FRIEZE GROUPS IN R 2

Transcription

1 FRIEZE GROUPS IN R 2 MAXWELL STOLARSKI Abstract. Focusing on the Eucidean pane under the Pythagorean Metric, our goa is to cassify the frieze groups, discrete subgroups of the set of isometries of the Eucidean pane under the Pythagorean metric whose transation subgroups are infinite cycic. We begin by deveoping a norma form for representing a isometries. To simpify the probem of composing isometries written in norma form, we sha discuss the properties of compositions of refections across different axes. Such a discussion wi naturay ead to a cassification of a isometries of the pane. Using such knowedge, we can then show that there are ony seven geometricay different types of frieze groups. Contents 1. Metric Spaces, Groups, and Isometries 1 2. E and the Norma Form Theorem 3 3. Generators and Reations 5 4. Compositions of Refections Across Different Axes 5 5. Cassification of Frieze Groups 9 Acknowedgments 13 References Metric Spaces, Groups, and Isometries Before we can begin to tak about frieze groups, we must discuss certain necessary preiminary concepts. We begin by formaizing the concept of distance between points in a set. Definition 1.1. A metric on a set X is a map d : X X R such that i) d(x, y) 0 for a x, y X and d(x, y) = 0 if and ony if x = y; ii) d(x, y) = d(y, x) for a x, y X; iii) d(x, y) + d(y, z) d(x, z) for a x, y, z X. A set X with a metric d is caed a metric space, written (X, d). We are specificay interested in distance preserving maps from a metric space to itsef and so we present Definition 1.2. Definition 1.2. An isometry of a metric space (X, d) is a bijection u : X X such that, for a x, y X, d(x, y) = d(xu, yu). (Note that for a map u : X Y and an eement x X we denote the image of x by xu instead of u(x).) We et Isom(X, d) denote the set of isometries of a metric space (X, d). In other words, Isom(X, d) is the set of functions from a set to itsef that preserve distance. Whie there are many such sets, we sha dea amost excusivey with 1

2 2 MAXWELL STOLARSKI the set of isometries of the Eucidean pane under the Pythagorean metric. We sha refer to this set as E, formay defined as E := Isom(R 2, d) where d(x, y) = 2 i=1 (x i y i ) 2, x = (x 1, x 2 ), and y = (y 1, y 2 ). Diverging from the topic of metric spaces, we present the foowing definitions, formaizations of fundamenta agebraic concepts: Definition 1.3. A group is a set X with a binary operation such that i) there is an identity eement 1 X such that, for a x X, 1x = x1 = x; ii) the binary operation is associative, i.e. for a x, y, z X, (xy)z = x(yz); iii) for every x X, there exists some y X (caed the inverse of x and written x 1 ) such that xy = yx = 1. Definition 1.4. A subgroup is a subset of a group G that forms a group under the same binary operation as G. We denote a subgroup H of G by writing H G. Definition 1.5. A subgroup H of a group G is said to be norma if Hx = xh for a x G. We denote a norma subgroup H of G by writing H G. There is, however, an equivaent definition of a norma subgroup that wi be usefu for ater proofs. Theorem 1.6. H G if and ony if x 1 Hx = H for a x G. Proof. Let x G and h H. Assume H G. By definition, Hx = xh, so hx = xh for some h H. Right mutipying both sides by x 1 yieds x 1 hx = h H, from which it foows that x 1 Hx H. Because x G is arbitrary, xhx 1 is aso a subset of H. Thus, H = x 1 xhx 1 x x 1 Hx and we have proved the theorem in one direction. For the other direction, assume x 1 Hx = H for a x G. Let x G and h H. By the assumption, x 1 hx = h H and right mutipication by x yieds hx = xh xh. Thus, Hx xh. Because x 1 Hx = H for a x G, it foows that xhx 1 = h H. By simiar ogic, xh Hx. Thus, xh = Hx and H G. The concepts of isometries and groups now combine in the foowing theorem. Note that, buiding off the previous notation, we write composition of maps f g as gf and (f g)(x) = f(g(x)) = xgf. Theorem 1.7. The set of isometries of a set X, i.e. under composition of maps. Isom(X), forms a group Proof. First, we need to show the existence of an identity eement. Consider the identity map 1 : X X, x x. The identity map 1 is obviousy a bijection. Furthermore, for a x, y X, d(x1, y1) = d(x, y) as x1 = x and y1 = y. Thus, 1 Isom(X). Let u Isom(X) and x X. Because x1u = xu = xu1, the identity map 1 Isom(X) is the identity eement such that, for a u Isom(X), u1 = 1u = u. Composition of maps is obviousy associative for x(tu)v = ((xt)u)v = xt(uv). Next, we need to show the existence of inverses, so et u Isom(X). Because u Isom(X), u is a bijection. Thus, there exists a map u 1 such that uu 1 = u 1 u = 1. Let x, y X. Because d(xu 1, yu 1 ) = d(xu 1 u, yu 1 u) = d(x1, y1) = d(x, y), u 1 Isom(X). Thus, for every u Isom(X), there exists u 1 Isom(X) such that uu 1 = u 1 u = 1.

