MATH 131 SOLUTION SET, WEEK 12

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1 MATH 131 SOLUTION SET, WEEK 12 ARPON RAKSIT AND ALEKSANDAR MAKELOV 1. Normalisers We first claim H N G (H). Let h H. Since H is a subgroup, for all k H we have hkh 1 H and h 1 kh H. Since h(h 1 kh)h 1 = k, it follows that hhh 1 = H, proving the claim. That H is in fact a normal subgroup of N G (H) is immediate from the definition of N G (H). Now let G := D 8 and H := f. Since f 2 = id G we have H = {id G, f. Thus g N G (H) {g id G g 1, gfg 1 = {id G, f {id G, gfg 1 = {id G, f gfg 1 = f for g G. By the relations defining G, any g G can be written uniquely as r i f j where 0 i 3 and 0 j 1, whence gfg 1 = (r i f j )f(f j r i ) = r i fr i = r 2i f. Since r 2i = id G precisely when i {0, 2 it follows that N G (H) = {id G, f, r 2, r 2 f. Then it s clear that N G (H)/H = {[id G ], [r 2 ], where [ ] denotes the corresponding coset of H. 2. Cayley graph for D 8 (i). As in the case of D 6 S 3 we discussed in lecture, the covering space E corresponding to φ : F 2 D 8 is the Cayley graph of D 8 with generators r, f, pictured below. (Here red arrows correspond to r and blue to f.) f rf id r r 3 r 2 r 3 f r 2 f Date: December 7,

2 (ii). Also analogous to the case of D 6 S 3, the covering space E corresponding to φ 1 ( f ) is given by quotienting the Cayley graph from (i) by the action of f, that is, by left-multiplication by f. This gives us the following graph. id r r 3 r 2 (iii). Recall a Galois automorphism is determined uniquely by where it sends any point. Let s consider the vertex id in E above. Its f loop must be sent to an f loop, so the only vertex it can be sent to is r 2. And indeed rotation by π, which of course has order 2, gives such a automorphism. Thus this and the identity are the two elements of Gal(E /B). 3. F n as a subgroup of F 2 Alex likes graphs, so here s a fun geometric solution. Recall the following facts: If p : (E, e) (B, b) is a covering, π 1 (E, e) p π 1 (E, e) π 1 (B, b). The fundamental group of a graph G is a free group on E(G) V (G) + 1 generators. Since π 1 (S 1 S 1 ) F 2, if we can show a covering graph G of S 1 S 1 with E(G) V (G) + 1 = n, we ll be done. Now consider the covering space G n of S 1 S 1 obtained by a directed a cycle with n 1 vertices that overlaps with a directed b cycle with n 1 vertices, for all n 3. This space is illustrated below in the cases of n = 4 (here red arrows correspond to a and blue to b): Observe that E(G n ) V (G n ) + 1 = 2(n 1) (n 1) + 1 = n, so π 1 (G n ) F n, as desired. Note thatthe generating set of loops for π 1 (G n ) F n maps to a generating set of loops for F n π 1 (S 1 S 1 ) F 2. So pick some base point e n G n and consider a spanning tree T n G n that consists of going counterclockwise along an a edge n 2 times (notice that G n has n 1 vertices, so the tree has n 2 edges). Recall that T n is deformation retractable to e n, and this deformation retract induces a homotopy equivalence between G n and a bouquet of n circles wedged together at e n. It s easy to see that the paths in 2

3 G n that are homotopy equivalent to the n circles are given by the following n loops based at e n in G n : a b a a b a 1,. a... a b a 1... a 1, n 2 n 3 a... a, n 1 a... a n 2 b 1 The above generating set of F n maps to a generating set of F n as a subgroup of F 2, given by S = {ab, a 2 ba 1,..., a n 2 ba (n 3), a n 1, a n 2 b Isometries and free products It suffices to show that there is no relation between a and b except for a 2 = b 2 = id G. So, assume that a n1 b m2... a n k b m k = id G for some integers n i, m i. Since a 2 = b 2 = id G, we can assume without loss of generality n i {0, 1, m i {0, 1. Thus, we have four cases: (1) ab... ab n = id G for some n > 0. Observe that ab : x x 1. Thus, x n = x for all x R, which is a contradiction. (2) ab... ab a = id G for some n 0. This means that x n = x for all x R, n which is a contradiction again. (3) The other two cases can be deduced in a similar manner. 5. Galois groups more generally This is just [Munkres, Theorem 81.2]. (Sorry Arpon is such a bad CA, but he s a bit pressed for time.) Recall the following facts: 6. Degree 5 Galois covers of S 1 S 1 The set of reasonable Galois covering spaces of S 1 S 1 with Galois group G is in bijection with the set of kernels of surjective homomorphisms ψ : F 2 G. For a connected Galois covering space p : (E, e) (B, b), the Galois group Gal(E/B) is in bijection with the fiber p 1 (b) Thus, since we re looking for degree 5 covers and 5 is prime, the Galois group can only be Z 5. So we want to count the number of different subgroups of F 2 which arise as kernels of surjective homomorphisms ψ : F 2 Z 5. 3

