Algebra. Übungsblatt 5 (Lösungen)

Transcription

1 Fakultät für Mathematik Sommersemester 2017 JProf. Dr. Christian Lehn Dr. Alberto Castaño Domínguez Algebra Übungsblatt 5 (Lösungen) Aufgabe 1. Für eine Gruppe G und ein g G sei der Zentralisator von g als definiert. Z(g) = {h G hg = gh} 1. Bestimmen Sie für n N die Zentren der symmetrischen Gruppe S n sowie der Diedergruppe D n 2. Was ist der Zentralisator von (2 3 4) in S 5 bzw. in D 5? 3. Was ist der Zentralisator von (1 2 3)(4 5 6) in S 7? Proof. 1. Let us begin with the symmetric group. If n = 2, S n = C2, which is abelian, so Z(S 2 ) = S 2 in this case, so let us assume that n 3. From exercise 1.1 of sheet 1, we know that for any permutation σ and any k-cycle π = (x 1 x k ), both in S n, σπσ 1 = (σ(x 1 ) σ(x k )). (0.1) We will use that fact (or variants of it) in the rest of the sheet. For the moment, take a permutation π S n and let 1 i, j n two different indexes such that π(i) = j. Now choose any transposition σ = (j k), where k i, j (it exists, because n 3). Then, σπσ 1 (i) = σπ(i) = σ(j) = k, whereas π(i) = j k. This shows that for any permutation we chose we can always find some conjugate to it that does not send all the elements of {1,..., n} to the same elements, so it cannot be the same permutation. In conclusion, Z(S n ) = {e} for all n 3. The case of the dihedral group is simpler to handle but more interesting. Again, when n = 2 everything is quite obvious: D 2 = V4, which is again abelian, so Z(D 2 ) = D 2. Assume then from now on that n 3. Remember that the dihedral group could be presented as d, s d n = s 2 = e, sd = d 1 s.

2 We will also use that several times afterwards. Take then an element d a s b D n, and let us see whether it belongs to the center of not. Let then d j s k D n be any other element from the group and impose that d a s b d j s k = d a+( 1)bj s b+k = d j+( 1)ka s b+k = d j s k d a s b. That would happen if and only if a + ( 1) b j j + ( 1) k a mod n, for every (j, k) {0, 1,..., n 1} {0, 1}. If b = 1, then we would have that a j j + ( 1) k a, which cannot happen for every pair j, k: if k = 0 then j j. Then b should vanish, but then we would have that a ( 1) k a, that is to say, a must be congruent modulo n to a, which happens if and only if n is even and a = n/2. Summing up, the center of the dihedral group is { d Z(D n ) = n/2 n even {e} n odd. 2. An equivalent way of conceiving the centralizer of a permutation π is as the set of permutations σ that leave π invariant after conjugation, that is, σπσ 1 = π. Let τ = (2 3 4). By formula 0.1, for any σ S n στσ 1 = (σ(2) σ(3) σ(4)) =: τ σ. If we impose that τ = τ σ, then necessarily σ(2), σ(3) and σ(4) must be 2, 3 and 4 in the same order up to some cyclic permutation. Apart from those values, σ can be arbitrary on the rest, so it can exchange 1 and 5. In other words, σ (1 5), (2 3 4), so Z ( (2 3 4) ) (1 5), (2 3 4), but the opposite inclusion is trivial, so we have the equality. There is no 3-cycle in D 5. We can see it with the presentation of D n as a subgroup of S n given in class, with generators ( ) 1 2 n d = (1 2 n), s = = (2 5)(3 4; 1 n 2 all the elements of D 5 are either 5-cycles or product of two disjoint transpositions. However, to compensate the short answer, let us calculate the centralizer of any element in D 5. In order to do so we will consider separate cases according to the expression of such elements. Case 0: Obviously Z(e) = D 5. Case 1: Z(s). Let d a s b D 5 such that sd a s b = d a s b s. That is equivalent to say d a s b+1 = d a s b+1, or d a = d a. Since 5 is odd, a must vanish and so Z(s) = s.

