Group Theory and Applications MST30010

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1 Group Theory and Applications MST30010

2 ii

3 Contents 1 Material from last year 1 2 Normal subgroups and quotient groups 3 3 Morphims 7 4 Subgroups of quotient groups 13 5 Group actions Size of orbits Special case: The action by inner automorphisms The Sylow theorems (Sylow, 1872) 23 7 Application to combinatorics The question An answer, using group actions The crystalographic / wallpaper groups Isometries of the plane Complex numbers and isometries Point groups of isometry groups Crystallographic groups iii

4 iv CONTENTS 7 The Sylow theorems 6 An application to combinatorics 5 Group actions 8 Wallpaper groups (at the end) 4 Subgroups of quotient groups 3 Morphisms 2 Normal subgroups and quotients 1 Material from last year Structure of the course

5 Chapter 1 Material from last year We consider a group (G, ). By this we mean that G is a group, and the group operation on G is denoted by. The identity element of G will be denoted by e, sometimes by e G if we want to be very precise (for instance when dealing with several groups at the same time, to indicate which identity element we are considering). We will also (most of the time) not bother with writing the, i.e. we will simply write ab instead of a b. It is important to remember that the element e is the only element in G with the property that e x = x (or x e = x) for every x G. Practically, it means that if we want to show that some element z is equal to e, one way to proceed is to show that z x = x (or x z = x) for all (for some would be enough) x G. A subset H of G is called a subgroup of G if H, equipped with the operation, is a group with identity element e G. The following criterion is almost always used to check that a subset is a group Proposition 1.1. A subset H of G is a subgroup if and only if 1. H ; 2. for every a, b H, a b H; 3. for every a H, a 1 H. The sets a H := {a h h H} for a G are called the left cosets of H in G, while the sets H a := {h a h H} 1

6 2 CHAPTER 1. MATERIAL FROM LAST YEAR for a G are called the right cosets of H in G. We will mostly consider left cosets, and we will simply call them cosets of H, but every statement about left cosets can be formulated as a similar statement for right cosets. Cosets are interesting because they have some remarkable properties, that we sum up in the following proposition. Proposition 1.2. Let H be a subgroup of G. For every x, y G: 1. x H xh = H. 2. x xh. 3. xh = yh h H x = yh. 4. xh yh xh = yh (in other words: xh yh = or xh = yh). 5. The map H xh h xh is bijective. In particular, if H is finite the sets H and xh have the same cardinality. Thus, any two cosets of H are either equal or disjoint (in other words, if two cosets of H have non-empty intersection, they are equal), and every element of G belongs to at least one coset. In particular, if you consider all the different cosets of H, their union will be G, and every element of G will belong to exactly one of them. Since they all have the same number of elements, Lagrange s theorem follows: Theorem 1.3 (Lagrange). Let H be a subgroup of the the group G, with G finite. Then H divides G. The quotient G / H is called the index of H in G, and is denoted by [G : H]. The proof of Lagrange s theorem makes clear that [G : H] is the number of different cosets of H.

7 Chapter 2 Normal subgroups and quotient groups We consider a group (G, ), and we denote by G/H the collection of all cosets of H: G/H = {xh x G}. We want to define a product on G/H that will turn it into a group. The most obvious definition is (ah) (bh) = (ab)h. However, this definition has a serious problem: Take a, a, b, b G such that ah = a H and bh = b H. By definition of the product in G/H we have (ah)(bh) = (ab)h and (a H)(b H) = (a b )H. But since ah = a H and bh = b H, we must have (ah)(bh) = (a H)(b H), so (ab)h = (a b )H It means that for this definition of product to make sense, it is necessary to have, for every a, a, b, b G: (ah = a H and bh = b H) (ab)h = (a b )H. (2.1) The serious problem mentioned above is that this property does not always hold (but it does in many cases, for instance if G is Abelian). So we will have to distinguish those subgroups H of G for which property (2.1) holds. 3

8 4 CHAPTER 2. NORMAL SUBGROUPS AND QUOTIENT GROUPS Definition 2.1. A subgroup H of G is called normal if a G aha 1 = H, where aha 1 := {axa 1 x H}. We write H G to denote that H is a normal subgroup of G. Remark 2.2. (easy exercise). aha 1 = H aha 1 H ah = Ha If H is normal in G then property 2.1 holds: (a b )H = a (b H) = a (bh) = a (Hb) = (a H)b = (ah)b = a(hb) = (ab)h. In this case (H normal subgroup of G) we define as above a product on G/H. With this product G/H becomes a group, called the quotient group of G by H, and we have (the verifications are easy). e G/H = e H = H and (xh) 1 = x 1 H Exercise 2.3. If G is Abelian, then every subgroup H of G is normal and G/H is always Abelian. The following definition provides a very important example of normal subgroup. Definition 2.4. Let (G, ) be a group. The centre of G is Z(G) = {g G gx = xg x G}. Exercise 2.5. Show that Z(G) is a subgroup of G, that Z(G) is a normal subgroup of G, and that Z(G) is Abelian. We can also easily compute the order of G/H (have a look at the proof of Lagrange s theorem): G/H = {xh x G} = G / H = [G : H]. Why is it interesting to define the quotient group of G by H? A first easy answer is because it contains less elements than G. Therefore it could be easier to study, and what we learn about G/H can then be used to

9 learn something about the properties of G. Another answer requires first a simple observation: Let a H. Using that H is a subgroup, it is easy to check that ah = H = e H. In other words, the coset of H defined by a is equal to the identity element of G/H. From this point of view, when we go from G to G/H and replace each element a by a H, if a H we identify the element a H with e H = e G/H. In other words, we collapse all the elements of H to the identify of G/H. Intuitively, we make H disappear, and G/H should be simpler to study than G. Following this philosophy when studying a group G, we can try to do this: 1. Find a well chosen normal subgroup H of G. 2. Study G/H, which we hope is easier to study than G. 3. Use the information we obtain about G/H to say something about the properties of G. We will actually use this strategy in class. It explains why it is always interesting to find normal subgroups. We finish this section with two important facts about the order of an element. Remark Let G be a group and let g G. Assume g = k (for some g e), and g m = e for some m N. Then k divides m. Proof: By definition of k we have m k, so we can write m = qk + r for some integers q, r with 0 r < k. It follows that e = g m = g qk+r = (g k ) q g r = g r. Since r < k we must have (by definition of k), r = 0, so m = qk. 2. Assume G is a finite group, and let g G. Then the order of gh in G/H divides the order of g in G. Proof: Let k = g and l = gh. Then (gh) k = g k H = eh = H (which is the identity in G/H), so by the previous result we have l k. 5

10 6 CHAPTER 2. NORMAL SUBGROUPS AND QUOTIENT GROUPS

11 Chapter 3 Morphims We begin with an example. We consider the group (Z/2Z, +). The table of the operation + is and we see that the identity element of (Z/2Z, +) is 0. We now define a group (G, ) of order 2 by G = {a, b} and by giving the table for the operation (it is easy, but tedious, to check that (G, ) is a group): a b a a b b b a It should be clear that the groups (Z/2Z, +) and (G, ) are the same, but with + written, 0 written a and e written b. We want to give a more concrete definition for what it means to be the same. Definition 3.1. Let (G, ) and (H, ) be two groups. A map f : G H is called an isomorphism (of groups) if the following holds 1. f is bijective. 2. For every a, b G, f(a b) = f(a) f(b). We say that the groups (G, ) and (H, ) are isomorphic if there is an isomorphism f : G H, and we write G = H. 7