3 FRIEZE GROUPS IN R 2 3 To compete the proof, we need to show that Isom(X) is cosed under composition of maps. In other words, we need to show that, for a u, v Isom(X), uv Isom(X). Let u, v Isom(X) and et x, y X. By definition, d(x, y) = d(xu, yu) and xu, yu X. It foows that d(xu, yu) = d(xuv, yuv). By the transitivity of equaity, d(x, y) = d(xuv, yuv). Therefore, uv Isom(X). 2. E and the Norma Form Theorem This section deas excusivey with E, the set of isometries of the Eucidean pane under the Pythagorean metric. The most we-known isometries of the pane are perhaps transations, rotations, and refections. We begin by formay defining these isometries. Later, the Norma Form Theorem wi show that every isometry of R 2 can be written as a composition of a refection, rotation, and transation. Definition 2.1. A transation t is a map that moves every point a fixed distance in a fixed direction. In symbos, for some x R 2 with Cartesian coordinates (x 1, x 2 ) t : (x 1, x 2 ) (x 1 + a 1, x 2 + a 2 ) where a 1 and a 2 are constants that define a constant vector a = (a 1, a 2 ). We denote such a transation by writing t = t(a) and ca the direction of a the axis of t. Transations are orientation preserving (OP) and have no fixed points, uness, of course, the transation is the trivia transation through the zero vector. Transations compose according to the foowing rue: If a = (a 1, a 2 ) and b = (b 1, b 2 ) are constant vectors in R 2, then t(a)t(b) = t(a + b) where t maps a point (x 1, x 2 ) in the pane to (x 1 + a 1 + b 1, x 2 + a 2 + b 2 ). It foows from this fact that the set of a transations in E form a subgroup of E. Definition 2.2. A rotation s of the pane is a map that moves every point through a fixed ange about a fixed point, caed the center. Taking the center O to be the origin of a poar coordinate system, s : (ρ, θ) (ρ, θ + α) where (ρ, θ) are the poar coordinates of an arbitrary point in R 2 and α is a fixed ange. We denote such a rotation by writing s = s(o, α). Rotations are order preserving and non-trivia rotations have ony one fixed point, namey the center. Rotations with the same center compose according to the foowing rue: s(o, α)s(o, β) = s(o, α + β) where s(o, α + β) : (ρ, θ) (ρ, θ + α + β) and α and β are constant anges. It foows that rotations about the same center O form a subgroup of E. Definition 2.3. A refection r is a map that moves every point in the pane to its mirror image across a ine. This ine is caed the axis of r and we denote such a refection by writing r = r(). In symbos, given a point P R 2, if P then P r = P. If P /, then P r is the unique point in R 2 such that is the perpendicuar bisector of P and P r. Refections are orientation reversing and fix ony the points on the axis of the refection. The inverse of a refection is itsef, i.e. r() 2 = 1. The composition

4 4 MAXWELL STOLARSKI of refections across different axes, as we as the composition of rotations about different centers, is the topic of discussion in Section 4. The proofs that transations, rotations, and refections are isometries of R 2 and satisfy their described properties regarding orientation, fixed points, and composition are somewhat trivia and wi not be discussed here. Having formaized the notions of transation, rotation, and refection, we seek to show that every isometry of the pane can be written as a composition of a refection, rotation, and transation. With that goa in mind, the foowing emma, that every isometry of R 2 is determined by its effect on 3 non-coinear points, wi be a key first step towards proving the Norma Form Theorem. Lemma 2.4. Let O, P, Q be 3 non-coinear points of R 