I am reading chapter 2 of Tinkham's Group Theory and Quantum Mechanics and am completely baffled by his Chapter 2-9, and perhaps that's because the terminology/notation seems so ancient. I've attached the relevant passage at the end, but essentially, letting $G_i$ denote an arbitrary conjugacy class of some group $G$, Tinkham writes a formula like $$G_iG_j = \sum_k c_{ijk}G_k.$$ Now I suppose the left-hand side I can interpret as a product of subsets, but the right-hand side? What could the "addition" of subsets mean? If anyone has a reference which makes this all more rigorous I would really appreciate it.
The one illegible bit in the below should read $\mathscr{R}=\mathscr{H}$. Tinkham also seems weirdly to be referring to some multiset type of structure even though "classes" were defined earlier as conjugacy classes (sets).
We can do this a little more carefully by doing the following. Define the group ring of the group over the integers by taking all formal linear combinations of the group elements with integer coefficients. That is, an element of the group ring is given by $$ \sum_jn_jg_j \,, $$ where $j$ runs over the group elements $g_j$ and $n_j\in\mathbb{Z}$. Addition of elements in this ring is given by $$ \sum_jn_jg_j+\sum_jm_jg_j = \sum_j(n_j+m_j)g_j \,, $$ and multiplication is given by $$ \left(\sum_in_ig_i\right)\left(\sum_jm_jg_j\right) = \sum_{i,j}(n_jm_j)(g_ig_j) \,, $$ where the products in parentheses are the normal products in the respective algebraic structures (multiplication of integers and multiplication of elements in the group, respectively). Finally, (integer) scalar multiplication via $$ m\sum_jn_jg_j = \sum_j(mn_j)g_j\,, $$ makes this into a $\mathbb{Z}$-algebra.
Now, consider a set of special elements of the group ring, $$ \mathcal{C}_j=\sum_{g\in C_j}g\,, $$ where $C_j$ is a conjugacy class of the group. This element uniquely corresponds to the conjugacy class in the sense that we can interpret the expansion $\sum_in_ig_i$ as "counting" the number $n_i$ of times $g_i$ shows up in a multiset of elements of $G$. This element is in the center of the group ring, and in fact the collection of $\mathcal{C}_j$'s is a basis for the center of the group ring (see here). For instance, since $g^{-1}C_jg=C_j$, we can easily see that $$ g^{-1}\mathcal{C}_jg = g^{-1}\left(\sum_{h\in C_j}h\right)g = \sum_{h\in C_j}g^{-1}hg =\sum_{ghg^{-1} \in C_j} h =\sum_{h \in C_j}h = \mathcal{C_j}\,, $$ and therefore any formal $\mathbb{Z}$-linear combination of elements of the group will commute with $\mathcal{C_j}$.
Suppose that we multiply two of these elements $\mathcal{C}_j$ by each other. Since $$ g^{-1}\mathcal{C}_i\mathcal{C}_jg = g^{-1}\mathcal{C}_igg^{-1}\mathcal{C}_jg =\mathcal{C}_i\mathcal{C}_j\,, $$ which means that $\mathcal{C}_i\mathcal{C}_j$ is in the center of the group ring. As such, it can be expanded as an integer linear combination of the $\mathcal{C}_i$'s. Finally, we can make contact with the notation from Tinkham:
We then have that $$ \mathcal{C}_i\mathcal{C}_j = \sum_ka_{ijk}\mathcal{C}_k\,. $$ for some coefficients $a_{ijk}$ since the $\mathcal{C}_k$ are a basis of the center of the group ring (in which $\mathcal{C}_i\mathcal{C}_j$ resides).
Our goal is to explain why $a_{ijk} = c_{ijk}$, where the $c$'s are defined in the excerpt from Tinkham in the OP as "the integer telling how often the complete class $\mathcal{C}_k$ appears in the product $\mathcal{C}_i\mathcal{C}_j$".
By explicitly computing the product in the algebra, we get $$ \mathcal{C}_i\mathcal{C}_j = \left(\sum_{g\in C_i}g\right)\left(\sum_{h\in C_j}h\right) = \sum_{g\in C_i,h\in C_j}gh\,. $$ If we expand out this double sum, we can see that each product of one element from $C_i$ and one element from $C_j$ shows up exactly once, and so we get exactly the (multi-)set of elements arrived at by multiplying the conjugacy classes $C_i$ and $C_j$, interpreted as a formal sum. Because of that, we can rearrange the sum as $$ \mathcal{C}_i\mathcal{C}_j = \sum_{k}c_{ijk}\mathcal{C}_k \,, $$ thereby arriving at Tinkham's formal expression for the product of conjugacy classes expanded as a "sum" of conjugacy classes.
Here is an example, using the symmetric group $S_3$ on three elements. A conjugacy class in a symmetric group is made up of all elements with the same cycle structure, so for $S_3$, we have $$ C_0 = {e}\,,~~~~~~C_1={(12),(23),(13)}\,,~~~~~~C_2 = {(123),(132)}\,. $$ Let's form the special formal linear combinations of the elements from these conjugacy classes, i.e., $$ \mathcal{C}_0 = e\,,~~~~~~\mathcal{C}_1=(12)+(23)+(13)\,,~~~~~~\mathcal{C}_2 = (123)+(132)\,. $$ Now, we multiply these elements. (We have to choose a convention for cycle multiplication: we will choose the one in which the permutation on the left is performed first.) The result is \begin{align} \mathcal{C}_0\mathcal{C}_0 &= (e)\cdot(2) = e \\&=\mathcal{C}_0 \,,\\ \mathcal{C}_0\mathcal{C}_1 &= (e)\cdot((12)+(23)+(13)) = (12)+(23)+(13) \\&=\mathcal{C}_1 \,,\\ \mathcal{C}_0\mathcal{C}_2 &= (e)\cdot((123)+(132)) = (123)+(132) \\&=\mathcal{C}_2 \,,\\ \mathcal{C}_1\mathcal{C}_1 &= ((12)+(23)+(13))\cdot((12)+(23)+(13)) \\ &= (12)(12)+(12)(23)+(12)(13) +(23)(12)+(23)(23)+(23)(13) \\&\qquad\mbox{} +(13)(12)+(13)(23)+(13)(13) \\&= e+(132)+(123)+(231)+e+(213)+(132)+(123)+e \\&=3e + 3(132)+3(123) \\&=3\mathcal{C}_0+3\mathcal{C}_2\,, \\ \mathcal{C}_1\mathcal{C}_2 &= ((12)+(23)+(13))\cdot((123)+(132)) \\&= (12)(123)+(12)(132)+ (23)(123)+(23)(132)+ (13)(123)+(13)(132)+ \\&= (13) + (23) + (21) + (31) + (23) + (12) \\&= 2(12) + 2(13) + 2(23) \\&= 2\mathcal{C}_1\,, \\ \mathcal{C}_2\mathcal{C}_2 &= ((123)+(132))\cdot((123)+(132)) \\&= (123)(123)+(123)(132)+(132)(123)+(132)(132) \\&= (132) + e + e + (123) \\&= 2\mathcal{C}_0 + 2\mathcal{C}_2\,. \end{align} Through this exhaustive calculation, we can see that on each penultimate line, we have a list of elements that break up naturally into full copies of conjugacy classes. The ``addition'' here is really just the same as a disjoint union of copies of the conjugacy classes. Alternatively, these can be written as mulitsets. But this representation in terms of an algebra really captures the important information, allows us to represent the info in a nice notational way, and allows us to use algebra to make further conclusions.