12 8 CHAPTER 3. MORPHIMS We can see that, in the above example, the map f : Z/2Z G, f(0) = a, f(1) = b is an isomorphism. Intuitively, two groups are isomorphic when they are the same group, the only differences being that the elements and the operation are called differently (but their behaviour is the same, as stated in the second property in the definition of isomorphism). In practise it is not always easy to find isomorphisms, but many maps that are not quite isomorphisms are still very interesting and will provide much information about groups. Definition 3.2. Let (G, ) and (H, ) be two groups. A map f : G H is called an morphism (or homomorphism) (of groups) if the following holds for every a, b G, f(a b) = f(a) f(b). An isomorphism is then a bijective morphism. An isomorphism from G to G is called an automorphism. Lemma 3.3. Let f : G H be a morphism of groups. Then 1. f(e G ) = e H, 2. for every x G, f(x 1 ) = f(x) 1. Proof. Exercise. Examples and remarks: 1. f isomorphism implies f 1 isomorphism. 2. Let f : G H and ϕ : H K be morphisms of groups. Then ϕ f : G K is a morphism of groups. 3. Let k Z. The map (Z, +) (Z, +), x kx is a morphism of groups. 4. det : (GL n (R), ) (R \ {0}, ) is a morphism of groups, because det(ab) = det A det B. 5. exp : (R, +) (R >0, ) is an isomorphism of groups, because it is bijective and e a+b = e a a b.

13 9 6. If H G then the map π : G G/H a ah is a morphism of groups, since π(ab) = (ab)h = (ah)(bh) = π(a)π(b). 7. Let (G, ) be a groups. For a G we define τ a : G G g aga 1 τ a is an automorphism of G, called the inner automorphism determined by the element a. Proof that τ a is an isomorphism: Morphism: τ a (xy) = axya 1, and τ a (x)τ a (y) = axa 1 aya 1 = axya 1. They are equal. Injective: Assume τ a (x) = τ a (y). Then axa 1 = aya 1. Multiplying this equality on the left by a 1 and on the right by a we obtain a 1 axa 1 a = a 1 aya 1 a, i.e. x = y. Surjective: Let g G. Then g = a(a 1 ga)a 1 = τ a (a 1 ga). It is also easy to check that (τ a ) 1 = τ a 1. Definition 3.4. Let f : G H be a morphism of groups. The kernel of f is ker f = {g G f(g) = e H }. The image of f is Im f = {f(g) g G}. Exercise (important): Show that ker f is a normal subgroup of G and that Im f is a subgroup of H. Lemma 3.5. Let f : G H be a morphism of groups. Then Proof. Exercise. f is injective ker f = {e G }. As suggested at the beginning of this chapter, two groups are isomorphic if they are the same. In practise it is not always easy to find an isomorphism between two given groups, but it can be easier to find morphisms (possibly between other, but related, groups). The following very important result provides a way to produce isomorphisms out of morphims.

14 10 CHAPTER 3. MORPHIMS Theorem 3.6 (First isomorphism theorem). Let G, H be groups and let f : G H be a morphism of groups. Then the map f : G/kerf Im f, x ker f f(x) is well-defined and is an isomorphism of groups. (Recall that G/ ker f = {x ker f x G}.) Before seeing the proof, a few words about what well-defined means: The map f is defined on elements of G/ ker f. Suppose we have a, b G such that a ker f = b ker f. The definition of f gives f(a ker f) = f(a) and f(b ker f) = f(b). Since a ker f = b ker f we must have f(a ker f) = f(b ker f), i.e. f(a) = f(b). This means that for our definition of f to make sense, we must have a ker f = b ker f implies f(a) = f(b). This is what is meant by f is well-defined. In other words: f is defined on the set G/ ker f, i.e. the set of all cosets of ker f. But the explicit definition of f(a ker f) uses the element a. It is this that creates the problem since different elements can define the same coset of ker f. Proof. We show that f is well-defined: Let a, b G be such that a ker f = b ker f. We know that it is equivalent to a = bh for some h ker f, and it follows that f(a) = f(bh) = f(b)f(h) = f(b) (since h ker f, so f(h) = e). We show that f is a morphism of groups: Let x ker f, y ker f G/ ker f. Then f((x ker f)(y ker f)) = f((xy) ker f) = f(xy), and f(x ker f) f(y ker f) = f(x)f(y) = f(xy). We show that f is bijective: It is clear that f is surjective, so we only have to show that it is injective. Let x ker f, y ker f G/ ker f be such that f(x ker f) = f(y ker f), i.e. f(x) = f(y). Then f(y) 1 f(x) = e, i.e. f(y 1 x) = e, i.e. y 1 x ker f. Let h ker f be such that y 1 x = h. Then x = yh, which implies x ker f = y ker f. Remark 3.7. Let H be a subgroup of G. Then H is a normal subgroup of G if and only if H is the kernel of some morphism of groups. Proof: Let π : G G/H, π(x) = xh. We know that π is a morphism of groups, and is easy to check that ker π = H. We already know this. Optional: The second isomorphism theorem If A and B are subsets of a group G, we define AB = {ab a A, b B}.

15 Proposition 3.8. Let A and B be subgroups of a group G, with A G. Then AB = BA and AB is a subgroup of G. Proof. We first show that AB = BA. This is an equality between sets, so we have to check that AB BA and BA AB. We only show the first inclusion (the second is similar). Let ab AB (with a A and b B). We want to show ab = b a with b B and a A. Take b = b, then solve the equation ab = ba to find a. Finally use that A G to show that a A. We now show that AB is a subgroup of G, and for this we use proposition 1.1. Checking the different parts of it is easy using that AB = BA. Theorem 3.9 (Second isomorphism theorem). Let H and N be subgroups of a groups G, with N G. Then (N H) G, N HN, and the groups H/(H N) and NH/N are isomorphic. More precisely the isomorphism is given by the map H/(H N) HN/N h(h N) hn Proof. The part about the normal subgroups is left as an exercise, and we only show the part about the isomorphism. The strategy to find isomorphism between two groups, when one of them is a quotient, is often to use the first isomorphism theorem. So we have to find a map f such that the map f given by the first isomorphism theorem is the isomorphism we are looking for. Here we take: f : H HN/N h hn We then show that 1. f is a morphism of groups (exercise). 2. f is surjective (exercise). In other words, Im f = HN/N. 3. ker f = H N. For this we show both inclusions ker f H N and H N ker f. 11

