DIHEDRAL GROUPS KEITH CONRAD


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1 DIHEDRAL GROUPS KEITH CONRAD 1. Introduction For n 3, the dihedral group D n i defined a the rigid motion 1 of the plane preerving a regular ngon, with the operation being compoition. Thee polygon for n = 3, 4, 5, and 6 are pictured below. The dotted line are line of reflection: reflecting the polygon acro each line bring the polygon back to itelf, o thee reflection are in D 3, D 4, D 5, and D 6. In addition to reflection, a rotation by any multiple of 2π/n radian around the center carrie the polygon back to itelf, o D n contain ome rotation. We will look at elementary apect of dihedral group: liting it element, relation between rotation and reflection, conjugacy clae, the center, and commutato. Throughout, n Finding the element of D n The point in the plane that have a pecified ditance from a given point i a circle, o the point with pecified ditance from two given point are the inteection of two circle, which i two point. For intance, the blue point in the figure below have the ame ditance to each of the two black point. Lemma 2.1. Every point on a regular polygon i determined, among all point on the polygon, by it ditance from two adjacent vertice of the polygon. Proof. In the picture above, let the black dot be adjacent vertice of a regular polygon. Since they are adjacent, the line egment connecting them i an edge of the polygon and the polygon i entirely on one ide of the line through the black dot. That how the two 1 A rigid motion i a ditancepreerving tranformation, like a rotation, a reflection, or a tranlation, and i alo called an iometry. 1
2 2 KEITH CONRAD blue dot can t both be on the polygon, o a point on the polygon i ditinguihed from all other point on the polygon (not from all other point in the plane!) by it ditance from the two adjacent vertice. Theorem 2.2. The ize of D n i 2n. Proof. Our argument ha two part: an upper bound and then a contruction of enough rigid motion to achieve the upper bound. Step 1: #D n 2n. Pick two adjacent vertice of a regular ngon, and call them A and B a in the figure below. An element g of D n i a rigid motion preerving the polygon, and it mut carry vertice to vertice (how are vertice unlike other point in term of their ditance relationhip with all point on the polygon?) and g mut preerve adjacency of vertice, o g(a) and g(b) are adjacent vertice of the polygon. B g(b) A g(a) P For any point P on the polygon, the location of g(p ) i determined by g(a) and g(b), becaue the ditance of g(p ) from the adjacent vertice g(a) and g(b) equal the ditance of P from A and B, and therefore g(p ) i determined on the polygon by Lemma 2.1. To count #D n it thu uffice to find the number of poibilitie for g(a) and g(b). Since g(a) and g(b) are a pair of adjacent vertice, g(a) ha at mot n poibilitie (there are n vertice), and for each choice of that g(b) ha at mot 2 poibilitie (one of the two vertice adjacent to g(a)). That give u at mot n 2 = 2n poibilitie, o #D n 2n. Step 2: #D n = 2n. We will decribe n different rotation and n different reflection of a regular ngon. Thi i 2n different rigid motion ince a rotation and a reflection are not the ame: a rotation fixe jut one point, the center, but a reflection fixe a line. A regular ngon can be rotated around it center in n different way to come back to itelf (including rotation by 0 degree). Specifically, we can rotate around the center by 2kπ/n radian where k = 0, 1,..., n 1. Thi i n rotation. To decribe reflection preerving a regular ngon, look at the picture on the fit page: for n = 3 and n = 5 there are line of reflection connecting each vertex to the midpoint of the oppoite ide, and for n = 4 and n = 6 there are line of reflection connecting oppoite vertice and line of reflection connecting midpoint of oppoite ide. Thee decription of the poible reflection work in general, depending on whether n i even or odd: For odd n, there i a reflection acro the line connecting each vertex to the midpoint of the oppoite ide. Thi i a total of n reflection (one per vertex). They are different becaue each one fixe a different vertex. For even n, there i a reflection acro the line connecting each pair of oppoite vertice (n/2 reflection) and acro the line connecting midpoint of oppoite ide (another n/2 reflection). The number of thee reflection i n/2+n/2 = n. They are