2 and et u 1, u 2 E such that Ou 1 = Ou 2, P u 1 = P u 2 and Qu 1 = Qu 2. Then u 1 = u 2. Proof. Define u = u 1 u 1 2. Thus, O = Ou, P = P u, and Q = Qu. E is a group by Theorem 1.7, so u E. Let R R 2. Because u E, d(o, R) = d(ou, Ru) = d(o, Ru). Thus, Ru ies on circe C 1 center O radius d(o, R). By simiar ogic, Ru ies on circe C 2 center P radius d(p, R). Thus, R, Ru C 1 C 2. Because O, P, Q are non-coinear, O P. It foows that C 1 C 2 {1, 2}. If C 1 C 2 = 1, then R = Ru and we are done. Otherwise, C 1 C 2 = 2, C 1 C 2 = {R, R }, and Ru {R, R }. It foows that ine OP is the perpendicuar bisector of RR. Equivaenty, for any point S R 2, d(s, R) = d(s, R ) if and ony if S OP. Because O, P, Q are non-coinear, Q / OP, so d(q, R) d(q, R ). Because d(q, R) = d(qu, Ru) = d(q, Ru), R Ru. Therefore, R = Ru for a R R 2 and we have proved the emma. Q R O P R Theorem 2.5. (Norma Form Theorem) Fix a point O and a ine R 2 such that O. Any u E can be written uniquey as u = r ɛ st where r is a refection over axis, ɛ {0, 1}, s is a rotation about center O, and t is a transation. Proof. Let t be a transation such that Ou = Ot. Thus, Out 1 = O. Let P such that P O. It foows that 0 < d(o, P ) = d(out 1, P ut 1 ) = d(o, P ut 1 ). Thus, both P and P ut 1 ie on the circe center O with radius d(o, P ) = d(o, P ut 1 ). It foows that there exists a rotation s about O such that P s = P ut 1. Thus, P = P ut 1 s 1. Because s is a rotation about O, Os = O, from which it foows that O = Out 1 s 1. Let Q /. Furthermore, d(o, Q) = d(out 1 s 1, Qut 1 s 1 ) = d(o, Qut 1 s 1 ) and d(p, Q) = d(p ut 1 s 1, Qut 1 s 1 ) = d(p, Qut 1 s 1 ). By simiar ogic as in Lemma 2.4, Qut 1 s 1 equas either Q or the refection of Q over ine. In the case of the former, ɛ = 0. In the case of the atter, ɛ = 1. In

5 FRIEZE GROUPS IN R 2 5 both cases, O and P remain fixed. It foows that, since O, P, Q are non-coinear, u = r ɛ st by Lemma 2.4. To prove uniqueness, suppose r ɛ st = r δ s t where ɛ, δ {0, 1}, s and s are rotations about O, and t and t are transations. If this isometry is orientation preserving, then ɛ = δ = 0. Otherwise, this isometry is orientation reversing and ɛ = δ = 1. Right mutipying both sides by r, if necessary, yieds st = s t. Further mutipication yieds s 1 s = t t 1 which is both a transation and a rotation about O. Because the rotation fixes O and ony the trivia transation has fixed points, s 1 s = t t 1 must equa the identity map 1. It foows that s = s and t = t. 3. Generators and Reations We now take a short break from isometries of the Eucidean pane to discuss the notation that wi be used to describe and cassify the frieze groups. We write G = X R where the symbos X, R, G have the foowing meaning: The set X of generators consists of symbos, usuay finite in number, say x 1,..., x n, n N {0}. We think of the symbos x ±1 i, 1 i n, as etters in an aphabet X ± from which words can be formed. The ength of a word is the number of its etters, assumed to be finite, and we aow the empty word e of ength zero. A word is reduced if it does not invove the etters x ±1 i in adjacent paces for any i {1,..., n}, and the set of a reduced words is denoted by F (X). The set R of defining reations consists of reations, that is, equations between words, usuay finite in number, say u i = v i, where u i, v i F (X), i {1,..., m}, m N {0}. We say that X R is a presentation of a group G or, equivaenty, G = X R, if the foowing three conditions are satisfied: i) every eement of G can be written as a word in X ± ; ii) the equations in R a hod in G; iii) any equation between words in X ± that hods in G is a consequence of the reations in R. Before