Problem Set 1 Solutions Math 109

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1 Problem Set 1 Solutions Math 109 Exercise 1.6 Show that a regular tetrahedron has a total of twenty-four symmetries if reflections and products of reflections are allowed. Identify a symmetry which is not a rotation and not a reflection. Check that this symmetry is a product of three reflections. Solution. How many ways are there to arrange the lels a, b, c, and d at the four vertices of the tetrahedron? There are 4 options for where to place a, and once it s placed there are 3 options for where to place b, then 2 options for c, and then d goes on whichever vertex is left. Thus there are ways to lel the vertices of the tetrahedron. Any symmetry sends vertices to vertices, and in fact is determined by where it sends the vertices. So if we start the tetrahedron with our favorite leling of the vertices, and apply products of reflections, then the result will give us one of the 24 leling options. Thus there are at most 24 distinct reflections and products of reflections. Given any two vertices, there is a reflection which swaps those two vertices and leaves the other two fixed - for example, in Figure 1.4 of Armstrong, to switch vertex 2 and 3, you reflect through the plane which goes through the center point of the line segment 23 and is normal to it. These swapping reflections are called a transpositions, and they allow us to get to any leling, a.k.a permutation, of the vertices by working one vertex at a time. Start with the vertices in a preferred leling, and take any of the 24 possible permutations as the goal. Do the following steps: 1. If vertex a is in the correct position, go to the next step. If not, apply the transposition that swaps a into its correct position. 2. If vertex b is in the correct position, go to the next step. If not, swap b into its correct position (because of step 1, a is not in b s correct position, so this leaves a fixed). 3. If vertex c is in the correct position, go to the next step. If not, swap c into its correct position (neither a nor b are in c s correct position, so this leaves them fixed). 4. a, b, and c are in the correct position, so d must be too. We re done! This gives a method for reaching any permutation of the vertices by a sequence of reflections, so there are 24 symmetries if products of reflections are allowed. Suppose the tetrahedron starts with its lels positioned as (1,2,3,4), and a reflection swaping 1 and 4 is performed to get (4,2,3,1), another to get (4,3,2,1), and a final to get (3,4,2,1). The symmetry (1,2,3,4) (3,4,2,1) can t be just a reflection, since all of the vertices have moved. For the same reason, it can t be a rotation which fixes one vertex. 1

2 Problem Set 8 2 of 8 Finally, it can t be a rotation which swaps a with b and c with d, for example because 1 goes to 4 s place, but 4 does not go to 1 s place. Exercise 2.3 Which of the following collections of 2 2 matrices with real entries form groups under matrix multiplication? (i) The set S 1 of matrices of the form bc for which ac b 2. (ii) The set S 2 of matrices of the form ca for which a2 bc. (iii) The set S 3 of matrices of the form for which ac 0. (iv) The set S 4 of matrices which have nonzero determinant and whose entries are integers. Solution. (i) and are matrices in S 1, but their product is not. Therefore S 1 is not closed under matrix multiplication and thus is not a group with this operation. (ii) and are matrices in S 2, but their product is not. S 2 is not closed under matrix multiplication and thus is not a group with this operation. (iii) closure: For any and a b in S3, the product a b aa +bc S c 3. Therefore S 3 is closed under matrix multiplication. associativity: Matrix multiplication is always associative, but it is easier to check the formula for matrices in our set S 3 than for general matrices: ( a ) b a b aa +bc a b c aa a aa b + c +bc c c c a a a b +b c c ( a b a b ) identity: is an element of S 3, and we know cd cd cd for any matrix cd, so there is an identity element in S3. inverses: For all in S3, ac 0 (and so a 0 c), so the matrix 1/a b/(ac) is in S 0 1/c 3 and is an inverse for because 1/a b/(ac) /c 01. 1/a b/(ac) 0 1/c

3 Problem Set 8 3 of 8 This shows that S 3 is a group under matrix multiplication. (iv) The matrix has nonzero determinant and integer entries, but we can see that it does not have an inverse with integer entries. Indeed, if cd 10 01, then we must have a 1/2 c, but there is no such matrix in S 4. Therefore S 4 does not have multiplicative inverses and cannot be a group. Exercise 2.5 A function from the plane to itself which preserves the distance between any two points is called an isometry. Prove that an isometry must be a bijection and check that the collection of all isometries of the plane forms a group under composition of functions. Solution. Suppose that f : R 2 R 2 is an isometry. We denote the distance between points a and b as d(a, b). f is injective: Let a, b R 2. If f(a) f(b), then 0 d(f(a),f(b)) d(a, b), so a b. f is surjective: Suppose p R 2 is a point in the codomain of f. We wish to show that there is some point in R 2 which is mapped to p by f. Choose two distinct points a, b R 2 in the domain of f, and let f(a) a, f(b) b. Looking in the codomain, let d 1 d(a,p ) and d 2 d(b,p ). Now draw circles of radius d 1 around a and d 2 around b in the domain, and draw corresponding circles of radius d 1 around a and d 2 around b in the codomain. These circles are exactly the points of distance d 1 from a, d 2 from b, etc., so in particular p lies in the intersection of the circles in the codomain. Notice that f must map the intersection of the circles in the domain to the intersection of the circles in the codomain. Indeed, if x is an intersection point of the two circles in the domain, then d(a, x) d 1 and d(b, x) d 2, and since f is an isometry, this means d(a,f(x)) d 1 and d(b,f(x)) d 2. But this last statement just says that f(x) must lie in the intersection of the two circles in the codomain. Using the fact that f is an isometry and the triangle inequality, we see that d(a, b) d(a,b ) d(a,p )+d(p, b )d 1 + d 2. If this inequality is an equality, then the circles in the codomain intersect in exactly one point, as do the circles in the domain (this happens when p lies on the line segment between a and b ). Calling p the point in the intersection of circles in the domain, we see that f(p) must be the intersection point in the codomain, i.e. f(p) p and we are done.

