I have been thinking about this for a while, but now got to the point where I got stuck. I don't know if it might be considered as a research level question, but I would be very happy if somebody knew of any direction on how to move on:


Define the functions $t\_k(x) = \sum\_{n=0}^{\infty}{\frac{x^{3n+k}}{(3n+k)!}}$ for $k=0,1,2$. The functions then satisfy:
$
\begin{pmatrix}\exp(x) \\ \exp(\omega x) \\ \exp(\omega^2 x)\end{pmatrix} = 
\begin{pmatrix}1 & 1 & 1\\1 & \omega & \omega^2 \\1 & \omega^2 & \omega\end{pmatrix} \cdot \begin{pmatrix}t\_0(x) \\ t\_1(x) \\ t\_2(x)\end{pmatrix}$
where $\omega = \exp(2\pi i / 3)$. 


As might be known, the matrix in the equation is the character table of the cyclic group $C\_3$ and also a Vandermonde matrix. Using this last matrix equation one can prove the following addition theorem:


$\begin{pmatrix}t\_0(x+y) \\ t\_1(x+y) \\ t\_2(x+y)\end{pmatrix} = \begin{pmatrix}t\_0(x) & t\_2(x) & t\_1(x) \\ t\_1(x) & t\_0(x) & t\_2(x)\\ t\_2(x) & t\_1(x) & t\_0(x)\end{pmatrix} \cdot \begin{pmatrix}t\_0(y)\\t\_1(y) \\ t\_2(y)\end{pmatrix}$


As might be known the matrix in the last equation is a circulant matrix, which is also the group matrix $(x\_{gh^{-1}})\_{g,h\in G}$ as defined by Dedekind.
Now it is clear how to do this construction for every cyclic group $C\_n$, which we just did for the cyclic group $C\_3$, namely define the functions $f\_g(x)$, $g\in C\_n$using the character table of $C\_n$. 


If one takes the determinant of the last matrix, one can show (using the theory of circular matrix) that this is equal to one:
$t\_0(x)^3+t\_1(x) ^3+t\_2(x)^3-3t\_0(x)t\_1(x)t\_2(x) = 1$ for all $x$.
I am able to prove that the determinant is equal to $1$ for every cyclic group. Notice also that the determinant is the group determinant of $C\_3$ as defined by Frobenius and Dedekind .


(1) Is it true, that for a general cyclic group the functions defined fullfill the "addition theorem" which is given by the Dedekind group matrix?


(2) How does one proceed with arbitrary finite groups (for example the Klein four group and the symmetric group on three elements)?


(3) Does somebody know of any other context, where the specific functions $t\_k$ $k=0,1,2$ appear?



Connection between cyclic group and exponential function
The natural generalization to a product of cyclic group is having generating functions on several variables. Take the Klein four group as an example. Define
$$t\_{k\_1,k\_2}(x\_1,x\_2) = \sum\_{n\_1,n\_2=0}^\infty \frac{x\_1^{2n\_1+k\_1} x\_2^{2n\_2+k\_2}}{(2n\_1+k\_1)!(2n\_2+k\_2)!}.$$
These functions satisfy
$$
\begin{pmatrix}
\exp(x\_1+x\_2) \\ \exp(x\_1-x\_2) \\ \exp(-x\_1+x\_2) \\ \exp(-x\_1-x\_2)
\end{pmatrix} =
\begin{pmatrix}
1 & 1 & 1 & 1 \\
1 & -1 & 1 & -1 \\
1 & 1 & -1 & -1 \\
1 & -1 & -1 & 1 
\end{pmatrix}
\begin{pmatrix}
t\_{0,0}(x\_1,x\_2) \\
t\_{0,1}(x\_1,x\_2) \\
t\_{1,0}(x\_1,x\_2) \\
t\_{1,1}(x\_1,x\_2)
\end{pmatrix}
$$
This implies the following addition theorem, where $x,y$ are vectors of length 2:
$$
\begin{pmatrix}
t\_{0,0}(x+y) \\
t\_{0,1}(x+y) \\
t\_{1,0}(x+y) \\
t\_{1,1}(x+y)
\end{pmatrix} =
\begin{pmatrix}
t\_{0,0}(x) & t\_{0,1}(x) & t\_{1,0}(x) & t\_{1,1}(x) \\
t\_{0,1}(x) & t\_{0,0}(x) & t\_{1,1}(x) & t\_{1,0}(x) \\
t\_{1,0}(x) & t\_{1,1}(x) & t\_{0,0}(x) & t\_{0,1}(x) \\
t\_{1,1}(x) & t\_{1,0}(x) & t\_{0,1}(x) & t\_{0,0}(x)
\end{pmatrix}
\begin{pmatrix}
t\_{0,0}(y) \\
t\_{0,1}(y) \\
t\_{1,0}(y) \\
t\_{1,1}(y)
\end{pmatrix}
$$
(I haven't verified the last claim, but it should follow in the same way as your corresponding claim.)


Extending this to the non-abelian case will probably be more interesting.


