Connection between cyclic group and exponential function 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?
 A: 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.
A: $\DeclareMathOperator\Tr{Tr}$This answer is maybe five years too late, but better late than never:
It seems that the "right" generalisation for finite groups is given by:
For a finite group $G$ we might search for / have:
$$t_g(x+y) = \sum_{h \in G} t_{gh^{-1}}(x) t_h(y).$$
We have the Fourier transform of $t_x:G \rightarrow \mathbf{C}$, $g \mapsto t_g(x)$ at the representation $\rho$ as:
$$\widehat{t_x}(\rho) := \sum_{g \in G} t_g(x) \cdot \rho(g).$$
The convolution is given by and satisfies:
$$(t_x \ast t_y)(g) = \sum_{h \in G} t_{gh^{-1}}(x) t_h(y) =^{!} t_g(x+y) = t_{x+y}(g). $$
By the formula above of the Fourier transform of a convolution, we have:
$$
\widehat{t_{x+y}}(\rho) = \widehat{t_x \ast t_y}(\rho) = \widehat{t_x}(\rho)\widehat{t_y}(\rho).
$$
The Plancherel formula translates in our case to:
$$\sum_{g \in G} t_{g^{-1}}(x) t_g(y) = \frac{1}{|G|}\sum_{ \rho \text{ irred. }} d_{\rho} \Tr \left ( \widehat{t_x}(\rho) \widehat{t_y}(\rho) \right)\\
= \frac{1}{|G|}\sum_{ \rho \text{ irred. }} d_{\rho} \Tr \left ( \widehat{t_{x+y}}(\rho) \right).
$$
This does not answer though how to construct those functions $t_g$ indexed by $G$ which satisfy the convolution identity above.
Edit:
Let me give an answer for the "general case":
Let $G=\langle S\rangle = \langle s_1,\dotsc,s_r\rangle$ be a finite group generated by $S$ with $1 \notin S $. For each $s_i$ we introduce a variable $x_{s_i}$ and the vector $x := (x_{s_i})_{s_i \in S}$.
Let us define the Fourier transform:
$$\widehat{t_{x}}(\rho)  := \widehat{t^{(S)}_{x}}(\rho) := \mathbf{1}_{d_{\rho}} \exp( \frac{1}{d_{\rho}} \sum_{s \in S}\chi_{\rho}(s) x_s )$$
where $\chi_{\rho}$ is the character of $\rho$ and $\mathbf{1}_{d_{\rho}}$ is the identity matrix of the same dimension as $\rho$ and $d_{\rho}$ is the dimension of $\rho$.
Then since $\exp(a+b) = \exp(a)\exp(b)$ we have:
$$\widehat{t^{(S)}_{x+y}}(\rho) = \widehat{t^{(S)}_{x}}(\rho) \widehat{t^{(S)}_{y}}(\rho). $$
But then we get (abbreviate $t := t^{(S)}$) :
$$\widehat{t_{x+y}}(\rho) = \widehat{t_x }(\rho)\widehat{t_y }(\rho) = \widehat{t_x \ast t_y}(\rho).$$
And applying the inverse Fourier transform to this last equation yields:
$$t_{g}(x+y) = t_{x+y}(g) = (t_x \ast t_y)(g) = \sum_{h\in G} t_{gh^{-1}}(x)t_h(y)$$
and the addition formula / convolution identity is proved for arbitrary finite groups. :-) Finally after five years! :-)
The inverse Fourier transform is given by:
$$t_g(x) = \frac{1}{\lvert G\rvert}\sum_{ \rho \text{ irred. }} d_{\rho} \Tr \left ( \rho(g^{-1})\widehat{t_{x}}(\rho) \right)$$
and plugging in $\widehat{t_{x}}(\rho) = \mathbf{1}_{\rho}\exp(\sum_{s \in S}\chi_{\rho}(s) x_s)$ we get
$$t_g(x) = \frac{1}{|G|}\sum_{ \rho \text{ irred. }} d_{\rho} \Tr \left ( \rho(g^{-1})\exp( \frac{1}{d_{\rho}} \sum_{s \in S}\chi_{\rho}(s) x_s)\right).$$
Second edit:
The determinant of the group matrix evaluated at the defined functions is also, equal to $1$, because:
If $T_G := (t_{gh^{-1}})_{g,h \in G}$ is the group matrix defined for the functions defined above:
$$\widehat{t_{x}}(\rho) := \mathbf{1}_{d_{\rho}} \exp( \frac{1}{d_{\rho}}  \sum_{s \in S}\chi_{\rho}(s) x_s )$$
where $S$ (with $1 \notin S$) generates the finite group $\left< S \right > = G$.