4 Problem Set 8 4 of 8 Now suppose that d(a, b) d(a,b ) <d 1 + d 2. In this case the circles in the domain intersect at exactly two points, as do the circles in the codomain. Lel the intersection points of the domain x and y; one of the intersection points in the codomain is p, so lel the other one q. If f(x) p, we are done. If not then f(x) q, and since f is injective, f(y) q so we must have f(y) p. This completes the proof of surjectivity. isometries form a group: closure: If f, g are isometries of R 2, f g : R 2 R 2, and for all a, b R 2, d(a, b) d(g(a),g(b)) d(f(g(a)),f(g(b))), so f g is an isometry. associativity: Function composition is always associative when it s defined: (f g) h(x) f(g(h(x))) f (g h). identity: The identity function on R 2 is certainly distance-preserving. inverses: Since we have proved that any isometry f is a bijection, we know that the function f 1 exists, and we have d(f 1 (a),f 1 (b)) d(ff 1 (a), ff 1 (b)) d(a, b). Therefore f 1 is an isometry. This shows that isometries of the plane form a group. Exercise 2.6 Show that the collection of all rotations of the plane out a fixed point P forms a group under composition of functions. Is the same true of the set of all reflections of lines which pass through P? What happens if we take all the rotations and all the reflections? Solution. Without loss of generality, we may suppose that P is the origin. Any rotation of the plane can be specified by the angle θ through which it rotates the positive x-axis, and it s clear that the composition of rotations θ, θ is the rotation θ + θ. As ove, function composition is always associative, and the identity function on R 2 ( rotation by 0 ) is an identity element. The inverse of rotation by θ is just rotation by θ. Therefore rotations of the plane out a fixed point form a group. The set of reflections through lines through P does not form a group under composition of functions. To see this intuitively, draw a triangle in the plane and lel the vertices a, b, c in a clockwise order. You can see that any reflection will reverse the orientation of the leling, so that when you read off a, b, c, it will now be in counterclockwise order. When you look at the image of the triangle after applying two reflections along different lines, the orientation of the leling will be clockwise again. Thus the composition of these two reflections can t be equal to a single reflection. Reflections are not closed under

5 Problem Set 8 5 of 8 function composition, so they don t form a group. We will see this more rigorously below. Notice that we can already say that the collection of reflections, rotations, and any of their compositions forms a group under composition of functions. Associativity and existence of an identity are as ove. Because the composition of a reflection with itself is the identity, and rotations have inverses, any composition of reflections and rotations has an inverse - just write the composition backwards and put inverses everywhere. For example, it s easy to see (r 1 r 2 s 1 r 3 s 2 s 3 ) 1 s 1 3 s 1 2 r 1 3 s 1 1 r 1 2 r 1 1. However we can actually show that any composition of rotations and reflections fixing the origin can be written as a single reflection or a single rotation (contrast with Exercise 1.6!). This shows that the collection of reflections and rotations is a group - the statement and any of their compositions ove was unnecessary. Recall from linear algebra that rotation counterclockwise by θ is a linear map, and its matrix with respect to the standard basis of R 2 is cos θ sin θ R θ, sin θ cos θ and that the matrix corresponding to reflection across a line making an angle of φ with the positive x-axis is cos 2φ sin 2φ S φ. sin 2φ cos 2φ (If you didn t remember these, you could produce them yourself: recall that the first column should be the image of the vector 0, 1 and the second is the image of 1, 0.) Using the trigonometric identities cos(s + t) cos s cos t sin s sin t and sin(s + t) sin s cos t + cos s sin t liberally, we compute the matrix for the following compositions by matrix multiplication: R θ S φ S φ R θ S φ S ψ cos(θ +2φ) sin(θ +2φ) sin(θ +2φ) cos(θ +2φ) cos(2φ θ) sin(2φ θ) sin(2φ θ) cos(2φ θ) cos(2φ 2ψ) sin(2φ 2ψ) sin(2φ 2ψ) cos(2φ 2ψ) S θ+2φ S 2φ θ R 2φ 2ψ In particular we can see explicitly that a composition of reflections is not a reflection.