3 Case 2: Z(d k ). Let again d a s b D 5 such that d k d a s b = d a s b d k. In other words, d a+k s b = d a+( 1)bk s b, or equivalently, d k = d ( 1)bk. Using again that 5 is odd the only possibility is that b = 0, so Z(d k ) = d for any k 0. Case 3: Z(d k s). As before, consider an element d a s b D 5 such that d k sd a s b = d a s b d k s. It is the same saying that d k a s b+1 = d a+( 1)bk s b+1, or d k a = d a+( 1)bk. Once again, because of 5 being odd, k a cannot be congruent modulo 5 to a k, so b must be zero, but then again we reach to a contradiction unless a = 0 (a a mod 5), so Z(d k s) = {e} for all k Define τ = (1 2 3)(4 5 6). As done in the previous paragraph, thanks to formula 0.1 we know that for any permutation σ S n, στσ 1 = (σ(1) σ(2) σ(3))(σ(4) σ(5) σ(6)) =: τ σ. As before, we need that τ = τ σ, but now we have somewhat more freedom, because σ can exchange the indexes of τ 1 := (1 2 3) and τ 2 = (4 5 6) and not only leave some of them invariant. A similar argument as above tells us that both τ 1 and τ 2 are in the centralizer, but also any other permutation π which exchanges the two 3-cycles, up to a cyclic reordering of both of them, before or after applying. The basic one is π = (1 4)(2 5)(3 6), and any other permutation exchanging τ 1 and τ 2 is a product of π, τ 1 and τ 2 : as we said, reordering using the τ i before and after exchanging them, we can assume that we are exchanging the elements placed in the same positions of the permutations, that is to say, applying π. In the end we get that Z(τ) = (1 2 3), (4 5 6), (1 4)(2 5)(3 6). In fact, the discussion above shows only the inclusion from the left to the right, but the opposite one is trivial. Aufgabe 2. Wir beschäftigen uns mit den Konjugationsklassen von S n. Nach Satz II.1.9 aus der Vorlesung lässt sich ein σ S n (bis auf Reihenfolge) eindeutig als Produkt elementfremder Zykel schreiben. Es sei σ = σ 1... σ k eine solche Produktdarstellung, wobei σ i ein r i -Zykel sei. Durch Umsortieren können wir erreichen, dass r 1 r 2... r k gilt. Wir nennen (r 1,..., r k ) den Zykeltyp von σ. Es gilt selbstverständlich r = r r k n. Wir ordnen dem Zykeltyp die Partition (r 1,..., r k, 1,..., 1) von n zu, wobei wir n r Einsen ergänzen. Es gilt also: r r k + } 1 +. {{ } = n. (n r)-mal Zeigen Sie die folgenden Aussagen:

4 1. Zwei Permutationen in S n sind genau dann konjugiert, wenn sie denselben Zykeltyp haben. 2. Die obige Zuordnung definiert eine Bijektion von der Menge der Zykeltypen von S n auf die Menge der Partitionen von n. 3. Es bezeichne P (n) die Anzahl der Partitionen von n. Dann gilt: P (n)x n 1 = 1 x k Proof. n=0 k=1 (Sie brauchen sich nicht um Konvergezfragen zu kümmern.) 1. On one hand, assume that a permutation σ is conjugate to some other τ via π. Say that τ is of cycle type (r 1,..., r k ) and write τ as a decomposition of disjoint cycles: Then, by formula 0.1 again, τ = (a 11 a 1r1 ) (a k1 a krk ). σ = (π(a 11 ) π(a 1r1 )) (π(a k1 ) π(a krk )). Since all the cycles above were disjoint and permutations are bijective maps of a given finite set, the cycles below will also be disjoint, so the cycle type of σ is also (r 1,..., r k ). Now assume that both of them have the same cycle type (r 1,..., r k ), and write τ = (a 11 a 1r1 ) (a k1 a krk ), σ = (b 11 b 1r1 ) (b k1 b krk ). Now choose any bijection of {1,..., n} sending a ij to b ij. That can be thought of as a permutation of S n (call it π), and by formula 0.1, we would have that πτπ 1 = σ, so that τ and σ would be conjugate each other. 2. By construction any cycle type gives rise to a unique partition of n (the correspondence is a mapping), and for such partitions, there cannot be two different cycle types corresponding to it, because both are ordered in the same way (the mapping is injective). Let us see that such correspondence is also surjective. Indeed, for any partition (r 1,..., r m ) of n, where only the first k numbers are strictly bigger than 1, we can construct the permutation σ = (1 2 r 1 )(r r r 1 + r 2 ) (r r k r r k ), which clearly has cycle type (r 1,..., r k ). In fact, using the first point we can show that there is a bijective correspondence between the partitions of n and the conjugation classes of S n.

