I have some questions regarding the notation and solution to the following problem involving the direct product:
The diagram below shows the symmetry operations of an equilateral right triangular prism. The elements of $D_3$ are the identity $E$, $(\mathrm{a})$ the rotations $C_3$ and ${C_3}^2$, and $(\mathrm{b})$ the vertical reflection planes $\sigma_{v,1}$, $\sigma_{v,2}$, and $\sigma_{v,3}$, and those of $C_s$ are the identity $E$ and $(\mathrm{c})$ a horizontal reflection plane $\sigma_h$ midway between the two horizontal faces of the prism.
Show that the group $D_{3h}$ prism is given by the direct product $$D_{3h}=D_3 \otimes C_s$$ where $C_s=\{E, \sigma_h\}$
My first question is regarding the notation, according to this article on Wikipedia for the direct product of two groups, the correct notation is actually $D_{3h}=D_3 \oplus C_s$ and not $D_{3h}=D_3 \otimes C_s$. This is confusing me as I thought the '$\oplus$' was the notation reserved for the direct sum. Why then the inconsistency for the notation?
Looking across the literature, "Contemporary Abstract Algebra" by Joseph A Gallian, 10th edition, page 172 uses $\oplus$, yet "Group Theory - Application to the Physics of Condensed Matter", by Dresselhaus, 2008, page 101 uses $\otimes$ as does "Group Theory in Physics" by Wu-Ki Tung, 1985, page 25. Yet other books such as "Applications of Group Theory in Quantum Mechanics" by Petrashen and Trifonov, published in 2009 on page 49 use a regular "$\times$". Due to this I'm very confused which notation to use.
On to the question itself, using the definition from my notes:
DEFINITION $1.2.$ A direct product of two groups $H$ and $K$, denoted as $H \otimes K$, is defined as the set of all distinct ordered pairs $(h, k)$, where $h$ is an element of $H$ and $k$ is an element of $K$, with the binary composition law $$(h_1,k_1)(h_2,k_2) \equiv (h_1h_2,k_1k_2),$$ determined by the composition rules within $H$ and $K$. This direct product, also known as the external product, satisfies the group axioms.
Applying the procedure outlined above to the 2 groups I find that,
$$D_{3h}=D_3\otimes C_s =\left(E, C_3, {C_3}^2, \sigma_{v,1}, \sigma_{v,2}, \sigma_{v,3}\right)\left(E, \sigma_h\right)$$ $$=\left(E, C_3, {C_3}^2, \sigma_{v,1}, \sigma_{v,2}, \sigma_{v,3}, \sigma_h, C_3\sigma_h,{C_3}^2\sigma_h,\sigma_{v,1}\sigma_h,\sigma_{v,2}\sigma_h,\sigma_{v,3}\sigma_h\right)\tag{1}$$
The problem with this equation is that I don't know what the elements of $D_{3h}$ are, so I followed the prescription laid out by my notes but how does this prove that $D_{3h}=D_3 \otimes C_s$?
Here is the author's solution:
The group $D_3$ of the top and bottom faces has elements $$D_3=\{E, C_3, {C_3}^2, \sigma_{v,1}, \sigma_{v,2}, \sigma_{v,3}\}$$ The prism has an additional symmetry operation, a horizontal reflection plane $\sigma_h$, resulting from the equivalence of the top and bottom faces. The associated group is $C_s = \{E, \sigma_h\}$. Since $\sigma_h$ operates on vertical coordinates, and the elements of $D_3$ operate on horizontal coordinates, the elements of $C_s$ and $D_3$ commute. Hence, the group of the prism, which is denoted as $D_{3h}$, is $$D_{3h}=D_3 \otimes C_s.$$
Again, this solution doesn't really seem to 'prove' that $D_{3h}=D_3 \otimes C_s$, to me the author has just given a handwaving argument and I'm left confused.
So to summarize, is there a way to be more formal about this proof (like what I was trying to do in equation $(1)$), ie. utilizing the recipe given, that $(h_1,k_1)(h_2,k_2)=(h_1h_2,k_1k_2)$?
First, it might be useful to read this answer to see what is usually meant by direct sum and direct products of groups. In particular, you'll see that finite direct sums are the same as finite direct products.
Now, to answer your first question, Wikipedia rightly states that in the case of abelian groups the (finite) direct product is sometimes referred to as the direct sum, and denoted by $\oplus$. If you want an explanation for the notation, it has to do with the properties of the category $\operatorname{Ab}$ of abelian groups, i.e. the "collection" of all abelian groups and homomorphisms thereof. In $\operatorname{Ab}$, finite direct sums/products correspond to both the categorical concept of coproduct and that of product (it turns out that this seemingly innocuous formal property of $\operatorname{Ab}$ can be abstracted to more complicated categories, which by virtue of this formal similarity behave something like $\operatorname{Ab}$, an observation of extreme importance in modern mathematics), thus they are denoted the same. In the case of general groups (i.e. not necessarily abelian), this is no longer true (direct products still correspond to products, but coproducts correspond rather to the free product of groups), thus the direct product of groups is usually denoted with $\times$ (it is still a different concept that the direct sum $\oplus$, even if they coincide for finitely many terms).
As for $\otimes$, I don't think it is standard in mathematics to denote the direct product of groups with that symbol (which is reserved for other things); I suspect those books where you found such a notation reserve $\times$ for something else and use $\otimes$ to avoid confusion. Moreover, they seem to be written by physicists (I'm not being snarky here, it's just that physicists are interested in different things and might use different notations from mathematicians).
Now we come to your actual question. I agree that the solution given is not necessarily very clear as to why $D_{3h}$ is exactly the same as $D_3\times C_s$, but it has the right ideas to solve the problem. You do agree that $D_3\times C_s$ is a subgroup of $D_{3h}$, considering what the solution says? If you do (and I believe the solution is clear enough on this, but feel free to ask for clarifications), we just need to show that any element of $D_{3h}$, i.e. any symmetry of the prism, is an element of $D_3\times C_s$, i.e. a symmetry of the triangle plus possibly a reflection that swaps the two triangular faces. But this is clear: a symmetry of the prism is determined by whether it exchanges the two triangular faces and by how it exchanges the vertical edges (by virtue of symmetries being "rigid": they do not change angles and length; you might feel the need to make this more precise, which is annoying but not difficult), thus the cardinality of $D_{3h}$ is the same as the cardinality of its subgroup $D_3\times C_s$, which means they are the same.
Edit: let me add something more. The "recipe" for the group operation in a direct product of groups $H\times K$ is really just telling you that elements of the two factors (identified as pairs $(a,E)$ or $(E,b)$ in the product, for $a\in H$ and $b\in K$) are somehow independent of each other, i.e. they commute, and generate the group. The direct product $H\times K$ is uniquely determined by this property (you might want to think about why this is true: if any other group had this property it would be easy to find an isomorphism with $H\times K$). Thus, whenever you have two subgroups of a group with the elements of one commuting with those of the other, you can consider a bigger subgroup, the direct product of the two subgroups with the operation described.