3 DIHEDRAL GROUPS 3 different becaue they have different type of fixed point on the polygon: different pai of oppoite vertice or different pai of midpoint of oppoite ide. In D n it i tandard to write r for the counterclockwie rotation by 2π/n radian. Thi rotation depend on n, o the r in D 3 mean omething different from the r in D 4. However, a long a we are dealing the dihedral group for one value of n, there houldn t be any confuion. From the rotation r, we get n rotation in D n by taking powe (r ha order n): 1, r, r 2,..., r n 1. (We adopt common grouptheoretic notation and deignate the identity rigid motion a 1.) Let be a reflection acro a line through a vertex. See example in the polygon below. 2 Any reflection ha order 2, o 2 = 1 and 1 =. Theorem 2.3. The n reflection in D n are,,,..., r n 1. Proof. The rigid motion,,,..., r n 1 are different ince 1, r, r 2,..., r n 1 are different and we jut multiply them all on the left by the ame element. No r k i a rotation becaue if r k = r l then = r l k, but i not a rotation. The group D n conit of n rotation and n reflection, and no r k i a rotation, o they are all reflection. Since each element of D n i a rotation or reflection, there i no mixed rotationreflection : the product of a rotation r i and a reflection r j (in either order) i a reflection. The geometric interpretation of the ucceive reflection,,, and o on i thi: drawing all the line of reflection for the regular ngon and moving clockwie around the polygon tarting from a vertex fixed by we meet ucceively the line fixed by,,..., r n 1. In the polygon below, we let be the reflection acro the line through the rightmot vertex and then label the other line of reflection accordingly. Convince youelf, for intance, that i the next line of reflection if we follow them around counterclockwie. 2 The convention that fixe a line through a vertex matte only for even n, where there are ome reflection acro a line that doen t pa through a vertex, namely a line connecting midpoint of oppoite ide. When n i odd, all reflection fix a line through a vertex, o any of them could be choen a.
4 4 KEITH CONRAD r 3 r 4 r 3 r 5 r 4 r 3 Let ummarize what we have now found. Theorem 2.4. The group D n ha 2n element. A a lit, (2.1) D n = {1, r, r 2,..., r n 1,,,..., r n 1 }, In particular, all element of D n with order greater than 2 are powe of r. Watch out: although any element of D n with order greater than 2 ha to be a power of r, becaue any element that in t a power of r i a reflection, it i fale in general that the only element of order 2 are reflection. When n i even, r n/2 i a 180degree rotation, which ha order 2. Clearly a 180degree rotation i the only rotation with order 2, and it lie in D n only when n i even. 3. Relation between rotation and reflection The rigid motion r and do not commute. Their commutation relation i a fundamental formula for computation in D n, and goe a follow. Theorem 3.1. In D n, (3.1) 1 = r 1. Proof. Since every rigid motion of a regular ngon i determined by it effect on two adjacent vertice, to prove 1 = r 1 in D n it uffice to check 1 and r 1 have the ame value at ome pair of adjacent vertice. Recall i a reflection fixing a vertex of the polygon. Let A be a vertex fixed by and write it adjacent vertice a B and B, with B appearing counterclockwie from A and B appearing clockwie from A. Thi i illutrated in the figure below, where the dahed line through A i fixed by. We have r(a) = B, r 1 (A) = B, (A) = A, and (B) = B. B A The value of 1 and r 1 at A are ( 1 )(A) = ()(A) = r((a)) = r(a) = (B) = B and r 1 (A) = B, B
5 DIHEDRAL GROUPS 5 while their value at B are ( 1 )(B) = ()(B) = r((b)) = r(b ) = (A) = A and r 1 (B) = A. Since 1 and r 1 are equal at A and at B, they are equal everywhere on the polygon, and therefore 1 = r 1 in D n. Equivalent way of writing 1 = r 1 are (ince 1 = ) (3.2) r = r 1, = r 1. What thee mean i that when calculating in D n we can move r to the other ide of by inverting it. By induction (or by raiing both ide of (3.1) to an integral power) check (3.3) r k = r k, r k = r k for any integer k. In other word, any power of r can be moved to the other ide of by inveion. For example, conitent with r k being a reflection we can check it quare i trivial: (r k ) 2 = r k r k = r k r k = 2 = 1. The relation (3.2) involve a particular rotation and a particular reflection in D n. In (3.3), we extended (3.2) to any rotation and a particular reflection in D n. We can extend (3.3) to any rotation and any reflection in D n : a general reflection in D n i r i, o by (3.3) In the other order, (r i )r j = r i r j = r j r i = r j (r i ). r j (r i ) = r i r j = r i r j = (r i )r j. Thi ha a nice geometric meaning: when multiplying in D n, any rotation can be moved to the other ide of any reflection by inverting the rotation. Thi geometric decription make uch algebraic formula eaier to remember. Let put thee formula to work by computing all the commutato in D n. Theorem 3.2. The commutato in D n are the ubgroup r 2. Proof. The commutator [r, ] i r 1 1 = r 1 = r 2, o r 2 i a commutator. More generally, [r i, ] = r i r i 1 = r i r i 1 = r 2i, o every element of r 2 i a commutator. To how every commutator i in r 2, we will compute [g, h] = ghg 1 h 1 when g and h are rotation or reflection and check the anwer i alway a power of r 2. Cae 1: g and h are rotation. Since any two rotation commute, ghg 1 h 1 i trivial. Cae 2: g i a rotation and h i a reflection. Write g = r i and h = r j. Then h 1 = h, o ghg 1 h 1 = ghg 1 h = r i r j r i r j = r i+j r (j i) = r 2i. Cae 3: g i a reflection and h i a rotation. Since (ghg 1 h 1 ) 1 = hgh 1 g 1, by Cae 2 the commutator hgh 1 g 1 i a power of r 2, o paing to it invee tell u that ghg 1 h 1 i a power of r 2.