5 3. This fact is quite known, going back to Euler. Express each factor at the right as an infinite series, 1 1 x = x a k. k Then k=1 a k =0 1 1 x = k k=1 a k =0 x ka k. With this expression and some results from analysis it is clear that the product converges in the disk x < 1, because every factor is a convergent series in the same domain with independent term 1, such that the product at the right-hand side has only a finite amount of summands corresponding to a given power of x. Now pay attention to the right-hand side above. If we write it as a series n p nx n it is clear that p n will be the sum of all the possible ways of writing n as a sum a 1 + 2a ka k + of positive integers. But each sum like that one represents one partition: say that n = a 1 + 2a Na N. Then we get (N, (a N )..., N,..., 2, (a.. 2)., 2, 1, (a.. 1)., 1). Since the coefficients of the series we multiply are all one, we add one to p n for each partition of n, so in the end p n = P (n). Aufgabe 3. Man betrachte die Diedergruppe G = D n und zeige, dass G für gerades n genau n+6 und für ungerades n genau n+3 Konjugationsklassen enthält. 2 2 Hinweis: Bestimmen Sie die Mächtigkeit der Konjugationsklassen getrennt für Spiegelungen und Drehungen. Unterscheiden Sie hierbei die Fälle: n gerade, n ungerade. Proof. Let n be even and start with the symmetry s. Take then any element of D n of the form x = d a s b and consider d a s b ss b d a, which is d 2a s. Since n is even, 2a only covers half of the possible exponents for d. Let us see that there are no more elements in that conjugation class. Indeed, take another d j s k D n. Then d j s k d 2a ss k d j = d 2j+2( 1)ka s, which can also be written as d a s for an even exponent a. Now if x = ds, all its conjugates can be written as d a s b dss b d a = d 2a+( 1)b s, where the exponent of d will always run along {1, 3,..., 2n 1} because of n being even. Analogously as before, now taking any member of that conjugation class, say d a s for an odd a, their conjugates lie also in the same class, since d j s k d a ss k d j = d 2j+( 1)ka s. Since a was odd, 2j + ( 1) k a so would be for any value of j.

6 The next elements we can think of are those of the form d a. Their conjugates will be d j s k d a s k d j = d ( 1)ka, that is, d a itself or its inverse d a. It is clear that the class contains only those two elements, which are the same in the case of a being n/2. Summing up, when n is even, we have one class for the neutral element, one for s and the d a s of even powers of d, another for the d a s of odd powers of d and n/2 classes coming from the powers of d alone. In total they are n/2 + 3 = (n + 6)/2. And now the work is done; if n is odd following the same argument as before we could argue that the conjugation class of s also contains all the elements of the form d a s. Counting all the classes we get 1 for the neutral element, another for the symmetry s and (n 1)/2 for the powers of d. They add up to (n 1)/2 + 2 = (n + 3)/2. Aufgabe 4. Schauen Sie sich die Definition der Platonischen Körper in Ihrem Lieblingsbuch 1 an. Ist P einer der Platonischen Körper, so bezeichnen wir mit S(P ) seine Symmetriegruppe (Drehungen und Spiegelungen) und mit D(P ) seine Drehgruppe. 1. Warum gilt S(P ) = S(P ) und D(P ) = D(P ), falls (P, P ) = (Oktaeder, Hexaeder) bzw. (Ikosaeder, Dodekaeder)? 2. Zeigen Sie, etwa durch Durchnummerieren der Ecken, dass S(P ) = S 4, D(P ) = A 4, falls P das Tetraeder ist. Geben Sie Erzeuger der Gruppen und deren geometrische Interpretation an. 3. Zeigen Sie, dass S(P ) = S 4 Z/2Z, D(P ) = S 4, falls P das Hexaeder ist. Geben Sie Erzeuger für D(P ) an. Hinweis: Zeigen Sie, dass S(P ) von D(P ) und der Punktspiegelung am Ursprung (d.h., am Zentrum des Hexaeders) erzeugt wird. 4. Zeigen Sie, dass S(P ) = A 5 Z/2Z, D(P ) = A 5, falls P das Ikosaeder ist. Proof. The symmetry groups of the Platonic solids are those isometries of R 3 (with the usual Euclidean distance) that leave the figure invariant. It can be proved that all those isometries are linear, consisting of turns, who preserve the orientation of the solid, and turns and symmetries, changing or not the orientation. Since we are already given that here, let us take advantage from it. 1. If we take an hexahedron and focus on the middle points of its six faces, we can form the convex polyhedron having them as vertices. That new polyhedron is regular, and is an octahedron. If we do the same with the octahedron, we obtain an inscribed hexahedron this time. This is the duality of the pair octahedron, hexahedron. All this is represented in figure 1. 1 Wir gehen davon aus, dass Ihr Lieblingsbuch eine Definition der Platonischen Körper enthält.

