Given that ${\bf r}_a, {\bf r}_b, \dots$ are irreducible representations of a finite group $G$, the k-fold Kronecker product of $\bf{r}_a$ on a symmetric subspace ${\rm Sym}^k(V)$ is expanded in terms of $\bf{r}_b$. Here, $V$ is an associated space with $\bf{r}_a$. Then the generating function of coefficients of the expansion is given by the Molien's formula \begin{align} M({\bf r}_b, {\bf r}_a, t) = \frac{1}{|G|} \sum_{C_i} |C_i| \frac{\bar{\chi}(g_i)}{{\rm det}(1 - t A_i^{{\bf r}_a})}, \end{align} with $C_i$ being a conjugacy class of $G$ and $A_i$ is an arbitrary element of $C_i$.
Then, if we calculate for some finite groups like $S_3$ or $A_4$, we notice that for ${\bf r}_b = {\bf 1}$ (${\chi}=1$), this formula reduces to
\begin{align} M({\bf 1}, {\bf r}_a, t) = \frac{1 + \sum_{k} d_k t^k}{(1-t^{a_1})^{n_1}(1-t^{a_2})^{n_2} \dots} \end{align} where $a, d, n$ are positive integers.
However, how can I prove this result?
I think the way of expanding the determinant in the denominator is important, but I have not figured out so far.
In the Mukai's textbook, we see that the Hilbert series can always expressed in the above form. However, I'd like to know how we can derive the above form from the Molien's formula.
P.S. I tried to keep this post short because the previous ones were long, but I apologize if it is difficult to understand.