).$$

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