For a finite group and a field of characteristic p, the group algebra is a symmetric Frobenius algebra via the coefficient of the identity

On basis elements one has $⟨[g],[h]⟩=\{\begin{array}{ll}1,& h=g^{-1},\\ 0,& h\mathrm{\ne }{g}^{-1}\mathrm{.}\end{array}$ Thus the matrix of the form in the basis from [F1] is a permutation matrix, so the form is nondegenerate.

Associativity is immediate from the definition, because $⟨xy,z⟩$ and $⟨x,yz⟩$ are both the coefficient of $[e]$ in $xyz$. Symmetry holds because the coefficient of $[e]$ in $[g][h]$ is $1$ exactly when $h=g^{-1}$, which is equivalent to $g=h^{-1}$ and hence to the coefficient of $[e]$ in $[h][g]$ being $1$. Bilinearity is clear from the coefficient functional. Therefore the form is associative, symmetric, and nondegenerate.

A finite-dimensional algebra equipped with such a bilinear form is symmetric Frobenius. Hence $k[G]$ has the stated structure.