1. Linear algebra and group theory
  2. Finite groups
  3. 9b. Dihedral groups

9b. Dihedral groups

[math] \newcommand{\mathds}{\mathbb}[/math]

In order to have now some theory going, we obviously have to impose some conditions on the groups that we consider. With this idea in mind, let us work out some examples, in the finite group case. The simplest possible finite group is the cyclic group [math]\mathbb Z_N[/math]. There are many ways of picturing [math]\mathbb Z_N[/math], both additive and multiplicative, as follows:

Definition

The cyclic group [math]\mathbb Z_N[/math] is defined as follows:

  • As the additive group of remainders modulo [math]N[/math].
  • As the multiplicative group of the [math]N[/math]-th roots of unity.

The two definitions are equivalent, because if we set [math]w=e^{2\pi i/N}[/math], then any remainder modulo [math]N[/math] defines a [math]N[/math]-th root of unity, according to the following formula:

[[math]] k\to w^k [[/math]]

We obtain in this way all the [math]N[/math]-roots of unity, and so our correspondence is bijective. Moreover, our correspondence transforms the sum of remainders modulo [math]N[/math] into the multiplication of the [math]N[/math]-th roots of unity, due to the following formula:

[[math]] w^kw^l=w^{k+l} [[/math]]

Thus, the groups defined in (1,2) above are isomorphic, via [math]k\to w^k[/math], and we agree to denote by [math]\mathbb Z_N[/math] the corresponding group. Observe that this group [math]\mathbb Z_N[/math] is abelian.


Another interesting example of a finite group, which is more advanced, and which is non-abelian this time, is the dihedral group [math]D_N[/math], which appears as follows:

Definition

The dihedral group [math]D_N[/math] is the symmetry group of the regular polygon having [math]N[/math] vertices, which is called regular [math]N[/math]-gon.

Here are some basic examples of regular [math]N[/math]-gons, at small values of the parameter [math]N\in\mathbb N[/math], and of their symmetry groups:


\underline{[math]N=2[/math]}. Here the [math]N[/math]-gon is just a segment, and its symmetries are the identity [math]id[/math] and the obvious symmetry [math]\tau[/math]. Thus [math]D_2=\{id,\tau\}[/math], and in group theory terms, [math]D_2=\mathbb Z_2[/math].


\underline{[math]N=3[/math]}. Here the [math]N[/math]-gon is an equilateral triangle, and the symmetries are the [math]3!=6[/math] possible permutations of the vertices. Thus we have [math]D_3=S_3[/math].


\underline{[math]N=4[/math]}. Here the [math]N[/math]-gon is a square, and as symmetries we have 4 rotations, of angles [math]0^\circ,90^\circ,180^\circ,270^\circ[/math], as well as 4 symmetries, with respect to the 4 symmetry axes, which are the 2 diagonals, and the 2 segments joining the midpoints of opposite sides.


\underline{[math]N=5[/math]}. Here the [math]N[/math]-gon is a regular pentagon, and as symmetries we have 5 rotations, of angles [math]0^\circ,72^\circ,144^\circ,216^\circ,288^\circ[/math], as well as 5 symmetries, with respect to the 5 symmetry axes, which join the vertices to the midpoints of the opposite sides.


\underline{[math]N=6[/math]}. Here the [math]N[/math]-gon is a regular hexagon, and we have 6 rotations, of angles [math]0^\circ,60^\circ,120^\circ,180^\circ,240^\circ,300^\circ[/math], and 6 symmetries, with respect to the 6 symmetry axes, which are the 3 diagonals, and the 3 segments joining the midpoints of opposite sides.


We can see from the above that the various dihedral groups [math]D_N[/math] have many common features, and that there are some differences as well. In general, we have:

Proposition

The dihedral group [math]D_N[/math] has [math]2N[/math] elements, as follows:

  • We have [math]N[/math] rotations [math]R_1,\ldots,R_N[/math], with [math]R_k[/math] being the rotation of angle [math]2k\pi/N[/math]. When labelling the vertices of the [math]N[/math]-gon [math]1,\ldots,N[/math], the rotation formula is:
    [[math]] R_k:i\to k+i [[/math]]
  • We have [math]N[/math] symmetries [math]S_1,\ldots,S_N[/math], with [math]S_k[/math] being the symmetry with respect to the [math]Ox[/math] axis rotated by [math]k\pi/N[/math]. The symmetry formula is:
    [[math]] S_k:i\to k-i [[/math]]


Show Proof

This is clear, indeed. To be more precise, [math]D_N[/math] consists of:


(1) The [math]N[/math] rotations, of angles [math]2k\pi/N[/math] with [math]k=1,\ldots,N[/math]. But these are exactly the rotations [math]R_1,\ldots,R_N[/math] from the statement.