From this we get, since we know by Frobenius, the factorization of the group determinant :
$$\det(T_G) = \prod_{\rho \text{ irred.}} \det( \sum_{g \in G} t_g(x) \rho(g) )^{d_{\rho}} = \prod_{\rho \text{ irred.}} \det( \widehat{t_x}(\rho))^{d_{\rho}} = \prod_{\rho \text{ irred.}} \det( \mathbf{1}_{\rho} \exp \left ( \frac{1}{d_{\rho}}  \sum_{s \in S} \chi_{\rho}(s) x_s \right ) )^{d_{\rho}} $$
$$= \prod_{\rho \text{ irred. }} \exp( \sum_{s \in S} \chi_{\rho}(s) x_s)^{\deg(\rho)}$$
$$ =\exp\left( \sum_{\rho \text{ irred.}} \deg(\rho) \sum_{s\in S} \chi_{\rho}(s) x_s \right)$$
and which is equal to:
$$=\exp(\sum_{s \in S} x_s \cdot \left ( \sum_{\rho} \deg(\rho) \chi_{\rho}(s) \right ) )  = \exp(0)=1$$
where we have $\left ( \sum_{\rho} \deg(\rho) \chi_{\rho}(s) \right )=0$ for all $s\neq 1$, because the regular character equals $0$ for all $s \neq 1$.
Here are the $6$ functions for generating set $S = \{(1,2),(2,3)\}$.
Addition theorem satisfied :  0 == 0
Permutation $g =  () $
$$t_g(x_i)  =  \frac{1}{6} \, e^{\left(x_{0} + x_{1}\right)} + \frac{1}{6} \, e^{\left(-x_{0} - x_{1}\right)} + \frac{2}{3} $$
Addition theorem satisfied :  0 == 0
Permutation $g =  (1,3,2) $
$$t_g(x_i)  =  \frac{1}{6} \, e^{\left(x_{0} + x_{1}\right)} + \frac{1}{6} \, e^{\left(-x_{0} - x_{1}\right)} - \frac{1}{3} $$
Addition theorem satisfied :  0 == 0
Permutation $g =  (1,2,3) $
$$t_g(x_i)  =  \frac{1}{6} \, e^{\left(x_{0} + x_{1}\right)} + \frac{1}{6} \, e^{\left(-x_{0} - x_{1}\right)} - \frac{1}{3} $$
Addition theorem satisfied :  0 == 0
Permutation $g =  (2,3) $
$$t_g(x_i)  =  \frac{1}{6} \, e^{\left(x_{0} + x_{1}\right)} - \frac{1}{6} \, e^{\left(-x_{0} - x_{1}\right)} $$
Addition theorem satisfied :  0 == 0
Permutation $g =  (1,3) $
$$t_g(x_i)  =  \frac{1}{6} \, e^{\left(x_{0} + x_{1}\right)} - \frac{1}{6} \, e^{\left(-x_{0} - x_{1}\right)} $$
Addition theorem satisfied :  0 == 0
Permutation $g =  (1,2) $
$$t_g(x_i)  =  \frac{1}{6} \, e^{\left(x_{0} + x_{1}\right)} - \frac{1}{6} \, e^{\left(-x_{0} - x_{1}\right)} $$
Here is a small sanity check in Sagemath for $S_3$:
example Sagemath computation
A: One can start with finding $t_k$ explicitly. Note that $t_0$ satisfies the differential equation $t_0^{'''}=t_0$. Thus $t_0(x)=Ae^{x}+Be^{\omega x}+Ce^{\omega^2 x}$, with  $A$, $B$, $C$ to be found using initial values:
$t_0^{(k)}(0)$, $k=0,1,2$. In particular
\begin{aligned}
A+& B+C&=1\\
A+\omega &B+\omega^2 C&=0\\
A+\omega^2 &B+\omega C&=0
\end{aligned}
gives $$A=B=C=\frac{1}{3}, \qquad  t_0(x)=\frac{1}{3}\left(e^{x}+e^{\omega x}+e^{\omega^2 x}\right),$$
something that also follows from the 1st displayed equation in the question.
