I want to show that 1 is an eigenvalue of the linear transformation $\phi_\alpha:K[G] \rightarrow K[G], \phi_\alpha(\beta)=\alpha\beta$ where $\alpha = <h>$ for some $h \in G$, $K$ is a field and $G$ is a group.
So far, I have shown that the map $G \rightarrow G$, $g \mapsto hg$ is a bijection, and so given $\beta = \sum_{g \in G} r_g<g> \in K[G]$ (i.e. $r_g \in K$ and only finitely many of them are non-zero) we have $$\alpha\beta = <h> \cdot \beta = \sum_{g \in G}r_g <hg> = \sum_{g \in G} r_{hg} <hg>$$
So if we were to have $\alpha\beta=\beta$ we need $r_{hg} = r_g$ for all $g \in G$. Now to be an eigenvector, $\beta$ must be non-zero. But this is where I get stuck: we must have $r_{h^i} = 0$ for all $i \in \mathbb{Z}$ as otherwise $ r_h^i = r_{h^{i+1}} = r_{h^{i+2}} =...$ are all non-zero, but by the definition of $K[G]$ there can only be a finite number of non-zero coefficients.
I'm also not sure where the fact that $K$ is a field rather than just a ring comes into play. Any tips?
I suspect that this question is really supposed to be about finite groups. Otherwise, the notion of "eigenvalue" gets a bit tricky.
If that is indeed the case, defining $r_g = 1$ for all $g$ works.