6 Problem Set 8 6 of 8 Exercise 2.7 Let x and y be elements of a group G. Prove that G contains elements w and z which satisfy wx y and xz y, and show that these elements are unique. Solution. wx y if and only if wxx 1 yx 1, i.e. if and only if w we yx 1. Likewise xz y if and only if x 1 xz x 1 y, if and only if z x 1 y. Exercise 3.1 Show that each of the following collections of numbers forms a group under addition. (i) The even integers. (ii) All real numbers of the form a + b 2 where a, b Z. (iii) All real numbers of the form a + b 2 where a, b Q. (iv) All real numbers of the form a + bi where a, b Z. Solution. (i) The sum of two even integers 2n and 2m is 2(n + m), which is also even, so the set of even integers is closed under addition. Addition of integers is associative, and 0 is an even integer and is the additive identity element. The inverse of 2n is 2n, which is also an even integer. Therefore the even integers form a group under addition. (ii) For a, b, a,b Z, (a + b 2) + (a + b 2) (a + a ) + (b + b ) 2, and since (a + a ), (b + b ) Z, the given set is closed under addition. Addition of real numbers is always associative (though here we can get away with using only associativity for Z). The identity element is clearly , and the inverse of a + b 2 is a +( b) 2, which is in our set. This shows that the set is a group under addition. (iii) The proof of (iii) is the same as that of (ii) after replacing every occurrence of Z with Q. (iv) The proof of (iv) is the same as (ii) after replacing 2 with i. Exercise 3.8 Show that if a subset of {1, 2,..., 21} contains an even number, or contains the number 11, then it cannot form a group under multiplication modulo 22. Solution. We will show that even numbers and the number 11 cannot have multiplicative inverses modulo 22, which will show that a set containing an even number or 11 cannot be a group under multiplication mod 22. Let 2m be an even integer, and suppose that there were some integer a such that (2m)a 1 mod 22. This means there would be some integer b such that 2ma 1+(22)b. However, we could then write 2(ma + (11)b) 1, and this implies that 2 divides 1, a contradiction. Thus 2m cannot have a multiplicative inverse mod 22.

7 Problem Set 8 7 of 8 Very similarly, if there were an integer a such that 11a 1 mod 22, then there would be an integer b such that 11a b, which implies 11(a +2b) 1, and hence that 11 divides 1, a contradiction. (More generally, this method will show that any number sharing a divisor with n does not have a multiplicative inverse mod n.) Exercise 4.5 An element x of a group satisfies x 2 e precisely when x x 1. Use this observation to show that a group of even order must contain an odd number of elements of order 2. Solution. Let G be a group of even order. Define S 1 to be the set of elements in G of order 1, S 2 to be the set of elements of order 2, and S to be the set of elements of order higher than 2. These subsets partition G as a set, meaning that their union is G and the intersection of any two is empty, and so G S 1 + S 2 + S. Our goal is to show that S 2 is odd, and we know that G is even. There is exactly one element of G of order 1, namely the identity e, so S 1 1 is odd. Thus it will suffice to show that S is even. If S 0 we are done. Otherwise, lel the elements of S as {s 1,s 2,..., s n }. Where does the inverse of s 1 lie? It s easy to show that an element and its inverse have the same order (do this), so the inverse of s 1 lies in S. Furthermore, since the order of s 1 is higher than 2, we know that s 2 1 e, and hence that s 1 1 s 1. Therefore s 1 1 is an element in the set {s 2,..., s n }, and after releling the elements if necessary, we may assume that s 2 s 1 1. If n 2, we are done. Otherwise, there is some s 3, and we may ask where its inverse is. Again it must be in S, and it is not equal to s 1,s 2 or s 3, so there must be some s 4. Continue this process until reaching s n ; if n is odd we reach a contradiction, for s n will have no inverse. Thus S must be even, as desired. Additional Problems Exercise 1 Show that if a in a group G, then b e. Solution. a implies a 1 a 1 a implies eb e implies b e. Exercise 2 Show that the equation ax b has a unique solution in G. Solution. ax b in G if and only if a 1 ax a 1 b, i.e. if and only if x a 1 b. This shows that a 1 b satisfies the equation (a solution exists), and that if some element x of G

8 Problem Set 8 8 of 8 satisfies the equation, then x a 1 b (the solution is unique). Exercise 3 Let G be a group with multiplicative notation. Define the opposite group G op to be the same underlying set as G, but endowed with the operation a b ba. Prove that G op is in fact a group. Solution. binary operation: Since G is a group, ba G is an element of the set G op, so the star operation is well defined in G o p. associativity: For all a, b, c G op, a (b c )a (cb) (cb)a c(ba) by associativity in G (ba) c (a b ) c. identity: e G is also an identity for G op, since e a ae a ea a e for all a G. inverses: The inverse of an element a in G is also the inverse of a in G op, since a a 1 a 1 a e aa 1 a 1 a. This shows that G op is a group.