One also sees that $f'_1=f_0$, $f'_2=f_1$, so
$$f_1(x)=\frac{1}{3}\left(e^x+\frac{e^{\omega x}}{\omega}+\frac{e^{\omega^2 x}}{\omega^2}\right), \qquad
f_2(x)=\frac{1}{3}\left(e^x+\frac{e^{\omega x}}{\omega^2}+\frac{e^{\omega^2 x}}{\omega}\right).$$
A: Just to explain why the determinant equals $1$ (too long for a comment). Let me define
$$e_k(x)=\exp(\omega^kx),$$
so that
$$t_0=\frac13(e_0+e_1+e_2),\quad t_1=\frac13(e_0+\omega^2e_1+\omega e_2),\quad t_2=\frac13(e_0+\omega e_1+\omega^2e_2).$$
Then
$${\rm Circ}(t_0,t_1,t_2)=\frac13\begin{pmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{pmatrix}{\rm diag}(e_0,e_1,e_3)\begin{pmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega \\ 1 & \omega & \omega^2 \end{pmatrix}.$$
Hence the determinant equals
$$-\frac1{27}(3\omega(1-\omega))^2e_0e_1e_2=e_0e_1e_2\equiv1.$$
Remark. The identity $T(x+y)=A(x)T(y)$ mentionned in the Question, where
$$T=\begin{pmatrix} t_0 \\ \vdots \\ t_{n-1} \end{pmatrix},\qquad A=\begin{pmatrix} t_0 & t_{n-1} & \ldots & t_1 \\ t_1 & t_0 & & t_2 \\ \vdots & \ddots & \ddots & \vdots \\ t_{n-1} & \ldots & t_1 & t_0 \end{pmatrix},$$
implies actually (because $A(x)$ is circulant and its first column is $T(x)$) the identity $A(x+y)=A(x)A(y)$. This tells us that $A(x)=\exp(xA'(0))$. This is actually obvious, where
$$A'(0)=\begin{pmatrix} 0 & \ldots & \ldots & 0 & 1 \\ 1 & \ddots & & & 0 \\ 0 & \ddots & \ddots &  & \vdots \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \ldots & 0 & 1 & 0 \end{pmatrix}.$$
A: First, let me apologize for posting this second answer. But it gives a definitive answer; at least I think so.
The problem is to find functions $t_g(x)$ for $g\in G$ (a finite group) and $x\in{\mathbb R}$ (or ${\mathbb C}$), such that
$$\sum_{h\in G}t_{gh^{-1}}(x)t_h(y)=t_g(x+y).$$
What seems reasonable is to impose the restriction that each $t_\bullet(x)$ be a class function (it depends only upon the conjugacy class of $g$).
As noted by stackExchangeUser, by Fourier transform, this amounts to finding functions $\widehat{t_x}(\rho)$ satisfying
$$\widehat{t_x}(\rho)\widehat{t_y}(\rho)=\widehat{t_{x+y}}(\rho),$$
where now $\rho$ runs over complex representations of $G$. Actually, it is enough to verify the identity above for irreducible representations.
What remained un-noticed is that the latter identity can be rewritten
$$M_\rho(x+y)=M_\rho(x)M_\rho(y),$$
where $M_\rho(x)$ is another way to write $\widehat{t_x}(\rho)$. Now, we see that $M_\rho$ is nothing but an exponential:
$$M_\rho(x)=\exp(xA_\rho),\qquad A_\rho:=M_\rho'(0)=\sum_{g\in G}t_g'(0)\rho(g).$$
This shows how to find the most general solution $T=(t_g)_{g\in G}$ of the problem: choose any class function $a:G\to{\mathbb C}$. Consider the regular representation $R$, acting over ${\mathbb C}[G]=:V$. Form $A:=\sum_{g\in G}a_gR(g)\in{\rm End}(V)$. Then define $M(x)=\exp(xA)$. Because the image $R(G)$ is closed under composition, $M(x)$ belongs to the linear subspace of ${\rm End}(V)$ spanned by $R(G)$. It even belongs to the subspace spanned by the class-sums $R_s:=\sum_{h\in c}R(h)$ ($c$ are conjugacy classes). Therefore $M(x)$ decomposes as
$$M(x)=\sum_{\rm classes}m_s(x)R_s=\sum_{g\in G}t_g(x)R(g),$$
where $t_\bullet(x)$ are class functions.