we forget the definition of generators, we sha present the foowing emma that wi ater be usefu in the cassification of frieze groups. Lemma 3.1. Let T G, t be a generator of T, and r G. Then, r 1 tr is a generator of the group r 1 T r = {r 1 xr x T }. Proof. Let x r 1 T r. Because t is a generator of T, it foows that x = r 1 t r for some Z. Note that (r 1 tr) = r 1 trr 1 tr...r 1 tr = r 1 t r. Thus, x = (r 1 tr). Because x r 1 T r is arbitrary, r 1 tr is a generator of r 1 T r. 4. Compositions of Refections Across Different Axes Returning once again to isometries of the pane, we now take a ook at compositions of refections across different axes. Doing so provides a powerfu too with which to compose and cassify isometries in E and wi furthermore simpify the cassification of frieze groups. As different axes of refection can easiy become confused, many diagrams have been incuded in this section to iustrate the various axes and ines. We now begin with the composition of two refections across different axes. Theorem 4.1. Let r, r E be refections across distinct ines,, respectivey. If, then rr is a rotation about the point of intersection of and through

6 6 MAXWELL STOLARSKI an ange twice that from to. If, then rr is a transation in the direction norma to and through distance twice that from to. Proof. Let r and r be refections across ines and, respectivey. Assume. Either or. Case 1: Suppose. Thus, = {O} where O is the point of intersection of ines and (see figure). Let P R 2. Taking O as the origin and as the axis, we can write P in terms of poar coordinates P = (ρ, θ). Thus, r : (ρ, θ) (ρ, θ). Let P r = P = (ρ, φ). Because bisects P OP, θ+φ 2 = α and it foows that φ = 2α θ. Thus, rr : (ρ, θ) (ρ, 2α + θ) and, therefore, rr = s(o, 2α). In other words, rr is a rotation about O through twice the ange from to. Case 2: Suppose. Let P R 2. Taking as the x-axis, we can write P in terms of Cartesian coordinates P = (x, y). Thus, r : (x, y) (x, y) and r : (x, y) (x, z). Note that y+z 2 = a, so z = 2a y. It foows that rr : (x, y) (x, 2a + y). Therefore, rr = t(0, 2a), i.e. a transation through twice the distance between ines and. It is worth noting that Theorem 4.1 aso indicates that any transation or rotation can be written as the composition of two refections. In the case of a rotation s about some center O, simpy draw two ines,, both containing O, such that and the ange from to is haf the ange of rotation. It foows from Theorem 4.1 that r()r( ) = s. Simiary, in the case of a transation t through a vector a, draw two parae ines, such that and are perpendicuar to a and the distance between and equas haf the magnitude of a. By Theorem 4.1, r()r( ) = t. α O Case 1: a Case 2: Before continuing, we introduce a new isometry that wi appear in the next theorem. Definition 4.2. Given a pair of distinct points P, P on a ine, the isometry q(p, P ) = r()t( P P ) is caed a gide refection. (Note that t( P P ) is the unique transation such that P t = P.) Unike the isometries defined in Section 2, gide refections are both orientation reversing and have no fixed points. As the foowing theorem shows, gide refections aso arise from the composition of three refections across distinct axes. Theorem 4.3. The product of 3 refections in E is either a refection or gide refection according as the number of points of intersection of distinct axes is ess than or greater than 3 2. Proof. Let r 1, r 2, r 3 E be refections across distinct ines 1, 2, 3 respectivey. Because two distinct ines can have at most 1 point in common, it foows that the tota number n points of intersection of 3 distinct ines is either 0,1,2, or 3.