16 12 CHAPTER 3. MORPHIMS

17 Chapter 4 Subgroups of quotient groups Let G be a group and let H be a normal subgroup of G. Let π : G G/H g gh We know that π is a morphism of groups. Lemma 4.1. Let f : G 1 G 2 be a morphism of groups and let A be a subgroup of G 1. Then f 1 (A) (= {x G 1 f(x) A}) is a subgroup of G 1. Proof. Easy exercise. Let A be a subgroup of G/H, and define B = π 1 (A). By the previous lemma, we know that B is a subgroup of G. Moreover H B (since H = ker π = π 1 (e H/H ) and e G/h A), and π(b) = A (exercise). Theorem Every subgroup A of G/H is of the form π(b) for some subgroup B of G with H B. Note that in this case H B and π(b) = {bh b B} = B/H. 2. More precisely, the following two maps are bijections and inverse of each other: {A A subgroup of G/H} λ µ {B B subgroup of G, H B} A π 1 (A) B/H = π(b) B 13

18 14 CHAPTER 4. SUBGROUPS OF QUOTIENT GROUPS Proof. We just proved 1. For 2 we just have to check that λ and µ are inverse of each other, i.e. λ µ = Id and µ λ = Id. We start with λ µ = Id. Let B be a subgroup of G with H B. Then π(b) is a subgroup of G/H and λ µ(b) = π 1 (π(b)) and we want to show that it is equal to B. It is an equality between sets, so we show both inclusions: It is rather clear that B π 1 (π(b)) (just look at the definition of π 1 of a set). The other inclusion: Let x π 1 (π(b)), i.e. π(x) π(b), i.e. xh = bh for some b B. Therefore x = bh for some h H, and bh B since H B and B is a subgroup. We now show µ λ = Id: Let A be a subgroup of G/H. Then µ λ(a) = µ(π 1 (A)) = π(π 1 (A)) = A. Remark 4.3. The idea behind this result is simply that a map f from X to Y can be used to move information / objects / etc. between X and Y. Here we have the map π from G to G/H. We use it to move a subgroup of B of G from G to G/H by computing π(b). We also use it to move a subgroup A of G/H from G/H to G by computing π 1 (A).

19 Chapter 5 Group actions Group actions are historically the motivation behind the concept of group, and occur when the elements of a group G are equipped with the ability to move elements of a set X. Historically the only groups that were considered were groups for which this property is rather obvious, as in the following example. We consider the group S n of all bijections from {1,..., n} to {1,..., n}. Consider an element σ S n. By definition σ can move the elements of {1,..., n}, more precisely σ sends k {1,..., n} to σ(k). Therefore we have a map S n {1,..., n} {1,..., n} (σ, k) σ(k) Group actions are a description of that kind of behaviour of groups, and are an essential tool in the study of groups (and in other areas of mathematics). Definition 5.1. Let G be a group and X be a set. A (left) action of G on X is a function G X X (g, x) g x (the dot has nothing to do with the product from G and is just the way we denote this function; it is convenient to denote it as a product) with the following properties, for every x X and g, h G: 1. e x = x; 2. g (h x) = (gh) x. In such a case we say that G acts on X (on the left). 15

20 16 CHAPTER 5. GROUP ACTIONS Definition 5.2. Let G act on the set X and let x X. (under this action) is O x = {g x g G}. Examples As seen above, S n acts on {1,..., n} by S n {1,..., n} {1,..., n} (σ, k) σ(k) The orbit of x Let k {1,..., n}. We show that O k = {1,..., n}. That O k {1,..., n} is clear. Let l {1,..., n} and let σ = (k l). Then σ(k) = l, which shows that l O k. 2. The group GL n (R) acts on R n by We compute the different orbits. GL n R n R n (A, u) Au O 0 = {0}, because A0 = 0 for every matrix A. Let u R n \ {0}. Then O u = R n \ {0}: It is clear that O u R n, and for every A GL n (R), Au 0 (since A represents an injective linear map). Let v R n, to show that v O u we only have to find an invertible matrix B such that Bu = v, in other words we want to find a bijective linear map f : R n R n such that f(u) = v. Let {u, u 2,..., u n } be a basis of R n, and let {v, v 2,..., v n } be a basis of R n. We define f by f(u) = v and f(u i ) = v i for i = 2,..., n. Therefore we have two orbits: {0} and R n \ {0}. Notice that they are disjoint and that their union is R n. 3. Let G be the set of rotations of R 2 about 0. With the composition of maps as operation, G is a group. G acts on R 2 by G R 2 R 2 (f, u) f(u) The orbit of 0 is {0} since f(0) = 0 for every f G. Let u R 2 \ {0} with u = r. Then the orbit of u is the circle of centre 0 and radius r. Therefore we have an infinite number of orbits: All circles with centre 0 and radius r, with r 0. Observe again that the orbits are disjoint and that their union is R 2.

21 17 4. Let u R 2. (a) Let G = Ru = {au a R}. (G, +) is a group, and G acts on R 2 by G R 2 R 2 (au, x) x + au The orbit of x is the line containing x and parallel to u. (b) Let G = Zu = {au a Z}. (G, +) is a group, and G acts on R 2 by G R 2 R 2 (au, x) x + au The orbit of x is contained in the line containing x and parallel to u, but only consists of the point obtained from x by adding integer multiples of u. 5. Let G be a group and let H be a subgroup of G. Then H acts on G by left translation: H G G (h, g) hg product in G The orbit of g G is O g = {hg h H} = Hg. 6. Let H be a subgroup of G. H acts by conjugation on G as follows H G G (h, g) hgh 1 (terminology: the name comes from the fact that a group element of the form b 1 ab is called a conjugate of a). Proposition 5.4. Let G be a group acting on the set X. Then 1. For every x X, x O x ; 2. For every x, y X, O x = O y or O x O y = ; It follows that every element of X belongs to exactly one orbit. Proof. Exercise.

22 18 CHAPTER 5. GROUP ACTIONS If X is a set, we denote by S X the set of all bijections from X to X. It is a group for the operation composition of maps. (So according to this notation S n = S {1,...,n}.) Observe that every g G defines an element σ g S G. For this we simply take σ g : G G x gx We have to check that σ g S G, i.e. that σ g is a bijection. We first show that σ g is injective: Assume σ g (x) = σ g (y), so gx = gy. Multiplying both sides on the left by g 1 yields x = y. We show that σ g is surjective: Let x G. Then x = g(g 1 x), i.e. x = σ g (g 1 x). Therefore, this construction defines a map Σ : G S G g Σ(g) = σ g Lemma 5.5. The map Σ is an injective morphism of groups. Proof. We first check that it is a morphism of groups. Let g, h G. We have to show that Σ(gh) = Σ(g) Σ(h). This is an equality between elements S G, i.e. between functions from G to G. To check that Σ(gh) = Σ(g) Σ(h) we have to check that the two functions take the same value at each x G. Σ(gh)(x) = σ gh (x) = ghx, and (Σ(g) Σ(h))(x) = (σ g σ h )(x) = σ g (hx) = ghx. We now check that it is injective, and for that we use lemma 3.5. Let g ker Σ, i.e. Σ(g) = Id, i.e. gx = x for every x G. By taking x = e we obtain g = e, so ker Σ = {e}. We just proved the following. Theorem 5.6 (Cayley s Theorem). Let G be a group. injective morphism of groups Σ : G S G. In particular Then there is an 1. G is isomorphic to Σ(G), a subgroup of a group of permutations. 2. if G is finite, G is isomorphic to a subgroup of S n, where n = G. This theorem is interesting in the sense that it shows that everything that exists about groups already exists in groups of permutations. If you are less optimistic it also says that groups of permutations can be as complicated as it gets in group theory...