(2) The [math]N[/math] symmetries with respect to the [math]N[/math] possible symmetry axes, which are the [math]N[/math] medians of the [math]N[/math]-gon when [math]N[/math] is odd, and are the [math]N/2[/math] diagonals plus the [math]N/2[/math] lines connecting the midpoints of opposite edges, when [math]N[/math] is even. But these are exactly the symmetries [math]S_1,\ldots,S_N[/math] from the statement.

■

With the above description of [math]D_N[/math] in hand, we can forget if we want about geometry and the regular [math]N[/math]-gon, and talk about [math]D_N[/math] abstractly, as follows:

Theorem

The dihedral group [math]D_N[/math] is the group having [math]2N[/math] elements, [math]R_1,\ldots,R_N[/math] and [math]S_1,\ldots,S_N[/math], called rotations and symmetries, which multiply as follows,

[[math]] R_kR_l=R_{k+l} [[/math]]

[[math]] R_kS_l=S_{k+l} [[/math]]

[[math]] S_kR_l=S_{k-l} [[/math]]

[[math]] S_kS_l=R_{k-l} [[/math]]
with all the indices being taken modulo [math]N[/math].


Show Proof

With notations from Proposition 9.7, the various compositions between rotations and symmetries can be computed as follows:

[[math]] R_kR_l\ :\ i\to l+i\to k+l+i [[/math]]

[[math]] R_kS_l\ :\ i\to l-i\to k+l-i [[/math]]

[[math]] S_kR_l\ :\ i\to l+i\to k-l-i [[/math]]

[[math]] S_kS_l\ :\ i\to l-i\to k-l+i [[/math]]

But these are exactly the formulae for [math]R_{k+l},S_{k+l},S_{k-l},R_{k-l}[/math], as stated. Now since a group is uniquely determined by its multiplication rules, this gives the result.

■

Observe that [math]D_N[/math] has the same cardinality as [math]E_N=\mathbb Z_N\times\mathbb Z_2[/math]. We obviously don't have [math]D_N\simeq E_N[/math], because [math]D_N[/math] is not abelian, while [math]E_N[/math] is. So, our next goal will be that of proving that [math]D_N[/math] appears by “twisting” [math]E_N[/math]. In order to do this, let us start with:

Proposition

The group [math]E_N=\mathbb Z_N\times\mathbb Z_2[/math] is the group having [math]2N[/math] elements, [math]r_1,\ldots,r_N[/math] and [math]s_1,\ldots,s_N[/math], which multiply according to the following rules,

[[math]] r_kr_l=r_{k+l} [[/math]]

[[math]] r_ks_l=s_{k+l} [[/math]]

[[math]] s_kr_l=s_{k+l} [[/math]]

[[math]] s_ks_l=r_{k+l} [[/math]]
with all the indices being taken modulo [math]N[/math].


Show Proof

With the notation [math]\mathbb Z_2=\{1,\tau\}[/math], the elements of the product group [math]E_N=\mathbb Z_N\times\mathbb Z_2[/math] can be labelled [math]r_1,\ldots,r_N[/math] and [math]s_1,\ldots,s_N[/math], as follows:

[[math]] r_k=(k,1)\quad,\quad s_k=(k,\tau) [[/math]]

These elements multiply then according to the formulae in the statement. Now since a group is uniquely determined by its multiplication rules, this gives the result.

■

Let us compare now Theorem 9.8 and Proposition 9.9. In order to formally obtain [math]D_N[/math] from [math]E_N[/math], we must twist some of the multiplication rules of [math]E_N[/math], namely:

[[math]] s_kr_l=s_{k+l}\to s_{k-l} [[/math]]

[[math]] s_ks_l=r_{k+l}\to r_{k-l} [[/math]]

Informally, this amounts in following the rule “[math]\tau[/math] switches the sign of what comes afterwards”, and we are led in this way to the following definition:

Definition

Given two groups [math]A,G[/math], with an action [math]A\curvearrowright G[/math], the crossed product

[[math]] P=G\rtimes A [[/math]]
is the set [math]G\times A[/math], with multiplication as follows:

[[math]] (g,a)(h,b)=(gh^a,ab) [[/math]]

It is routine to check that [math]P[/math] is indeed a group. Observe that when the action is trivial, [math]h^a=h[/math] for any [math]a\in A[/math] and [math]h\in H[/math], we obtain the usual product [math]G\times A[/math].