7 FRIEZE GROUPS IN R 2 7 Suppose n = 0. It foows by Theorem 4.1 that r 1 r 2 is a transation in the direction norma to 1 through distance 2a where a is the distance between ines 1 and 2. By Theorem 4.1 there is a ine such that refection r across ine composed with r 3 is a transation in the direction norma to through distance 2a. Because 3 1, r 1 r 2 = rr 3. Thus, r 1 r 2 r 3 = rr 3 r 3 = r. Therefore, r 1 r 2 r 3 is refection r across ine. Simiary, for the case n = 1, we can draw the ine such that r 1 r 2 = rr 3 where r is a refection across ine. By simiar ogic, r 1 r 2 r 3 is refection r across ine a a 3 α α Consider the cases n 2. Note that rotating axes 1 and 2 about their point of intersection whie keeping 3 fixed does not change r 1 r 2. Thus, we can transform any case in which n 2 to the case n = 2 and 2 3 without affecting the composition r 1 r 2 r 3. So assume n = 2 and 2 3. Let O be the point of intersection of ines 1 and 2 and et Q be the point of intersection of ines 1 and 3 (see figure beow). Draw ine such that O and 2. Note that 3. Finay, et P be the point of intersection of ines and 3. It foows that either P = Q or P Q. Suppose P Q. Let O = Or 1 r 2 r 3 and simiary define Q. Because r 1 fixes O, O = Or 1 r 2 r 3 = Or 2 r 3. By simiar ogic, Q = Qr 2 r 3. By Theorem 4.1, O and Q are simpy the images of O and Q, respectivey, under transation in the direction of through distance 2a where a is the distance between ines 2 and 3. Because r 1 r 2 r 3 E, d(o, P ) = d(or 1 r 2 r 3, P r 1 r 2 r 3 ), d(o, Q) = d(or 1 r 2 r 3, Qr 1 r 2 r 3 ), and d(p, Q) = d(p r 1 r 2 r 3, Qr 1 r 2 r 3 ). It foows by Side-Side-Side that OP Q and O P Q are congruent. Thus, P r 1 r 2 r 3 can be one of two points as indicated by P and P in the figure beow. However, because r 1 r 2 r 3 is orientation reversing, it foows that P r 1 r 2 r 3 = P. Now, draw ine through P and P. Let q be the composition of transation t aong through distance d(p, P ) foowed by a refection r across axis. Note that q = tr = rt. Inspection of the figure beow reveas that O = Oq, P = P q, Q = Qq. By Lemma 2.4, q = r 1 r 2 r 3 = tr = rt. Therefore, r 1 r 2 r 3 is a gide refection. Suppose P = Q. Thus, = 1, 1 2, and 1 3. It foows that r 1 is the refection r across ine and r 1 r 2 is a transation t through distance 2a aong ine. As in the previous case, we have r 1 r 2 r 3 = rt = tr and r 1 r 2 r 3 is a gide refection. 2 1

8 8 MAXWELL STOLARSKI 2 3 α α Q Q α P α a a a O P O P After having discussed the compositions of two and three refections, the obvious next step is to cover the composition of four refections, which, in turn, sha describe the composition of any number of refections. In particuar, the composition of five refections aows for an aternative norma form as presented in the statement of the foowing theorem. Theorem 4.4. Every non-trivia isometry of R 2 is the product of at most 3 refections, and is either a rotation, transation, refection, or gide refection according to the foowing tabe: Fixed points? Yes No OP Rotation Transation OR Refection Gide Refection Proof. Consider the product u of four refections. By Theorem 4.1, u is either the composition of two transations, two rotations, or one of each. If u is the composition of two transations, then u is a transation. Suppose u = st where s = s(o, α) is a rotation and t = t(a) is a transation. Let be the ine through O perpendicuar to the direction a. Let be the ine through O such that the ange between and is α 2. Let be the perpendicuar bisector of O and Ot. It foows by Theorem 4.1 that s = r( )r() and t = r()r( ). Thus, u = st = r( )r()r()r( ) = r( )r( ). Because, u is a rotation.