23 5.1. SIZE OF ORBITS Size of orbits Definition 5.7. Let G act on X, and let x X. The stabilizer S(x) of x is the subgroup S(x) = {g G g x = x} of G. The fact that S(x) is indeed a subgroup is left as an exercise. We consider G/S(x) = {gs(x) g G} (it is not necessarily a group, since S(x) may not be a normal subgroup of G), and define ϕ : G/S(x) O x gs(x) g x 1. ϕ is well-defined: Assume gs(x) = hs(x), i.e. g = ha for some a S(x). Then g x = (ha) x = h (a x) = h x. 2. ϕ is surjective: Let g x O x. Then g x = ϕ(gs(x)). 3. ϕ is injective: Assume ϕ(gs(x)) = ϕ(hs(x)), i.e. g x = h x. Then h 1 g x = x, i.e. h 1 g S(x). Let a = h 1 g S(x). Then g = ha with a S(x), i.e. gs(x) = hs(x). So we just proved the following: Theorem 5.8. ϕ is a bijection from G/S(x) to O x. In particular both sets have the same cardinality, and if G is finite: (and O x divides G ). O x = G/S(x) = G / S(x) = [G : S(x)] Examples Let H be a subgroup of G and let H act on G by left multiplication: H G G (h, g) hg Let g G. We have so S(g) = {e} and, if G is finite h S(g) hg = g h = e, O g = H / {e} = H, which makes sense since O g = {hg h H} = Hg.

24 20 CHAPTER 5. GROUP ACTIONS 2. Let G = GL n (R) act on R n by The stabilizer of 0 is G R n R n (A, v) Av S(0) = {A GL n (R) A 0 = 0} = GL n (R). Let v R n \ {0}. The stabilizer of v is S(v) = {A GL n (R) A v = v}, the set of invertible matrices for which v is an eigenvector associated to the eigenvalue Special case: The action by inner automorphisms (Also called the action by conjugation.) Let G be a group and let a G. The map τ a : G G g a 1 ga is called the inner automorphism (or conjugation) determined by a. element a 1 ga is called a conjugate of g. The Definition The action of G on itself by inner automorphisms (conjugations) is defined by G G G (a, x) a 1 xa The orbit of x under this action: O x = {a 1 xa a G}, is called the conjugacy class of x. Two elements x and y are conjugates if they belong to the same conjugacy class ( x, y O z O x = O y (= O z ) x O y a G x = a 1 ya). Remark Since the conjugacy classes are orbits, they are all disjoints and cover G (in other words, they form what is called a partition of G).

25 5.2. SPECIAL CASE: THE ACTION BY INNER AUTOMORPHISMS The conjugacy class of x is equal to {x} if and only if x Z(G) (easy exercise). The stabilizer of x under this action: S(x) = {g G g 1 xg = x}, is called the centralizer of x and is denoted by C(x) (and since it is a stabilizer, it is a subgroup of G). Theorem 5.8 yields: Proposition Let G be a finite group and let x G. The number of conjugates of x in G is equal to the index of C(x) in G (= G / C(x) ). In particular it divides G. Proposition 5.13 (The class equation). Let G be a finite group an let C 1,..., C k be the different conjugacy classes in G, with C i = O xi for some x 1,..., x k G. Then G = C C k (5.1) = [G : C(x 1 )] + + [G : C(x k )]. (5.2) Moreover, if C 1,..., C l (with l k) are all conjugacy classes with at least two elements, then G = Z(G) + l C i (= Z(G) + i=1 l [G : C(x i )]). (5.3) i=1 Proof. By remark we have G = C 1 C k, and (5.1) follows, and (5.2) follows by proposition By remark there are as many conjugacy classes with one element as elements in Z(G) and (5.3) follows. We finish this chapter with an application. Definition Let p be a prime number. order is a power of p. A p-group is a group whose Proposition Let G be a p-group with G {e}. Then Z(G) {e}. Proof. Let n N, n 1 be such that G = p n. Using the terminology of proposition 5.13 we write G = Z(G) + l i=1 C i. Since C i divides G = p n and C i 2 we have C i = p n i for some n i 1. So p n = Z(G) + l i=1 pn i and Z(G) = p n l i=1 pn i. Since the right hand side is divisible by p, we obtain Z(G) p > 1.

26 22 CHAPTER 5. GROUP ACTIONS Lemma Let G be a group such that G/Z(G) is cyclic. Abelian. Then G is Proof. Exercise. Proposition Let G be a group of order p 2 with p prime. Then G is Abelian. Proof. Since Z(G) is a subgroup of G, we have Z(G) = 1, p or p 2. By proposition 5.15 we know that Z(G) = 1, so Z(G) = p or p 2. If Z(G) = p 2, then Z(G) = G and G is Abelian. If Z(G) = p, then G/Z(G) = G / Z(G) = p, so G/Z(G) is a cyclic group. By lemma 5.16, G is Abelian.

27 Chapter 6 The Sylow theorems (Sylow, 1872) Underlying idea: Lagrange s theorem says that the order of a subgroup of G divides the order of G. Conversely, if k divides G, does G have a subgroup of order k? In general the answer is no, but the Sylow theorems give a very precise positive answer for some values of k (exactly: when k is a power of a prime number). Lemma 6.1. Let G be a finite Abelian group and let p be a prime number. If p divides the order of G, then G has a subgroup of order p. Proof. By induction on G. If G = p (the smallest possibility since p G ), then G itself is the subgroup. Assume G > p and that we can find such a subgroup for every Abelian group of cardinality less than G. Let a G, a e. We consider two cases. Case 1: p divides a, i.e. a = mp for some m N. Then A m = p and a m is a subgroup of order p. Case 2: p does not divide a = a. Let A = a. Since G is Abelian, A is a normal subgroup of G, and the group G/A is defined. We have G/A = G / A, i.e. G = G/A A. Since G is divisible by p and A is not, G/A is divisible by p. But we also have G/A < G. Therefore by induction hypothesis G/A has a subgroup H of order p. Since H = p and p is prime, H is a cyclic subgroup of G/A, i.e. of the form ga for some ga G/A, where ga has order p in G/A. 23

28 24 CHAPTER 6. THE SYLOW THEOREMS (SYLOW, 1872) But by remark 2.6.2, the order of ga in G/A divides the order of g in G. So p divides g and we conclude as in the first case (with a = g). Theorem 6.2 (Sylow). Let G be a finite group and let p be a prime number. If p k divides G (for some k N {0}) then G has a subgroup of order p k. Proof. We proceed by induction on G. If G = 1. Then p k = 1 and G itself is a subgroup of order 1. If G > 1. We can assume p k > 1 (if p k = 1 we take the subgroup {e}). We consider 2 cases. 1. If p Z(G). Since Z(G) is Abelian, by lemma 6.1 Z(G) has a subgroup A of order p. Since A Z(G) we have A G (very easy exercise), so we can consider the group G/A. We have G/A = G / A and since A = p, we know that p k 1 G/A. Using the induction hypothesis, we obtain that G/A has a subgroup of order p k 1. By theorem 4.2 this subgroup is of the form B/A for some subgroup B of G with A B. Finally, B/A = B / A, so B = p k 1 p = p k. 2. If p G. By the class equation (proposition 5.13) we have G = Z(G) + l C i, i=1 where G is divisible by p k and Z(G) is not divisible by p. It follows that there is i {1,..., l} such that C i is not divisible by p. This C i is the conjugacy class of some element x i G and C i = [G : C(x i )] = G / C(x i ), i.e. G = C i C(x i ). Since p k divides G and C i is not divisible by p, we obtain that p k divides C(x i ). But C(x i ) < G (because C i 2, see proposition 5.13), so by induction C(x i ) has a subgroup of order p k, and this subgroup is also a subgroup of G. Corollary 6.3 (Cauchy s theorem). Let G be a finite group and let p be a prime number such that p G. Then G has an element of order p.

