guide:4e943cf45a

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

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As something more concrete now, which is a must-know, let us try to compute the dihedral group. This is a famous group, constructed as follows:

Definition

The dihedral group [math]D_N[/math] is the symmetry group of

[[math]] \xymatrix@R=12pt@C=12pt{ &\bullet\ar@{-}[r]\ar@{-}[dl]&\bullet\ar@{-}[dr]\\ \bullet\ar@{-}[d]&&&\bullet\ar@{-}[d]\\ \bullet\ar@{-}[dr]&&&\bullet\ar@{-}[dl]\\ &\bullet\ar@{-}[r]&\bullet} [[/math]]

that is, of the regular polygon having [math]N[/math] vertices.

In order to understand how this works, here are the basic examples of regular [math]N[/math]-gons, at small values of the parameter [math]N\in\mathbb N[/math], along with their symmetry groups:

\underline{[math]N=2[/math]}. Here the [math]N[/math]-gon is just a segment, and its symmetries are obviously the identity [math]id[/math], plus the symmetry [math]\tau[/math] with respect to the middle of the segment:

[[math]] \xymatrix@R=10pt@C=20pt{ &\ar@{.}[dd]\\ \bullet\ar@{-}[rr]&&\bullet\\ &} [[/math]]

Thus we have [math]D_2=\{id,\tau\}[/math], which in group theory terms means [math]D_2=\mathbb Z_2[/math].

\underline{[math]N=3[/math]}. Here the [math]N[/math]-gon is an equilateral triangle, and we have 6 symmetries, the rotations of angles [math]0^\circ[/math], [math]120^\circ[/math], [math]240^\circ[/math], and the symmetries with respect to the altitudes:

[[math]] \xymatrix@R=13pt@C=28pt{ &\bullet\ar@{-}[dddr]\ar@{-}[dddl]\ar@{.}[dddd]\\ \ar@{.}[ddrr]&&\ar@{.}[ddll]\\ \\ \bullet\ar@{-}[rr]&&\bullet\\ & } [[/math]]

Alternatively, we can say that the symmetries are all the [math]3!=6[/math] possible permutations of the vertices, and so that in group theory terms, 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:

[[math]] \xymatrix@R=26pt@C=26pt{ \bullet\ar@{-}[dd]\ar@{.}[ddrr]\ar@{-}[rr]&\ar@{.}[dd]&\bullet\ar@{-}[dd]\ar@{.}[ddll]\\ \ar@{.}[rr]&&\\ \bullet\ar@{-}[rr]&&\bullet } [[/math]]

Thus, we obtain as symmetry group some sort of product between [math]\mathbb Z_4[/math] and [math]\mathbb Z_2[/math]. Observe however that this product is not the usual one, our group being not abelian.

\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:

[[math]] \xymatrix@R=13pt@C=11pt{ &&\bullet\ar@{-}[ddrr]\ar@{-}[ddll]\ar@{.}[dddd]\\ &&&&\\ \bullet\ar@{-}[ddr]\ar@{.}[drrrr]&&&&\bullet\ar@{-}[ddl]\ar@{.}[dllll]\\ &&&&\\ &\bullet\ar@{-}[rr]\ar@{.}[uuurr]&&\bullet\ar@{.}[uuull]&& } [[/math]]

\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:

[[math]] \xymatrix@R=4pt@C=14pt{ &&\bullet\ar@{-}[ddrr]\ar@{-}[ddll]\ar@{.}[dddddddd]\\ &\ar@{.}[ddddddrr]&&\ar@{.}[ddddddll]\\ \bullet\ar@{-}[dddd]\ar@{.}[ddddrrrr]&&&&\bullet\ar@{-}[dddd]\ar@{.}[ddddllll]\\ &&&&\\ \ar@{.}[rrrr]&&&&\\ &&&&\\ \bullet\ar@{-}[ddrr]&&&&\bullet\ar@{-}[ddll]\\ &&&&\\ &&\bullet } [[/math]]

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 [math]1,\ldots,N[/math], the 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

Our group is indeed formed of [math]N[/math] rotations, of angles [math]2k\pi/N[/math] with [math]k=1,\ldots,N[/math], and then of 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.

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With the above result in hand, we can 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}\quad,\quad R_kS_l=S_{k+l} [[/math]]

[[math]] S_kR_l=S_{k-l}\quad,\quad S_kS_l=R_{k-l} [[/math]]

with all the indices being taken modulo [math]N[/math].

Show Proof

With notations from Proposition 4.24, the various compositions between rotations and symmetries can be computed as follows:

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

[[math]] S_kR_l\ :\ i\to l+i\to k-l-i\quad,\quad 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.

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

[[math]] s_kr_l=s_{k+l}\quad,\quad 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.

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Let us compare now Theorem 4.25 and Proposition 4.26. 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}\quad,\quad 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 [math](g,a)(h,b)=(gh^a,ab)[/math].

Now with this technology in hand, by getting back to the dihedral group [math]D_N[/math], we can improve Theorem 4.25, 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 4.26. 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 4.26, perturbed as in Definition 4.27. To be more precise, we have:

[[math]] \rho_k\rho_l=\rho_{k+l}\quad,\quad \rho_k\sigma_l=\sigma_{k+l} [[/math]]

[[math]] \sigma_k\rho_l=\sigma_{k+l}\quad,\quad \sigma_k\sigma_l=\rho_{k+l} [[/math]]

But these are exactly the multiplication formulae for [math]D_N[/math], from Theorem 4.25. 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.

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General references

Banica, Teo (2024). "Graphs and their symmetries". arXiv:2406.03664 [math.CO].