7 Figure 1: Octahedron inscribed in hexahedron and hexahedron inscribed in octahedron Figure 2: Dodecahedron inscribed in icosahedron and icosahedron inscribed in dodecahedron Now every turn or symmetry that leaves one of our figures invariant must then exchange the vertices of them. If they preserve the location of the vertices, they will also do the same with the edges, and the faces of the solids, and finally with the middle points of the faces, too. Therefore, any affinity preserving the hexahedron would leave its inscribed octahedron invariant as well, and viceversa, so their symmetry groups will be the same. The same situation appears when dealing with the dodecahedron and the icosahedron (figure 2), so the argument can be repeated mutatis mutandis. Note that we need both inscriptions; an octahedron can be inscribed in a tetrahedron such that the vertices of the octahedron are the middle points of the edges of the tetrahedron. However, there is no reciprocal inscription, so there are more isometries leaving the octahedron invariant than those fixing the tetrahedron. 2. The tetrahedron has four vertices, so every turn or symmetry of it will induce a permutation of those four points, that is to say, we will have a mapping S(T ) S 4, and also its restriction D(T ) S 4. Both of them will be in fact group homomorphisms just by construction, because the composition of two isometries must go to

8 Figure 3: A turn of the tetrahedron the composition of the corresponding permutations. In fact, both homomorphisms are injective: The four vertices of the tetrahedron form a reference system for the affine space R 3, so if there were an affine linear map in D(T ) or S(T ) such that all the vertices remained in their positions, that map should be the identity, because it would fix a reference system of the affine space. As a consequence, both D(T ) and S(T ) are subgroups of S 4. Let us prove that D(T ) = A 4 and S(T ) = S 4. Notice that every turn of angle ±120 o and of axis the line passing through one vertex and the middle point of the opposite face exchange cyclically the three remaining vertices (figure 3). Then all the 3-cycles are in D(T ); since A 4 is generated by the 3-cycles, then A 4 D(T ), so we will have the inclusions A 4 D(T ) S(T ) S 4. If we apply now a symmetry with respect to the plain passing through the green and blue vertices and the middle point of the edge joining the other two (figure 4), which can be seen as a 2-cycle, we obtain again the same tetrahedron but with a different orientation: note, for example, that the path green-red-yellow is clockwise at the left, whereas at the right is anticlockwise. On the other hand, every turn of the tetrahedron will preserve the orientation of its faces; we are just altering the point of view, not the figure itself. Then there is a permutation that belongs to S(T ) but not to D(T ). Since there is no intermediate group between A 4 and S 4 we conclude that D(T ) = A 4 and S(T ) = S 4. Regarding the generators, we have already studied them. Just for the sake of clearness, we repeat the information: D(T ) is generated by 3-cycles, namely the turns of ±120 o around the lines passing through one vertex and the middle point of the opposite face. On the other hand, S(T ) is generated by the transpositions, which are nothing else but the symmetries with respect the plane passing two vertices and the middle point of the edge joining the other two. With those generators we can compose the remaining turns and symmetries as in S A priori, following the same argument as before with the tetrahedron, we could claim