9 FRIEZE GROUPS IN R 2 9 Ot α 2 O Suppose u = ts. Thus, u 1 = s 1 t 1. By simiar ogic, u 1 is a rotation from which it foows that u is a rotation. Finay, suppose u is the composition of two rotations s(o, α) and s (O, α ). If O = O, then u is a rotation through α + α. Suppose O O. Let be the ine through O and O. Let be the ine through O such that the ange from to is α 2. Let be the ine through O such that the ange from to is α 2. It foows by Theorem 4.1 that s = r( )r() and s = r()r( ). Thus, u = r( )r( ) as r()r() = 1. There are now two cases: and. If, then u is a transation by Theorem 4.1 and this happens if and ony if the anges α α 2 and 2 are suppementary (i.e. sum up to ange π). Otherwise,. In this case, u is a rotation s(p, 2φ) where P = and 2φ = α + α. α 2 O α 2 O φ P α α O 2 2 O In summary, the product of four refections is either a transation or rotation which, in turn, is the product of two refections by Theorem 4.1. It foows that, when n 4, a product of n refections is a product of n 2 refections. By an obvious induction, it foows that when n 4, a product of n refections is a product of at most three refections. Now, consider the Norma Form Theorem from Section 2 which stated that any isometry in E coud be written as the composition of a refection, rotation, and transation. It foows from Theorem 4.1 that any isometry in E can then be written as the product of five refections, which, in turn, can be reduced to the product of at most three refections. The cases that then resut have aready been discussed in Theorems 4.1 and 4.3 and as such compete the proof of Theorem Cassification of Frieze Groups The effect of composing refections wi be enormousy usefu and often cited in our discussion of subgroups of E. Of the many such subgroups, we sha ony concern ourseves with those that are discrete, which we now define. Definition 5.1. A subgroup G of E is discrete if, for any point O R 2, every circe center O contains ony finitey many points in {Og g G}. Given a discrete subgroup G E, there is a transation subgroup T G defined as the set of a transations in G. Furthermore, there are, in fact, ony three types of transation subgroups of a discrete subgroup: the trivia subgroup of no transations, the free abeian group on two generators (i.e. t 1, t 2 t 1 t 2 = t 2 t 1 = Z 2 ),

10 10 MAXWELL STOLARSKI or the infinite cycic group. We sha imit our discussion to groups with the ast such type of transation subgroup. Definition 5.2. A frieze group is a discrete subgroup of E whose transation subgroup is infinite cycic, i.e. a subgroup of E whose transation subgroup is generated by ony one transation. There are, in fact, ony seven frieze groups. Before cassifying them, we sha need two emmas that show some usefu properties of transations. Lemma 5.3. Let r be a gide refection and et t be a transation such that the axes of r and t are parae. Then r and t commute. Proof. Taking the axis of r as the x-axis, we can write t = t(a) where a = (a 1, 0). Thus, rt and tr both map an arbitrary point with Cartesian coordinates (x, y) (x + a 1, y). It foows that r and t commute. Lemma 5.4. Let T G E where T is the transation subgroup of G. Then T is a norma subgroup of G. Proof. Let r be a refection and t = t(a) be a transation. Let P R 2 be an arbitrary point. Taking the axis of r as the x-axis we can describe the point P and the vector a in terms of Cartesian coordinates P = (x, y) and a = (a 1, a 2 ), respectivey. It foows that the image of P under