The identity $M(x+y)=M(x)M(y)$ writes $\sum_{g\in G}b_g(x)R(g)=0$ where $b_g(x):=t_g(x+y)-\sum_{h\in G}t_{gh^{-1}}(x)t_h(y)$. The latter identity implies classically $b\equiv0$, thus the functions $t_g$ solve the problem.
About the determinant. The matrix of $M(x)$ in the canonical basis of ${\mathbb C}[G]$ is the ``circulant''
$${\rm Circ}(T)=(t_{gh^{-1}})_{g,h\in G}.$$
We have
$$\det{\rm Circ}(T)=\det M(x)=\exp(x{\rm Tr}A)=\exp(|G|xa_e).$$
In particular $\det{\rm Circ}(T)\equiv1$ whenever $a_0=0$, which happens in the example given in OP's question.
Edit. As it has been known after Frobenius, $\det{\rm Circ}(T)$ factorizes as
$$\prod_{\chi{\rm irrep}} \Delta_\chi^{{\rm deg}\chi},$$
where $\Delta_\chi$ is a polynomial in $T$, homogeneous of degre ${\rm deg}\chi$. Each factor is the determinant of the restriction of ${\rm Circ}(T)$ on the $\chi$-component of $V$. Again this restriction is the exponential of $x\sum_Ga_gR(g)_\chi$, the linear combination of the corresponding restrictions of $R$. Thus
$$\Delta_\chi^{{\rm deg}\chi}=\det\left(x\sum_Ga_gR(g)_\chi\right)=\exp x\sum_Ga_g({\rm deg}\chi)\chi(g).$$
This yields
$$\Delta_\chi=\exp(x|G|\langle a,\chi\rangle).$$
A: The answers given are very good, yet I want to address the question of partial differential equations:
Let $G$ be a finite abelian group (written multiplicatively because the previous formulas are written this way). Then we have $d_{\rho} = 1$ and $\chi_{\rho}(g) = \rho(g)$ for every irreducible representaion $\rho$ of $G$.
We get:
$$\frac{\partial t_g(x)}{\partial x_{s_0}} = \frac{1}{|G|} \sum_{\rho \text{ irred. }} d_{\rho} \chi_{\rho}(g^{-1})\frac{1}{d_{\rho}} \chi_{\rho}(s_0) \exp \left( \frac{1}{d_{\rho}} \sum_{s \in S} \chi_{\rho}(s) x_s \right )$$
$$=\frac{1}{|G|} \sum_{\rho \text{ irred. }} \chi_{\rho}((s_0^{-1}g)^{-1})\exp \left( \sum_{s \in S} \chi_{\rho}(s) x_s \right )$$
$$=t_{s_0^{-1}g}(x)$$
Hence for a $h = s_1^{e_1} \cdots s_r^{e_r} \in G$ with $e = e_1 +  \cdots + e_r$ follows:
$$\frac{\partial t_g(x)}{\partial h} := \frac{\partial^e t_g(x)}{\partial x^{e_1}_{s_1} \cdots \partial x^{e_r}_{s_r}} $$
$$= t_{s_1^{-e_1} \cdots s_r^{-e_r} g}(x) = t_{h^{-1}g}(x)$$
So this means that the group $G$ operates on the set $T :=  \{ t_g(x) | g \in G \}$ via:
$$u \ast t_g(x) := \frac{\partial t_g(x)}{\partial u} = t_{u^{-1}g}(x)$$
We have already proven that:
$$1 = \det(T_G) = \det \left ( t_{gh^{-1}}(x))_{g,h \in G} \right )$$
So, because $G$ is abelian, we have $\frac{\partial t_g(x)}{\partial h} = t_{h^{-1}g}(x) =t_{g h^{-1}}(x)$ and we get one equation of parital derivatives satisfied by the group $G$:
$$1 = \det \left ( (\frac{\partial t_g(x)}{\partial h})_{g,h \in G} \right )$$
A: The three functions in the question appear as the solution of a partial differential equation given by the $\det(C_3)$:
http://portail.mathdoc.fr/JMPA/PDF/JMPA_1929_9_8_A8_0.pdf
P. Humbert. Sur une généralisation de l’équation de Laplace. Jour. Math. Pures
Appl., 94:145–159, 1929.
This information I have read in the following paper:
https://www.semanticscholar.org/paper/Partial-differential-equations-arising-from-Sebbar-Wone/1b43361d8ce2a2391abc4a894fc4452cc6cb7371