29 Proof. By theorem 6.2, G has a subgroup H of order p. This subgroup is cyclic and is generated by any a H \ {e}. Such an a as order p. Definition 6.4. Let p be a prime number. A p-group is a group of order p k for some k N. A p-subgroup is a subgroup that is a p-group. Corollary 6.5. Let G be a finite group and let p be a prime number. Then G is a p-group if and only if every element of G has order a power of p. Proof. We know that G = p k for some k N. Let a G. We know that a = a G and the result follows. Assume G is not a power of p. Then there is a prime number q p such that q G. By corollary 6.3, G has an element of order q, a contradiction. 25 Definition 6.6. Let p be a prime number and let G be a group of order p k m where p m (i.e. p k is the largest power of p dividing G ). A Sylow p-subgroup of G is a subgroup of G of order p k. Note that Sylow p-subgroups always exist by theorem 6.2. The next results show how Sylow p-subgroups are related to the different p-subgroups given by theorem 6.2 Theorem 6.7 (Sylow). Let G be a finite group and let p be a prime number such that p G. 1. Let H be a p-subgroup of G and let S be any Sylow p-subgroup of G. Then there is a G such that H asa 1. Note that asa 1 is itself a Sylow p-subgroup of G. 2. All Sylow p-subgroups of G are conjugate (i.e. if S 1 and S 2 are Sylow p-subgroup of G, then there is a G such that S 2 = as 1 a 1 ). 3. Let n p be the number of Sylow p-subgroups of G. Then (a) n p G; (b) n p = 1 mod p. The rest of this chapter is devoted to the proof of the different statements in this theorem, and will require the introduction of some new notions. We will also state some important corollaries. Lemma 6.8. Let G be a p-group and assume that G acts on a finite set X. We define X G = {x X g x = x g G}. Then X G = X mod p.

30 26 CHAPTER 6. THE SYLOW THEOREMS (SYLOW, 1872) Proof. If x X \X G then S(x) G (by definition of X G ), so [G : S(x)] > 1. Since [G : S(x)] = G / S(x), it is a power of p, which implies that p divides [G : S(x)]. We also have X = O(x 1 ) O(x l ) where O(x 1 ),..., O(x l ) are all the (different) orbits. Up to listing the orbits in a different order, we can assume X = O(x 1 ) O(x k ) O(x k+1 ) O(x l ), where O(x 1 ),..., O(x k ) are the orbits with exactly one element (therefore the others have at least 2 elements). By definition of X G, we have x X G if and only if O(x) = {x}, which means that X G = {x 1,..., x k }, so X G = k. Therefore X = X G + l i=k+1 O(x i ) = X G + l [G : S(x i )], i=k+1 and we already observed that [G : S(x i )] is divisible by p if x X G. The result follows. of theorem 6.7, part 1. Consider the following action of H on G/S: By lemma 6.8 and with H G/S G/S (h, gs) (hg)s (G/S) H = {gs G/S hgs = gs h H}, then (G/S) H = G/S mod p = q mod p (if G = qp k with p q). Note that q 0 mod p since p does not divide q. It follows that (G/S) H 0 mod p so in particular (G/S) H 0. Take gs (G/S) H. By definition of (G/S) H we have hgs = gs for every h H, i.e. g 1 hgs = S for every h H. But this means that g 1 hg S for every h H, i.e. h gsg 1 for every h H. To prove the other statements in Sylow s theorem, we need to introduce another group action. Let SG(G) be the set of all subgroups of G. If g G and H SG(G) then ghg 1 SG(G). More precisely, G acts on SG(G) by G SG(G) SG(G) (g, H) g 1 Hg

31 27 The orbit of H under this action is O(H) = {g 1 Hg g G}, the set of conjugates of H. By theorem 5.8, O(H) = G / S(H), where S(H) = {g G g 1 Hg = H}. Definition 6.9. S(H) is called the normalizer of H in G and is denoted N G (H) = {g G ghg 1 = H}. Lemma N G (H) is the largest subgroup of G in which H is normal. In other words 1. H N G (H); 2. It T is a subgroup of G such that H T, then T N G (H). Proof. Very direct exercise. Proposition Let G be a finite group and let p be a prime number such that p G. Let S be a Sylow p-subgroup of G. Assume S G. Then S contains every p-subgroup of G and S is the unique Sylow p-subgroup of G. Proof. Let T be a p-subgroup of G. By theorem there is g G such that T gsg 1 = S (since S G). Let T be a Sylow p-subgroup of G. By theorem there is g G such that T = gsg 1 = S. We are now in position to prove the last part of theorem 6.7. of theorem 6.7, part 2. Let S be a Sylow p-subgroup of G, and let S be the set of all Sylow p-subgroups of G. Note that if g G then g 1 Sg S. 1. Let G act on S by G S S (g, T ) gt g 1 By theorem 6.7.2, S = {gsg 1 g G}, in other words S is equal to the orbit of S. In particular, by theorem 5.8 so S divides G. S = G / S(S) = G / N G (S),

32 28 CHAPTER 6. THE SYLOW THEOREMS (SYLOW, 1872) 2. S acts on G by S S S (s, T ) st s 1 The orbit of S is {S} since sss 1 = S for every s S. Conversely, if T S and the orbit of T is {T }, then st s 1 = T for every s S, which means S N G (T ). So S and T are Sylow p- subgroups of N G (T ), and T N G (T ). Applying proposition 6.11 to N G (T ) (instead of G) we get S = T. Conclusion: {S} is the only orbit with one element. As usual, we know that S is the disjoint union of all orbits with exactly one element (there is just one: {S}) and all orbits with at least two elements. If O(A) is such an orbit, we have O(A) = S / S(A) so divides S and is at least 2, so is a multiple of p. It follows that S = 1 + some multiples of p = 1 mod p.

