Math 113, Summer 2016: Exam 1 Solution Instructor: James McIvor July 16, 2016 1. (20 points) (a) Let g be an element of a group G. Give the definition of C(g), the conjugacy class of g. Sk (b) If a subgroup H can be written as a union of conjugacy classes, say H = i=1 C(gi ), prove that H is normal in G. (c) Prove that the center Z(G) = fg 2 G j xg = gx for all x 2 Gg is a subgroup of G. Solution: (a) The conjugacy class of g is the orbit of g under the action of G on itself by conjugation. Equivalently, C(g) = fxgx −1 j x 2 Gg. (b) To prove H is normal, we need to show gHg −1 = H for any g. So let g be arbitrary. Then every element of gHg −1 has the form ghg −1 for some h 2 H. Since H is a union of conjugacy classes, and h 2 H, h 2 C(gi ) for some i. But then any conjugate of h must −1 also be in C(gi ), since conjugacy is an equivalence relation; hence ghg 2 C(gi ) ⊆ H, so gHg −1 ⊆ H. Moreover, we've seen that jgHg −1j = jHj, so they must be equal. (c) a) Z(G) contains the identity element because xe = ex for all x. b) Z(G) is closed under multiplication because if g, g 0 2 Z(G), we have, for any x, x(gg 0) = (xg)g 0 = (gx)g 0 = g(xg 0) = g(g 0x) = (gg 0)x, so gg 0 2 Z(G). c) Z(G) is closed under inversion because if g 2 Z(G), and x is arbitrary, xg −1 = (gx −1)−1 = (x −1g)−1 = g −1x, so g −1 2 Z(G). 2. True/False (20 points - 2 points each). No justification required. (a) If f : G ! H is a homomorphism, and H =∼ G= ker f , then f is surjective. True - G= ker f is isomorphic to imf , which is a subgroup of H. If in addition it's isomorphic to H itself, then the image must be all of H. (b) The permutations (2534)(8713) and (7534)(1628) are conjugate in S8. False - the first permutation is not written as a product of disjoint cycles, so we cannot determine its cycle type without simplifying. Expanding out the product gives (2534)(8713) = (1425387), which has cycle type 7,1. The other permutation has cycle type 4,4. So they are not conjugate. T (c) A group G is abelian if and only if every element of G is in g2G CentG (g). True - The intersection of the centralizers is just the center, the set of things that commute with everything else. If that center is the entire group, then everything commutes with everything, so G is abelian. (d) Every element in Sn an be written as a k-cycle, for some 1 ≤ k ≤ n. False - some permutations are products of disjoint cycles, e.g., (12)(34) in S4. 1 (e) If a group G acts on a set S, and the stabilizer of s 2 S is trivial, then there is only one orbit for this action. False - For example, let R× act on R2 n f0g by scaling. Then any nonzero vector has trivial stabilizer, but there are many orbits - they're lines through the origin (we saw this action in HW2 (the orbit space is P1). (f) There exists a surjective homomorphism D8 ! Z=4Z. False: Such a homomorphism would have a kernel of size two, and hence generated by a reflection or r 2 (the only elements of order two). The kernel would also be normal. The only normal subgroup of size two is that generated by r 2, since reflections don't commute with r (so, e.g., r < s > r −1 6=< s >). Thus any such homomorphism must map r 2 to 0.¯ But then there will be no element of order 4 in the image, so it couldn't be surjective. 2 (g) If N ⊂ D8 is the subgroup fe, r g, then D8=N is isomorphic to the Klein Four Group K4. True - This quotient will be a group of size four, but it won't contain an element of order four, since the only candidate for such an element would be rN (which is equal to r 3N), but (rN)2 = r 2N = N, so rN has order two. (h) In S4, there are five conjugacy classes. True - Conjugacy classes are determined by cycle types, and each cycle type must sum to four, so there are only five possibilities: 1,1,1,1 or 2,1,1 or 2,2 or 3,1 or 4. (i) An infinite group can have only finitely many finite subgroups. False - Consider the multi- × plicative group C . It has a subgroup µn of nth roots of unity for every n, each of which is finite (of order n). (j) Every cyclic group whose order is not prime contains a nontrivial proper subgroup. False - but I accepted both since it was a trick question Technically, a cyclic group of order one counts, and since it has only one subgroup (itself), it's a counterexample. A lot of you answered True, and that's good. If the group was assumed nontrivial, the reasoning is thus: By the structure theorem for cyclic groups it's enough to consider Z and Z=nZ (n > 1). In the case of Z, there's the subgroup 2Z, for instance. In the case of Z=nZ, for n composite, write n = km, for k, m 6= 1. Then the subgroup generated by k (or m) is proper and nontrivial. 3. (15 points) (a) Give the definition of an injective function. (b) Let f : K ! G, g : G ! H be homomorphisms. Prove that if g ◦ f is injective, then so is f . (c) For each of the following functions, say whether it is a) a homomorphism, b) injective, and c) surjective (you may assume all the functions are well-defined without checking). No justification required. i. Z ! Z=5Z; n 7! n −1 ii. D2n ! D2n; g 7! g iii. Z=pZ ! Z=p2Z; m 7! pm Solution: (a) Let f : S ! T . f is injective if f (s) = f (s0) implies s = s0. I also accepted \f is injective if its kernel is trivial" although technically that's incorrect since it only applies to homomorphisms, not arbitrary functions, and is not the definition but a consequence of the definition and the homomorphism property.. (b) We'll show the kernel of f is trivial. Pick k 2 ker f . Then f (k) = eG . Applying g to both sides gives (g ◦ f )(k) = g(f (k)) = g(eG ) = eH , since g is a homomorphism. Since (g ◦ f )(k) = eH and g ◦ f is injective, k = eK , which shows that ker f = feK g, so f is injective. 2 (c) i. This is a surjective homomorphism which is not injective. ii. This is not a homomorphism but it is injective and surjective. iii. This is an injective homomorphism which is not surjective. 4. (25 points) (a) Give the definition of a p-group. (b) Give an example of a 3-group which is not cyclic. (c) Let G be a group of order 45. Prove that there is exactly one Sylow 3-subgroup H and exactly one Sylow 5-subgroup K. (d) Let G be a group of order 99. There exists exactly one Sylow 3-subgroup H and exactly one Sylow 11-subgroup K. (You do not have to show this!) Prove that the action K × H ! H ;(k, h) 7! k · h = khk−1. is trivial.1 (e) (same assumptions as in (d)) Show that there exists x 2 K such that G=H = fx i H j i = 0, ... 10g. Deduce that K ⊂ Z(G), the center of G. Solution: (a) A p-group is a group whose order is a power of p. (b) Z=3Z × Z=3Z has order 9 = 32, so it's a p-group, but it's not cyclic because there are no elements of order 9. (c) By Sylow's theorems and corollaries, we have k3j5, so k3 = 1 or 5, and k3 ≡ 1 mod 3. Thus k3 = 1. Similarly, k5j9, so k5 = 1, 3, 9, and k5 ≡ 1 mod 5, so k5 = 1. (d) By the orbit stabilizer theorem, the size of each orbit must divide the order of K, 11, so each orbit has size 1 or 11. But by the orbit partition, the sum of the sizes of the orbits must be 9, the size of the set H being acted upon. Thus we cannot have any orbits of size 11, so each orbit Oh = fhg. This means that for any k and h, k · h, being in the orbit O, must be equal to h, so the action is trivial. (e) K has order 11 so is cyclic, generated by some x. We'll show that this generator has the desired property. First of all the set of cosets G=H has size 11 by Lagrange's Theorem, and the cosets H, xH, x 2H, ... x 10H (note that x 11H = eH = H are elements of this set of 11 cosets. So we'll be done if we can show that these 11 cosets are all distinct. Suppose that x i H = x j H for some 0 ≤ j < i ≤ 10 . Then x i−j 2 H. But x is a generator for K, so x i−j is also in K.