Now with this technology in hand, by getting back to the dihedral group [math]D_N[/math], we can improve Theorem 9.8, into a final result on the subject, as follows:

Theorem

We have a crossed product decomposition as follows,

[[math]] D_N=\mathbb Z_N\rtimes\mathbb Z_2 [[/math]]
with [math]\mathbb Z_2=\{1,\tau\}[/math] acting on [math]\mathbb Z_N[/math] via switching signs, [math]k^\tau=-k[/math].


Show Proof

We have an action [math]\mathbb Z_2\curvearrowright\mathbb Z_N[/math] given by the formula in the statement, namely [math]k^\tau=-k[/math], so we can consider the corresponding crossed product group:

[[math]] P_N=\mathbb Z_N\rtimes\mathbb Z_2 [[/math]]

In order to understand the structure of [math]P_N[/math], we follow Proposition 9.9. The elements of [math]P_N[/math] can indeed be labelled [math]\rho_1,\ldots,\rho_N[/math] and [math]\sigma_1,\ldots,\sigma_N[/math], as follows:

[[math]] \rho_k=(k,1)\quad,\quad \sigma_k=(k,\tau) [[/math]]

Now when computing the products of such elements, we basically obtain the formulae in Proposition 9.9, perturbed as in Definition 9.10. To be more precise, we have:

[[math]] \rho_k\rho_l=\rho_{k+l} [[/math]]

[[math]] \rho_k\sigma_l=\sigma_{k+l} [[/math]]

[[math]] \sigma_k\rho_l=\sigma_{k+l} [[/math]]

[[math]] \sigma_k\sigma_l=\rho_{k+l} [[/math]]

But these are exactly the multiplication formulae for [math]D_N[/math], from Theorem 9.8. Thus, we have an isomorphism [math]D_N\simeq P_N[/math] given by [math]R_k\to\rho_k[/math] and [math]S_k\to\sigma_k[/math], as desired.

■

As a third basic example of a finite group, we have the symmetric group [math]S_N[/math]. This is a group that we already met, when talking about the determinant:

Theorem

The permutations of [math]\{1,\ldots,N\}[/math] form a group, denoted [math]S_N[/math], and called symmetric group. This group has [math]N![/math] elements. The signature map

[[math]] \varepsilon:S_N\to\mathbb Z_2 [[/math]]
can be regarded as being a group morphism, with values in [math]\mathbb Z_2=\{\pm1\}[/math].


Show Proof

These are things that we know from chapter 2. Indeed, the group property is clear, and the count is clear as well. As for the last assertion, recall the following formula for the signatures of the permutations, that we know as well from chapter 2:

[[math]] \varepsilon(\sigma\tau)=\varepsilon(\sigma)\varepsilon(\tau) [[/math]]

But this tells us precisely that [math]\varepsilon[/math] is a group morphism, as stated.

■

We will be back to [math]S_N[/math] on many occasions, in what follows. At an even more advanced level now, we have the hyperoctahedral group [math]H_N[/math], which appears as follows:

Definition

The hyperoctahedral group [math]H_N[/math] is the group of symmetries of the unit cube in [math]\mathbb R^N[/math].

The hyperoctahedral group is a quite interesting group, whose definition, as a symmetry group, reminds that of the dihedral group [math]D_N[/math]. So, let us start our study in the same way as we did for [math]D_N[/math], with a discussion at small values of [math]N\in\mathbb N[/math]:


\underline{[math]N=1[/math]}. Here the 1-cube is the segment, whose symmetries are the identity [math]id[/math] and the flip [math]\tau[/math]. Thus, we obtain the group with 2 elements, which is a very familiar object:

[[math]] H_1=D_2=S_2=\mathbb Z_2 [[/math]]

\underline{[math]N=2[/math]}. Here the 2-cube is the square, and so the corresponding symmetry group is the dihedral group [math]D_4[/math], which is a group that we know well:

[[math]] H_2=D_4=\mathbb Z_4\rtimes\mathbb Z_2 [[/math]]

\underline{[math]N=3[/math]}. Here the 3-cube is the usual cube, and the situation is considerably more complicated, because this usual cube has no less than 48 symmetries. Identifying and counting these symmetries is actually an excellent exercise.