33 Chapter 7 Application to combinatorics Groups actions can be used to solce some combinatorial problems. We consider a typical such problem. Recall that D 12 is the group of symmetries of the plane that leave invariant the regular hexagon: An element σ of D e2 can be described by what it does to the vertices, i.e. by σ(1),..., σ(6). So we will consider σ as an element of S 6, and we will describe it using the notation for permutations. Let R be the rotation (around the centre of the hexagon) by π/3, and let F be the reflexion about the horizontal line containing vetices 1 and 4. R = F = ( ) = ( ), ( ) = (2 6)(3 5)

34 30 CHAPTER 7. APPLICATION TO COMBINATORICS The 12 elements of D 12 are Id R = ( )R 2 = (1 3 5)(2 4 6) R 3 = (1 4)(2 5)(3 6)R 4 = (1 5 3)(2 6 4) R 5 = ( ) F = (2 6)(3 5) RF = (1 2)(3 6)(4 5)R 2 F = (1 3)(4 6) R 3 F = (1 4)(2 3)(5 6)R 4 F = (1 5)(2 4) R 5 F = (1 6)(2 5)(3 4) We have D 12 = {R k F l 0 k 5, 0 l 1} with R 6 = Id, F 2 = Id and F R = R 5 F. 7.1 The question We want to colour the vertices of the hexagon. If we have n colours available, how many different colourings are possible? Other formulation: We have beads of n different colours and we want to make a necklace with 6 beads (we assume that there is no opening in the necklace). How many different necklaces can we make? The naive (and false) answer would be to say that twe have n choices for each bead, and therefore n 6 possibilities. The problem is that this method counts several times the same necklace / colouring. Here is an example with just two colours, R (red) and G (green): The colouring 3 2 c 1 = ( ) 6 R R G G G R

35 7.2. AN ANSWER, USING GROUP ACTIONS 31 is mapped by R (the element of D 12, not the colour) to 3 2 c 1 = ( ) 6 R R R G G G They are different colourings but they define the same necklace. So we need to find a way to count colouring, while identifying two colourings that only differ by the application of an element of D 1 2. In other words: D 12 acts on the set of colourings, and we want to count the number of orbits of this action. 7.2 An answer, using group actions We need to describe this more precisely: 1. A colouring is a map from {1,..., 6} (the set of beads) to {1,..., n} (the set of colours): c : {1,..., 6} {1,..., n} r c(r) the colour of bead r We can write c in the following way: c = ( ) c(1) c(2) c(3) c(4) c(5) c(6) 2. If σ D 12, σ maps c to σ c : {1,..., 6} {1,..., n} r c σ 1 In other words, σ c is the colouring c σ 1. An example should clarify why we use σ 1. Consider the colouring c 1

36 32 CHAPTER 7. APPLICATION TO COMBINATORICS from above: 3 2 c 1 = ( ) 6 R R G G G R 4 1 Applying R to c 1 gives us the colouring R σ R c 1 = ( ) 6 R R R G G G = c and a direct computation gives us 3. This defines a group action c 2 = c 1 R D 12 {colourings of {1,..., 6}} {colourings of {1,..., 6}} (σ, c) σ c := c σ 1 We want to count the number of orbits of this action. Theorem 7.1 (Orbit counting). Let G act on X, with G and X finite. For g G, let f(g) be the number of elements in X fixed by g: f(g) = {x X g x = x}. Then the number of orbits in X for this action is 1 f(g). G Proof. We define g G T := {(g, x) G X g x = x}. The trick consist in county the elements on T in two different ways.

37 7.2. AN ANSWER, USING GROUP ACTIONS For x X we define S(x) := {g G (g, x) T }. Therefore S(x) = {g G (g, x) T } and thus T = x X S(x). (7.1) 2. For g G we have by definition f(g) = {x X (g, x) T }, so T = g G f(g). (7.2) Putting (7.1) and (7.2) together, we obtain S(x) = f(g), g G so x X S(x) G x X }{{} ( ) = 1 G f(g). G But = O S(x) x the number of elements in the orbit of x. If O x = {x 1,..., x l }, then S(x) = 1 will appear l times in ( ) (one for G l each of x 1,..., x l ), which will bring 1 to the sum. Therefore, ( ) gives the number of orbits in X. We go back to the question about necklaces. Using theorem 7.1, to determine the number of orbits, we need to determine, for σ D 12, the number g G f(σ := the number of colourings fixed by the action of σ. We consider a few examples. σ = R 2 = (1 3 5)(2 4 6). Since σ(1) = 3 and σ(3) = 5, the vertices 1, 3 and 5 must have the same colour. Similarly, the vertices 2, 4 and 6 must have the same colour. So, if we have n colours available, we have n choices for the colour of {1, 3, 5} and n choices for the colour of {2, 4, 6}, so in total n 2 possibilities, i.e. f(r 2 ) = n 2. σ = F = (2 6)(3 5). As above, 2 and 6 must have the same colour, and 3 and 5 must have the same colour. We are also free to choose the colour of 1 and the colour of 4. Therefore there are n 4 possibilities, f(f ) = n 4.

38 34 CHAPTER 7. APPLICATION TO COMBINATORICS The general case is similar: Let σ D 12, and write σ as a product of t disjoint cycles, writing also the cycles of length one (which we usually do not bother doing, since they are the identity). Then f(σ) = n t (one choice of colour per cycle). We now apply theorem 7.1, and compute σ D 12 f(σ), by simply looking at all possible elements σ D 12 : There is one σ D 12 with six cycles (σ = Id). It contributes n 6 to the sum. There are three elements of D 12 with four cycles. They contribute 3n 4 to the sum. There are four elements of D 12 with 3 cycles. They contribute 4n 3 to the sum. There are two elements of D 12 with 2 cycles. They contribute 2n 2 to the sum. There are two elements of D 12 with 1 cycle. It contributes 2n to the sum. So the number of orbits (i.e. the number of different necklaces) is 1 12 (n6 + 3n 4 + 4n 3 + 2n 2 + 2n).

39 Chapter 8 The crystalographic / wallpaper groups 8.1 Isometries of the plane Definition 8.1. An isometry of R 2 is a map f from R 2 to R 2 such that, for every points A, B R 2, d(a, B) = d(f(a), f(b)). We denote by Isom(R 2 ) the set of all isometries of R 2. Examples: 1. Let u R 2. The translation of vector u (i.e. the map f(a) = A + u) is an isometry. 2. Let c R 2 and let α R. The rotation of centre c of angle α is an isometry. 3. Let L be a line in R 2. The reflection across L is an isometry. 4. The composition of a reflection across a line L and a translation by a vector parallel to this line is called a glide-reflection. It is also an isometry (it does not matter in what order you do the reflection and the translation, make a picture). We will see below that these are all the possible isometries of R 2. Corollary 8.2. Let f Isom(R 2 ). Then 1. f is bijective. 2. f 1 is an isometry. 35