4 Now, identify Z 5 = {0,..., 4. Every such homomorphism is uniquely determined by where it sends the generators a and b. Moreover, observe that postcomposing ψ with an automorphism of Z 5 gives us a different homomorphism, but leaves the kernel the same, since every automorphism of Z 5 fixes 0. Hence, without loss of generality, ψ sends a to either 0 or 1, and if it sends a to 0, it sends b to 1. Thus, there are several cases: (1) ψ(a) = 1, ψ(b) = j for j Z 5. The kernel is determined by the condition ψ(a n1 b m1... a n k b m k ) = 0 k k n i + j m i 0(mod 5) i=1 Clearly, for different j we get different kernels. (2) ψ(a) = 0, ψ(b) = 1. The kernel is determined by the condition i=1 ψ(a n1 b m1... a n k b m k ) = 0 k m i 0(mod 5) which is different from the kernels in the above case. Thus, the number of regular connected 5-covers of S 1 S 1 is 6. i=1 7. S 1 S 1 S 1 S 1 (i). Observe that if we take the image of the boundary of the square under the standard quotient map p : I 2 R 2 /Z 2, we get two circles glued together at a single point (the common image of the four vertices of the square). This gives us a natural homeomorphism from S 1 S 1 to p( I 2 ) π(i 2 ) = R 2 /Z 2. (ii). Suppose that there is a retraction r : R 2 /Z 2 p( I 2 ), and let i : p( I 2 ) R 2 /Z 2 be the canonical inclusion. Then, since r is a retraction, r i is the identity on p( I 2 ). Consequently, (r i) = r i is the identity on π 1 (p( I 2 )) = F 2, which means that the identity homomorphism on F 2 factors through π 1 (R 2 /Z 2 ) = Z 2. In particular, the homomorphism i is injective. However 7.1. Lemma. There is no injective homomorphism ψ : G H when G is nonabelian and H is abelian. Proof. Let ψ : G H be a homomorphism, and let a, b G be such that ab ba. Then, so ψ fails to be injective. ψ(ab) = ψ(a)ψ(b) = ψ(b)ψ(a) = ψ(ba) 8. S 1 S 1 is a torus with one point removed The point of the last problem was to give you a little hint of how this one might work. There, there was no retraction of the torus onto the image of the boundary of I 2 ; but in this case, it is easy to see that the torus without a point in fact deformation retracts to S 1 S 1. Probably the easiest way to convince yourself of this is by looking at the gluing diagram. Without loss of generality, the missing point is the center of the square 4

5 (why?). Then, we can expand this point to a continuously increasing square-shaped hole with sides parallel to the sides of I 2 and centered at the center of I 2, until we are left with the boundary. One can easily write down a concrete expression for this transformation, from which it is obvious that it is continuous. Let a = (1/2, 1/2) I 2. Thus, I 2 a deformation retracts to its boundary via some continuous homotopy H : [0, 1] (I 2 a) (I 2 a). Let p : I 2 R 2 /Z 2 be the standard quotient map that gives us the torus; then, it restricts to a quotient map p : I 2 a R 2 /Z 2 p(a). It is easy to see that p H is constant on fibers of id p, since the boundary I 2 is fixed throughout the homotopy H. Finally, id p is a quotient map, by the following general fact from Munkres: 8.1. Lemma (Munkres, Exercise 29.11). If p : X Y is a quotient map, and Z is locally compact and Hausdorff, then p id Z : X Z Y Z is a quotient map. Indeed, in our case Z = [0, 1] has the required properties (and perhaps it s not hard to see that the map in question is a quotient map from first principles). Hence, by the universal property of the quotient, we have a map h that makes the below diagram commute: I (I 2 a) id p I (R 2 /Z 2 a) p H h R 2 /Z 2 p(a) Since H is a deformation retract, h is also a deformation retract; finally, observe that the subspace to which R 2 /Z 2 p(a) retracts to is exactly the image of the boundary I 2, which is homeomorphic to S 1 S Continuation of 11.3 Thanks to Nat Mayer for writing good solutions to this one. For (x, y) R 2 denote by L(x, y) the line segment from (0, 0) to (x, y). (i). It suffices to show the segment L(q, p) is a fundamental domain for the line px = qy (since the boundary points are clearly identified, this then gives a circle on the torus). Suppose we have (x, y), (x, y ) on the line with (x x, y y ) = (n, m) Z 2. Then we must have pn = qm, and since p, q are relatively prime, this occurs if and only if n = kq and m = kp for some k Z. It s clear then the orbit of (0, 0) intersects the segment only at the boundary, and all orbits intersect the segment precisely once, as desired. (ii). Let a, b Z such that ap + bq = 1. Define the linear map F : R 2 R 2 by F (x, y) = (px qy, bx + ay), so det(f ) = 1. From problem 3 from last week, F is the lift of a homeomorphism f : T T. Clearly F takes the line segment L(q, p) to L(0, 1), whence f takes C(p, q) to C(1, 0). (iii). Applying the homeomorphism f from (ii) doesn t change the number of intersections so it suffices to find the number of intersections of C(1, 0) and C(as + br, ps qr). Lifting to the fundamental domains described in (i), it suffices to find the number of intersections of L(ps qr, as + br) {(ps qr, as + br) with the 5

6 Z 2 -translates of the y-axis, i.e., the numer of integer x-coordinates achieved. It s easy to see that as long as ps qr 0, this is precisely ps qr. 6