Symmetry group of rectangular cuboid

Question

In this Wikipedia page it is said that the symmetry group of the rectangular cuboid (a box with three unequal dimensions) is a dihedral group $D_8$ (well, some people may call it $D_4$...)

However, I don't see how it is the case. If the symmetry group is $D_8$, there is a rotation element $r$ that has order $4$, i.e., $r^4=e$, but there is no such element in the symmetry group of the rectangular cuboid that has order $4$. Every element in the symmetry group of the rectangular cuboid seems to have order $2$ (except for the identity, of course) and I think it should be $\mathbb{Z}_2 \times \mathbb{Z}_2\times \mathbb{Z}_2$.

Is there anything I misunderstand here?

Answer 1

You are correct.

If you label the vertices $1$ through to $8$, like so . . .

. . . you get the generators:

$$\begin{align} a&=(17)(28)(35)(46),\\ b&=(13)(24)(57)(68),\\ c&=(15)(48)(26)(37),\\ x&=(14)(23)(58)(67),\\ y&=(18)(27)(36)(45),\\ z&=(12)(34)(56)(78). \end{align}$$

Putting this in GAP, we get:

gap> G:=Group([(1,7)*(2,8)*(3,5)*(4,6), (1,3)*(2,4)*(5,7)*(6,8), (1,5)*(4,8)*(2,6)*(3,7), (1,4)*(2,3)*(5,8)*(6,7), (1,8)*(2,7)*(3,6)*(4,5), (1,2)*(3,4)*(5,6)*(7,8)]);
Group([ (1,7)(2,8)(3,5)(4,6), (1,3)(2,4)(5,7)(6,8), (1,5)(2,6)(3,7)(4,8), (1,4)(2,3)(5,8)(6,7), (1,8)(2,7)(3,6)(4,5), (1,2)(3,4)(5,6)(7,8) ])
gap> StructureDescription(G);
"C2 x C2 x C2"
gap>

This confirms your intuition.

---

You can confirm through direct calculation that the elements commute and each have order two. That is sufficient to conclude that it is indeed $\Bbb Z_2^3$ (as there is eight elements in total).

Answer 2

[Reposting my comment as an answer, per request in the comments]

I believe this apparent discrepancy is just a difference in notation, i.e., that what Wikipedia identifies as "$D_{2h}$" is indeed the abstract group $\mathbb{Z_2} \times \mathbb{Z_2} \times \mathbb{Z_2}$.

In particular, Wikipedia's "$D_{2h}$" may not be notation for the dihedral group of order $2h$ (or indeed any dihedral group), but Schoenflies notation, in which "$D_{kh}$," where $k$ is a parameter and $h$ is a decoration peculiar to this notation, is a group described in http://en.wikipedia.org/wiki/Schoenflies_notation (describing both "$D_n$" and "$D_{nh}$," and note that the latter is not just the former with $n$ replaced by $nh$).

As I interpret this description (which is admittedly somewhat opaque without a lot of background geometrical context), it looks as though "$D_{kh}$" as "defined" there is generally a group of order $4k$, and not necessarily the dihedral group of that order. Indeed, it seems possible that the somewhat vague verbal recipe for "$D_{kh}$" given on that Wikipedia page, if formalized more concretely via generators and relations, might lead to exactly the presentation that Shaun's answer identifies (or at least something close to it, and giving rise to the same abstract group) in the case $k=2$.

In partial confirmation of my interpretation, the Schoenflies notation is also used in http://en.wikipedia.org/wiki/Crystallographic_point_group, in particular in a table in the section labeled "Isomorphisms," which identifies "$D_{2h}$" as "$D_2 \times C_2$," where $C_2$ is elsewhere identified on that page as the cyclic group of order $2$ and "$D_2$" is elsewhere identified as the Klein 4-group (i.e., $C_2 \times C_2$, which is indeed the dihedral group of order $4$). More succinctly, this table identifies the group "$D_{2h}$" with $C_2 \times C_2 \times C_2$, as expected.