rtr proceeds as foows: (x, y) (x, y) (x + a 1, y + a 2 ) (x + a 1, y a 2 ). Therefore, P rtr = P t where t is transation by the vector a = (a 1, a 2 ). Thus, we have shown that if t is a transation and r is a refection then rtr is aso a transation. Now, et u G be an arbitrary isometry of R 2 and et t be an arbitrary transation. We caim that u 1 tu is a transation. If u is the identity, then u 1 tu is obviousy the transation t. Otherwise, u is the product of at most 3 refections according to Theorem 4.4. We can then rewrite u as the product of refections say u = r 1 r 2 r 3. It foows that u 1 tu = r 3 r 2 r 1 tr 1 r 2 r 3. We have aready shown that if r is a refection then rtr is a transation. It thus foows from associativity that r 3 r 2 r 1 tr 1 r 2 r 3 is aso a transation. By simiar ogic, if u is the product of one or two refections, then u 1 tu is a transation. Because u 1 tu is a transation, u 1 tu T. Therefore, by Theorem 1.6, T is a norma subgroup. At this point, we have a the toos we need to cassify the seven types of frieze groups. Theorem 5.5. If F is a frieze group, then F is one of the seven possibe groups: F 1 = t F 1 1 = t, r r 2 = 1, r 1 tr = t F 2 1 = t, r r 2 = 1, r 1 tr = t 1 F 3 1 = t, r r 2 = t, r 1 tr = t F 2 = t, s t s = t 1, s 2 = 1 F 1 2 = t, s,, r s 2 = 1, t s = t 1, r 2 = 1, t r = t, (sr) 2 = 1 F 2 2 = t, s, r s 2 = 1, t s = t 1, r 2 = t, t r = t, (sr) 2 = 1 Proof. Let F be a frieze group and et T be the transation subgroup of F. Because F is a frieze group, T = t where t is a transation that generates T. Case 1: Suppose F contains no non-trivia rotations. If F contains no refections, then F = F 1 = t. Otherwise, F contains a refection or a gide refection. Suppose

11 FRIEZE GROUPS IN R 2 11 F contains a refection r. By Lemma 3.1, r 1 tr generates r 1 T r. Note that T is a norma subgroup by Lemma 5.4. Thus, r 1 T r = T by Theorem 1.6. It foows that r 1 tr generates T. Therefore, r 1 tr = t ±1. Each case generates a different frieze group so either F = F1 1 = t, r r 2 = 1, r 1 tr = t or F = F1 2 = t, r r 2 = 1, r 1 tr = t 1. (Note: If F aso contained a gide refection q, then rq, being an orientation preserving isometry that has no fixed points, woud be a transation in F. It woud then foow that q woud aready be generated by r and t, and, therefore, not generate a new frieze group.) There does, however, remain the case in which F contains a gide refection but no refection. So, instead of a refection, suppose F contains a gide refection r. Thus, r 2 = t h for some h Z. Because r and t commute by Lemma 5.3, it foows that (rt k ) 2 = r 2 t 2k = t 2k+h for a k Z. Choose k Z such that 2k + h = 0 or 1 and define r = rt k. Because r 2 = t 2k+h, r 2 = 1 or t. Aso, because r = rt k, it foows that the group generated by r and t is the same group as that generated by r and t, i.e. r, t = r, t. If r 2 = 1, then the the frieze group F1 1 is generated. Otherwise, r 2 = t and we have a new frieze group F1 3 = t, r r 2 = t, r 1 tr = t. Case 2: Suppose F contains a non-trivia rotation s. By simiar ogic as in Case 1, s 1 ts = t ±1. Because s is non-trivia, s 1 ts = t 1. Thus, s is a rotation through π and, consequenty, s 2 = 1. Suppose there exists some rotation s F such that s s. By simiar ogic, s is a rotation through π. Because s s, it foows from our anaysis of composition of rotations in the proof of Theorem 4.4 that ss is a transation. Thus, ss T. Mutipying both sides by s, we see that s st and, thus, s is generated by