9 Figure 4: A symmetry of the tetrahedron that both D(H) and S(H) are subgroups of S 8, and we would not be wrong. Anyway, any finite group can be expressed as a subgroup of a big enough symmetric group, so that is not such a surprise and we should be more precise. For example, using the first point we could study the symmetry groups of the octahedron, which has only six vertices, so that D(H) and S(H) could also be seen as subgroups of S 6, yet we would be a bit far from our goal 2. Let us focus first on the group of turns D(H). The cube is a symmetric figure with respect the origin given by its middle point O, so any turn must leave that center invariant. Now consider the lines passing through a vertex and O. Since the hexahedron is symmetric with respect to O, those lines will pass in fact through two opposite vertices, but then, every turn will exchange the whole lines as well, not only the vertices. There are four couples of opposite vertices, and so four couples of lines. In the end, every turn exerts an action on the set of those lines, so following the argument from the last point, adapted to our case, we must have a group homomorphism D(H) S 4. The homomorphism above is again injective: we can choose four vertices of the cube so that their associated lines are different. Then they will form a reference system for the affine space R 3, so as in the previous paragraph, any turn, or even linear affine map reserving each individual point will necessarily be the identity. Then D(H) is indeed a subgroup of S 4. Let us show that it contains a transposition. Indeed, think of the turn of 180 o along the plane containing O and a pair of vertices opposite one to the other in a face (and then containing two lines of opposite vertices), like in figure 5. It will exchange the other remaining pair of couples of opposite vertices, so it will represent a transposition. In fact, any other transposition can be realised in such a way (there are ( 4 2) = 6 transpositions, corresponding to all the possible pairs of opposite vertices of the six faces, up to considering opposite vertices with respect to O), so D(H) will contain all the transpositions of S 4 and then D(H) = S 4. The rest of turns can be composed from those especial ones. For instance, it is easy to see 2 But not too far; S 6 is the smallest group in which we can embed S 4 C 2 = S4 (5 6)

10 Figure 5: A turn of the hexahedron Figure 6: Another turn of the hexahedron that it contains a 4-cycle, given by a ±90 o turn around a line passing through the middle points of opposite faces (figure 6), so we could also write D(H) as generated by a transposition and a 4-cycle. So that finishes the calculation of D(H). Let us go now for S(H). We know that the cube is symmetric with respect to O, so let us call ι that symmetry. With our notation, it will exchange couples of vertices of the same colour, but will not do anything else. Thanks to that, it will commute with every turn: any composition ισι (with σ being a turn) would change first the orientation of the lines joining vertices of the same colour, then exchange the lines as σ does and them reversing the orientation of those lines, that is to say, exchange the lines exactly as σ without changing the orientation of the cube. In other words, ισι = σ. This discussion tells us that S 4 C 2 is a subgroup of S(H) 3. Now think of any other isometry fixing the hexahedron. If it changes the orientation of the faces, by applying ι afterwards we would have another one preserving the orientation. Thanks to all we have done before, that must be an element of D(H). Therefore, S 4 C 2 = S(H) and we are done. 3 Note that the direct product comes from the commutativity of ι with the turns.

11 4. This point can be done as well, following similar ideas to the ones in the two previous paragraphs (for instance, every turn of the icosahedron would preserve the compound of five cubes that can be inscribed into it). Nevertheless, the exposition of the proof would lie a bit beyond the skills (especially those relative to drawing 4 ) and time availability of this humble Übungsleiter. If you, dear and interested reader, want to gain insight in this specific topic, two good references are this Wikipedia page and section 7.4 of wonderful Michael Artin s book Algebra. In general, the usual names of D(P ) and S(P ) for a polyhedron (or any other solid) P are groups of chiral (from the greek χειρ, hand ) and achiral symmetries, or group of symmetries and full group of symmetries. Moreover, one deals first with isometries, then proves that they must be affine linear maps and then that they must be composition of turns and reflections 5. Credits of the images: Livio Zefiro (1, 2). 4 Although, as you already know, geometry is the art of making beautiful mathematics with ugly drawings. 5 And in fact, every affine isometry is a composition of reflections by Cartan Dieudonné theorem.