6 6 KEITH CONRAD Cae 4: g and h are rotation. Write g = r i and h = r j. Then g 1 = g and h 1 = h, o ghg 1 h 1 = ghgh = (gh) 2 = (r i r j ) 2 = (r i j ) 2 = r 2(i j). 4. Conjugacy and Center While the number of reflection in D n (n 3) ha one formula for all cae, namely n, the geometric decription of reflection depend on the parity of n: for odd n all line of reflection look the ame, but for even n thee line fall into two type. See Figure 1 and 2. Figure 1. Line of Reflection for n = 3 and n = 5. Figure 2. Line of Reflection for n = 4 and n = 6. The different geometric decription of reflection in D n for even and odd n manifet themelve algebraically when we decribe the conjugacy clae of D n. Theorem 4.1. The conjugacy clae in D n are a follow. (1) If n i odd, the identity element: {1}, (n 1)/2 conjugacy clae of ize 2: {r ±1 }, {r ±2 },..., {r ±(n 1)/2 }, all the reflection: {r i : 0 i n 1}. (2) If n i even, two conjugacy clae of ize 1: {1}, {r n 2 }, n/2 1 conjugacy clae of ize 2: {r ±1 }, {r ±2 },..., {r ±( n 2 1) }, the reflection fall into two conjugacy clae: {r 2i : 0 i n 2 1} and {r 2i+1 : 0 i n 2 1}.
7 DIHEDRAL GROUPS 7 In word, the theorem ay each rotation i conjugate only to it invee (which i another rotation except for the identity and, for even n, the 180 degree rotation r n/2 ) and the reflection are all conjugate for odd n but break up into two conjugacy clae for even n. The two conjugacy clae of reflection for even n are the two type we ee in Figure 2: thoe whoe fixed line connect oppoite vertice (r even ) and thoe whoe fixed line connect midpoint of oppoite ide (r odd ). Proof. Every element of D n i r i or r i for ome integer i. Therefore to find the conjugacy cla of an element g we will compute r i gr i and (r i )g(r i ) 1. The formula r i r j r i = r j, (r i )r j (r i ) 1 = r j a i varie how the only conjugate of r j in D n are r j and r j. Explicitly, the baic formula r j 1 = r j how u r j and r j are conjugate; we need the more general calculation to be ure there i nothing further that r j i conjugate to. To find the conjugacy cla of, we compute r i r i = r 2i, (r i )(r i ) 1 = r 2i. A i varie, r 2i run through the reflection in which r occu with an exponent diviible by 2. If n i odd then every integer modulo n i a multiple of 2 (ince 2 i invertible mod n o we can olve a 2i mod n for i given any a). Therefore when n i odd {r 2i : i Z} = {r i : i Z}, o every reflection in D n i conjugate to. When n i even, however, we only get half the reflection a conjugate of. The other half are conjugate to : r i ()r i = r 2i+1, (r i )()(r i ) 1 = r 2i 1. A i varie, thi give u {, r 3,..., r n 1 }. The center i another apect where D n behave differently when n i even and n i odd. Theorem 4.2. When n 3 i odd, the center of D n i trivial. When n 3 i even, the center of D n i {1, r n/2 }. Proof. The center i the et of element that are in conjugacy clae of ize 1, and we ee in Theorem 4.1 that for odd n thi i only the identity, while for even n thi i the identity and r n/2. Example 4.3. The group D 3 ha trivial center. The group D 4 ha center {1, r 2 }. Geometrically, r n/2 for even n i a 180degree rotation, o Theorem 4.2 i aying in word that the only nontrivial rigid motion of a regular polygon that commute with all other rigid motion of the polygon i a 180degree rotation (when n i even).
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