Is the radical of an algebra a maximal ideal?

Question

Suppose we have a symmetric bilinear form $k[G] \times k[G] \rightarrow k$ for some finite group $G$ such that it is nondegenerate. My professor says that because it is nondegenerate we have that the Jacobson Radical of $k[G] = 0.$ I am confused as to how he concluded this. Because it is nondegenerate we have that the radical of $k[G], rad(k[G]) = 0.$ So is the radical a maximal ideal in $k[G]?$ If so, I understand. If not, I am not sure.

Answer 1

You must have misunderstood. Every group ring using a field and a finite group admits a nondegenerate bilinear form like this, since they are Frobenius, but many of them have nonzero Jacobson radicals.

For example example, if $F$ is the field of $p$ elements and $G$ is a finite $p$ group, $F[G]$ is local, so the Jacobson radical is maximal and nonzero.

Are you sure the instructor want referring to the radical of the bilinear form which would definitely be zero because that is the definition of nondegeneracy?