40 36 CHAPTER 8. THE CRYSTALOGRAPHIC / WALLPAPER GROUPS Proof. f injective: Let A B R 2. Then d(a, B) > 0 so d(f(a), f(b)) > 0 and thus f(a) f(b). f surjective: Let A B R 2. Let Z R 2. Let r 1 = d(z, f(a)) and r 2 = d(z, f(b)). What can you say about the intersection of the circles of centre f(a) and radius r 1 and of centre f(b) and radius r 2? About the circles of centre A and radius r 1 and centre B and radius r 2? Conclude. Let A, B R 2. We want d(f 1 (A), f 1 (B)) = d(a, B). Write A = f(c) and B = f(d) for some C, D. Then f 1 (A) = C and f 1 (B) = D, and since f is an isometry: d(c, D) = d(a, B), which is what we want. Corollary 8.3. Isom(R 2 ), with product the composition of maps, is a group. The identity element is the identity map. Recall that if A, B are two distinct points, then is the perpendicular bisector of [AB]. {P R 2 AP = BP } Proposition 8.4. Let f Isom(R 2 ) and let A, B, C R 2 be non-colinear such that f(a) = A, f(b) = B and f(c) = C. Then f = Id. Proof. Let D be another point in the place. Let r A = d(a, D), r B = d(b, D) and R C = d(c, D). Since d(f(a), f(d)) = d(a, D) and f(a) = A, f(d) belongs to the circle of centre A and radius r A. Similarly, it belongs to the circle of centre B and radius r B. The intersection of these two circles contains D, and possibly another point D, f(d) = D or D. If D = D we have f(d) = D. If D D, then the line (AB) is orthogonal to (DD ) and cuts [D, D ] in its middle (i.e. is the set of points at equal distance from D and D ). Since C is not on (AB), d(c, D) d(c, D ), so f(d) D. Corollary 8.5. If two isometries of R 2 are equal on 3 non-colinear points, then they are equal. Lemma 8.6. Let A, B R 2 with A B. Then there is exactly one reflection f such that f(a) = B. Proof. Take the reflection across L, where L is the perpendicular bisector of [AB]. Proposition 8.7. Let f Isom(R 2 ). Then f is the product of at most 3 reflections.

41 8.1. ISOMETRIES OF THE PLANE 37 Proof. Take A, B, C R 2 non-colinear. If we can find a map σ such that σ is a product of at most 3 reflections and σ(a) = f(a), σ(b) = f(b) and σ(c) = f(c), then we will have f = σ. Assume f(a) A. Then by lemma 8.6 there is a reflection σ 1 such that σ 1 (A) = f(a). Assume now σ 1 (B) f(b). Then there is a reflection σ 2 such that σ 2 (B) = Q. Observe that this is a reflection over the perpendicular bissector of [σ 1 (B)f(B)]. What about σ 2 (P )? Observe that d(f(a), f(b)) = d(a, B) = d(σ 1 (A), σ 1 (B)) = d(f(a), σ 1 (B)). Therefore f(a) is on the perpendicular bissector of [f(b)σ 1 (B)], so σ 2 (f(a)) = f(a). Therefore σ 2 σ 1 (A) = f(a), σ 2 σ 1 (B) = f(b). Let C = σ 2 σ 1 (C). Assume now that C f(c). Then there is a reflection σ 3 such that σ 3 (C ) = f(c). We can check as above that σ 3 (f(a)) = f(a) and σ 3 (f(b)) = f(b), and we have σ 3 σ 3 σ 1 (A) = A, σ 3 σ 3 σ 1 (B) = B, σ 3 σ 3 σ 1 (C) = C. The cases not covered are if A = f(a) or σ 1 (B) = f(b) or σ 2 σ 1 (C) = f(c). In each of these cases, we simply do not need the corresponding map σ 1, σ 2 or σ 3 (we can say that the identity map as the product of no reflection, or view it as the product of any reflection by itself). Proposition 8.8. The product of 2 reflections is 1. The identity, if the two reflections are equal. 2. A rotation if the reflections are over non-parallel lines. 3. A translation if the reflections are over parallel lines. The product s 1 s 2 s 3 of 3 reflections is 1. A reflection, if s 3 s 3 is a rotation. 2. A glide reflecion, if s 2 s 3 is a translation. Proof. Draw the pictures... Remark 8.9. Proposition 8.7 also gives the following (which we proved in a different way already): 1. Every isometry of R 2 is a bijection and its inverse is an isometry of R The set of all isometries of R 2, with operation the composition of maps, forms a group.

42 38 CHAPTER 8. THE CRYSTALOGRAPHIC / WALLPAPER GROUPS 8.2 Complex numbers and isometries It is sometimes convenient to use complex numbers to represent isometries Proposition Let r be the rotation of centre O of angle α. Then, for z C, r(z) = ze iα. 2. Let t be the translation of vector u C. Then t(z) = z + u. 3. Let f be the reflexion across the x-axis. Then f(z) = z. Proof. Immediate use of the properties of complex numbers. Corollary Let r be the rotation of centre u and of angle α. r(z) = (z u)e iα + u. Application: The product of two rotations is: a translation if they have equal and opposite angles, Then a rotation otherwise. Let r 1 be the first rotation and r 2 the second. We take for origin of the plane the centre of the rotation r 1. Then r 1 (z) = ze iα and r 2 (z) = (z u)e iθ + u, where u is the centre of r 2. Then r 1 r 2 (z) = [(z u)e iθ +u]e iα = (z u)e i(α+θ) + ue iα. How many fixed points does this isometry have? To find out, we solve r 1 r 2 (z) = z: (z u)e i(α+θ) + ue iα = z, i.e. z(1 e i(α+θ) ) = u(e i(α+θ) e iα ). If α β then there is a unique solution z 0. So r 1 r 2 is an isometry with a unique fixed point, i.e. a rotation (around z 0 ). If α = β then r 1 r 2 (z) = z u + ue iα is a translation of vector u(e iα 1). 8.3 Point groups of isometry groups We denote by O the point of R 2 of coordinates (0, 0). Lemma Let f Isom(R 2 ). Then there is a unique u R 2 and a unique isometry f 0 Isom(R 2 ) such that 1. f(a) = f 0 (A) + u for every A R 2 (i.e. f = t u f 0 ). 2. f 0 (O) = O. Proof. We first show that u and f 0 exist. Let u = f(o), and define f 0 = t u f. Then f 0 is an isometry and f 0 (O) = t u (f(o)) = O. By construction we have f = t u f 0. Uniqueness: If f = t v f 1 with f 1 (O) = O. Then f(o) = v and also f(o) = u, so u = v. Therefore f = t u f 0 = t u f 1 and f 0 = f 1.

43 8.4. CRYSTALLOGRAPHIC GROUPS 39 Since f 0 is uniquely determined by f, we can define a map λ : Isom(R 2 ) Isom(R 2 ) 0, f f 0, where Isom(R 2 ) 0 denote the subgroup of all isometries that fix the point O. Proposition The map λ is a morphism of groups and ker λ is the group of all translations. Proof. Exercise! Definition If G is a subgroup of Isom(R 2 ), then λ(g) is called the point group of G. 8.4 Crystallographic groups Definition Let P R 2. The symmetry group of P, denoted Sym(P ) is {f Isom(R 2 ) f(p ) = P }. It is a subgroup of Sym(R 2 ). Examples D 6 is the symmetry group of an equilateral triangle. It contains 6 elements (3 rotations and 3 reflections). 2. More generaly, D 2n is the symmetry group of a regular n-gon. It contains 2n elements (n rotations and n reflections). 3. If P is not symmetric, for instance if P is a kind of flag: then Sym(P ) = {Id}.

44 40 CHAPTER 8. THE CRYSTALOGRAPHIC / WALLPAPER GROUPS In the above examples the set P is bounded. It need not be the case, even in very natural situations.