s and t. It foows that if F contains no refections then F = F 2 = t, s t s = t 1, s 2 = 1. Suppose F contains a refection or gide refection r with axis. By simiar ogic as before, r 1 tr = t ±1. Thus, either or where is the axis of t. Suppose. It foows that (sr) 1 t(sr) = t and sr is an orientation reversing isometry (i.e. a refection or gide refection) with axis parae to. Because (sr) 1 t(sr) = t and rtr = t 1, it foows that rstst = r and tstr = (sr) Thus, we can rewrite the generators without changing the group, so et sr repace the generator r. This simpifies the cases or to ony the case as sr is an orientation reversing isometry with axis parae to. Now we caim that = where denotes the axis of the orientation reversing isometry r and sha prove the caim by contradiction. Suppose and consider the point Or and the isometry r 1 sr. It foows that Orr 1 sr = Osr = Or as O is the center of s. Since r 1 sr is orientation preserving and fixes Or, r 1 sr must be a rotation by Theorem 4.4. To show that r 1 sr is a non-trivia rotation, consider the some point P R 2 such that P O. The image of P r under r 1 sr is P sr. Because P O, P s P from which it foows that that P sr P r as the isometry r preserves distance. Thus, P r is not fixed under r 1 sr, so r 1 sr is a non-trivia rotation of the pane. We have aready estabished that Or is fixed and a non-trivia rotations in frieze groups are rotations through ange π so it foows that r 1 sr = s(or, π). Because O Or and π 2 and π 2 are suppementary, it foows from our anaysis of composition of rotations in Theorem 4.4 that s(o, π)s(or, π) is a transation τ. By ooking at the image of O under τ we see that τ = t(2 OOr). Because and, τ is a transation that is not in the frieze group. This contradiction proves the caim that =.

12 12 MAXWELL STOLARSKI It now foows that if r is a refection then sr is a refection by Theorem 4.3 with axis m at O. Thus, F = F 1 2 = t, s, r s 2 = 1, t s = t 1, r 2 = 1, t r = t, (sr) 2 = 1. Otherwise, r is gide refection. By simiar ogic as in Case 1, we can change generators to get r 2 = t. Because r 2 = t, is the axis of both r and t. Let O be the center of the rotation s. Let M be the midpoint of ine segment OOt and the M be the midpoint of ine segment OM. Let m the ine containing O such that m and et m be the ine containing M such that m as iustrated in the figure beow. m m O M M Ot It foows by Theorem 4.1 that s = r(m)r() and r = r()r(m)r(m ). Thus, sr = r(m ) and, consequenty, (sr) 2 = 1. This gives the fina frieze group F2 2 = t, s, r s 2 = 1, t s = t 1, r 2 = t, t r = t, (sr) 2 = 1. Having successfuy competed our goa, we concude with iustrations of the seven frieze groups, understanding the horizonta ine to represent the axis of transation and the images to continue indefinitey. F 1 = t F 1 1 = t, r r 2 = 1, r 1 tr = t F1 2 = t, r r 2 = 1, r 1 tr = t 1 F 3 1 = t, r r 2 = t, r 1 tr = t F 2 = t, s t s = t 1, s 2 = 1

13 FRIEZE GROUPS IN R 2 13 F 1 2 = t, s,, r s 2 = 1, t s = t 1, r 2 = 1, t r = t, (sr) 2 = 1 F 2 2 = t, s, r s 2 = 1, t s = t 1, r 2 = t, t r = t, (sr) 2 = 1 Acknowedgments. I woud ike to thank my mentor Daniee Rosso for his knowedge and guidance on the subject of the paper. Further thanks go out to Peter May for eading the REU and the Nationa Science Foundation for their support. Finay, I woud ike to thank Michae and Kim Stoarski for their generous support with regards to housing, food, and transportation. [1] D. L. Johnson. Symmetries. Springer References