All this looks quite complicated, but fortunately we can count [math]H_N[/math], at [math]N=3[/math], and at higher [math]N[/math] as well, by using some tricks, the result being as follows:

Theorem

We have the cardinality formula

[[math]] |H_N|=2^NN! [[/math]]
coming from the fact that [math]H_N[/math] is the symmetry group of the coordinate axes of [math]\mathbb R^N[/math].


Show Proof

This follows from some geometric thinking, as follows:


(1) Consider the standard cube in [math]\mathbb R^N[/math], centered at 0, and having as vertices the points having coordinates [math]\pm1[/math]. With this picture in hand, it is clear that the symmetries of the cube coincide with the symmetries of the [math]N[/math] coordinate axes of [math]\mathbb R^N[/math].


(2) In order to count now these latter symmetries, a bit as we did for the dihedral group, observe first that we have [math]N![/math] permutations of these [math]N[/math] coordinate axes.


(3) But each of these permutations of the coordinate axes [math]\sigma\in S_N[/math] can be further “decorated” by a sign vector [math]e\in\{\pm1\}^N[/math], consisting of the possible [math]\pm1[/math] flips which can be applied to each coordinate axis, at the arrival. Thus, we have:

[[math]] |H_N| =|S_N|\cdot|\mathbb Z_2^N| =N!\cdot2^N [[/math]]

Thus, we are led to the conclusions in the statement.

■

As in the dihedral group case, it is possible to go beyond this, with a crossed product decomposition, of quite special type, called wreath product decomposition:

Theorem

We have a wreath product decomposition as follows,

[[math]] H_N=\mathbb Z_2\wr S_N [[/math]]
which means by definition that we have a crossed product decomposition

[[math]] H_N=\mathbb Z_2^N\rtimes S_N [[/math]]
with the permutations [math]\sigma\in S_N[/math] acting on the elements [math]e\in\mathbb Z_2^N[/math] as follows:

[[math]] \sigma(e_1,\ldots,e_k)=(e_{\sigma(1)},\ldots,e_{\sigma(k)}) [[/math]]


Show Proof

As explained in the proof of Theorem 9.14, the elements of [math]H_N[/math] can be identified with the pairs [math]g=(e,\sigma)[/math] consisting of a permutation [math]\sigma\in S_N[/math], and a sign vector [math]e\in\mathbb Z_2^N[/math], so that at the level of the cardinalities, we have:

[[math]] |H_N|=|\mathbb Z_2^N\times S_N| [[/math]]

To be more precise, given an element [math]g\in H_N[/math], the element [math]\sigma\in S_N[/math] is the corresponding permutation of the [math]N[/math] coordinate axes, regarded as unoriented lines in [math]\mathbb R^N[/math], and [math]e\in\mathbb Z_2^N[/math] is the vector collecting the possible flips of these coordinate axes, at the arrival. Now observe that the product formula for two such pairs [math]g=(e,\sigma)[/math] is as follows, with the permutations [math]\sigma\in S_N[/math] acting on the elements [math]f\in\mathbb Z_2^N[/math] as in the statement:

[[math]] (e,\sigma)(f,\tau)=(ef^\sigma,\sigma\tau) [[/math]]

Thus, we are precisely in the framework of Definition 9.10, and we conclude that we have a crossed product decomposition, as follows:

[[math]] H_N=\mathbb Z_2^N\rtimes S_N [[/math]]

Thus, we are led to the conclusion in the statement, with the formula [math]H_N=\mathbb Z_2\wr S_N[/math] being just a shorthand for the decomposition [math]H_N=\mathbb Z_2^N\rtimes S_N[/math] that we found.

■

Summarizing, we have so far many interesting examples of finite groups, and as a sequence of main examples, we have the following groups:

[[math]] \mathbb Z_N\subset D_N\subset S_N\subset H_N [[/math]]

We will be back to these fundamental finite groups later on, on several occasions, with further results on them, both of algebraic and of analytic type.


General references

Banica, Teo (2024). "Linear algebra and group theory". arXiv:2206.09283 [math.CO].