45 8.4. CRYSTALLOGRAPHIC GROUPS 41 Figure 8.1: Some images from electron crytallography

46 42 CHAPTER 8. THE CRYSTALOGRAPHIC / WALLPAPER GROUPS Definition Let G be a subgroup of Isom(R 2 ). The translation group of G, tr(g) is the subgroup of G consisting of all the elements of G that are translations. Looking at the crystallography pictures, we see that we can repeat the patterns by translations to cover the whole plane. Moreover these translations seem to be all obtained by two linearly independent translations of minimal length. The same phenomenon occurs in some wallpaper patterns, or in some tillings: Figure 8.2: Alhambra palace Figure 8.3: China Figure 8.4: Egypt Figure 8.5: Pavement Definition A crystalographic group (or wallpaper group) is a subgroup

47 8.4. CRYSTALLOGRAPHIC GROUPS 43 G of Isom(R 2 ) such that the translation group of G is generated by two translations whose vectors are linearly independent. From now on, we assume that the plane is filled with a pattern P such that Sym(P ) is a crystallographic group G. We denote by T G the translation group of G: T G = {t m u t n v m, n Z} = {t mu+nv m, n Z}. where t u and t v are two translations with u and v linearly independent of minimal length. Observe that G acts on R 2 in a natural way: G R 2 R 2, (f, A) f(a). Definition The lattice of G, denoted L G, is the orbit of 0 under the action of the group T G : L G = {f(0) f T G } = {mu + nv m, n Z}. v O u Definition The parallelogram determined by O and the vectors u and v (with solid lines on the previous picture) is called the fundamental domain of G. Observe that moving the fundamental domain using the translations t u and t v will fill the plane. So to obtain the full pattern P you only need the information inside the fundamental domain. But there may be other symmetries of P, so that you only need to know part of the fundamental domain. We will describe the possible crytallographic groups using their lattices and their point groups. Lemma For every f G, f 0 (L G ) = L G.

48 44 CHAPTER 8. THE CRYSTALOGRAPHIC / WALLPAPER GROUPS Proof. We write f = t u f 0 as above. Let w L G. The the translation t w is in G, and so the map f t w f 1 is in G. But f t w f 1 (A) = f 0 (f 1 0 (A u) + w) + u = A u + f 0 (w) + u = A + f 0 (w), i.e. is equal to t f0 (w), so f 0 (w) = t f0 (w)(o) L G. Proposition 8.22 (The crystallographic restriction). Let f λ(g) be a rotation. Then the angle of f is π, 2π/3, π/2 or π/3. In other words: the order of f is 2, 3, 4 or 6. Proof. Let A L G be at minimal distance to O, and let d = d(a, O). By definition of L G, Ω := {B L G d(b, O) = d} is finite. Since d(a, O) = d(f(a), f(o)) = d(f(a), O), the successive images of A under f are all in Ω. Since Ω is finite there are i < j N such that f i (A) = f j (A), so f j i (A) = A, which show that f is of order j i finite. Choose now such an f with the highest possible order, i.e. with the smallest angle θ. As observed above, the order of f is finite so θ = 2π/n for some n N. By lemma 8.21, the point f(a) is in L G and therefore the point f(a) A is also in L G. So by choice of A we have f(a) A d (or f(a) A = 0 but this is not the case since f Id by choice). It means that θ π/3. f(a) O d θ A But θ = 2π/n π/3 implies that n {2, 3, 4, 5, 6}. Suppose that n = 5, i.e. θ = 2π/5. Let B = A + f 2 (A). Then B L G, but d(o, B) < d, contradiction. f 2 (A) O B = f 2 (A) + A 4π/5 A Lemma The set of all rotations in λ(g) is a cyclic subgroup of λ(g), generated by the rotation of smallest angle.

49 8.4. CRYSTALLOGRAPHIC GROUPS 45 Proof. All these rotations have centre O, so this set is closed under composition and under taking inverses. It also contains the identity, so it is a subgroup of λ(g). Let f be the rotation in λ(g) of smallest positive angle θ and let r be another rotation. Let α be the angle of r. Then α > θ and there is n N such that nθ α < (n + 1)θ. The map f n r 1 is a rotation in λ(g) of angle α nθ θ. Since θ is minimal positive, we have α nθ = 0, i.e. α = nθ, so r = f n. Proposition λ(g) is one of the following groups: C 1, C 2, C 3, C 4, C 5, C 6 the cyclic groups D 2, D 4, D 6, D 8, D 12 the dyhedral groups. Proof. Let f λ(g) be the rotation with the smallest positive angle θ (see previous lemma). If f generates λ(g), then λ(g) is a cyclic group of order f, i.e. of order 2, 3, 4, 5 or 6. If f does not generate λ(g), then λ(g) contains a reflexion h (by the previous lemma, λ(g) then contains at least one isometry that is not a rotation, but fixes O, so is a reflexion). Claim: Then λ(g) = {f n h i n Z, i {0, 1}}, with hf = f n 1 h where n = f. Let g λ(g). Then g is a rotation or a reflexion (since it fixes O). If g is a rotation we saw in the previous lemma that g = f n for some n N. If g is a reflection, then gh is a rotation of centre O, so gh = f n for some n and thus g = f n h (recall that h = h 1. The relation hf = f n 1 h is true for every rotation f of order n around O and every reflection h across a line containing O. This description of λ(g) is the descrition of a dihedral group, and the cardinality comes from the crystallographic restriction. We finaly observe that there are several different kinds of lattices L G. (1) If there is a rotation in λ(g) with non-zero angle. Take such a rotation f of minimal angle θ, and A L G such that A is minimal non-zero. If θ = π/2, then A and f(a) generate the lattice, so the square with vertices O, A, f(a) is a fundamental region, and L G is called the square lattice:

50 46 CHAPTER 8. THE CRYSTALOGRAPHIC / WALLPAPER GROUPS f(a) A + f(a) O A If θ = π/3 or θ = 2π/3, then A and f(a) generate the lattice, so thewhich is called the hexagonal lattice: f(a) A + f(a) O A (2) If λ(g) = {id}, then G only contains translations and the fundamental domain is the rectangle determined by the vectors u and v. The lattice L G is called rectangular. (3) If the only rotation in λ(g) is the identity, but G {Id}. Then λ(g) contains a reflections, and the fundamental domain will be a square, or a parallelogram with equal sides (a rhombus). The lattice is then called square or rhombic, accordingly. We can then describe all the finite crystallographic groups, according to the type of their lattice and to what their point group is. There are 17 different possible combinations. This result is originally due to Fedorov, Barlow (crystallogrphers) and Schoenflies in the late 19th century. (These groups are often described using notation from the international tables for X-ray crystallography, but we will skip this.) Here are some example of the patterns produced by the different crystallographic groups.

51 8.4. CRYSTALLOGRAPHIC GROUPS 47 Figure 8.6: Point group: C 4, rectangular lattice Figure 8.7: Point group: D 6, hexagonal lattice Figure 8.8: Point group: C 4, rectan-figurgular lattice nal 8.9: Point group: C 3, hexago- lattice Figure 8.10: Point group: C 3, rhombic lattice Figure 8.11: Point group: D 3, rectangular lattice

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