# FINAL EXAM MATH 150A FALL 2016

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1 FINAL EXAM MATH 150A FALL 2016 The final exam consists of eight questions, each worth 10 or 15 points. The maximum score is 100. You are not allowed to use books, calculators, mobile phones or anything else besides your writing utensils and the paper provided to you. You should write the answer to a question in the empty space on the same page. Please write your name clearly below and which discussion section you re officially enrolled in (Tuesday 5 : 10 6 : 00 or Tuesday 6 : 10 7 : 00.) NAME: SECTION: You have 2 hours to work on the 8 questions. If you don t know how to solve a question, don t spend all your time on it you can return later to it, after you ve worked through the other questions. You are free to use without proof any result we ve talked about in class. If you have questions about the statement of a problem, just raise your hand. Good luck! 1

2 2 MATH 150A Question 1. [15 points] (a) [5 points] What are the possible values of a perfect square n 2 mod 5? (b) [5 points] What are the possible values of a perfect fourth power n 4 mod 5? (c) [5 points] What are the possible values of a perfect hundredth power n 100 mod 5? Solution. (a) Answer: 0, 1, 4. There are 5 cases to check. (b) Answer: 0, 1. If you use (a) and n 4 = (n 2 ) 2, there are only 3 cases to check, 2 of which are obvious. (c) Answer: 0, 1. Of course, 0 and 1 are possible because = 0 and = 1. However, any perfect hundredth power is also a perfect fourth power, so, by part (b), there are no other possibilities.

3 FINAL EXAM 3 Question 2. [15 points] (a) [10 points] Use the Euclidean algorithm to compute the gcd of 216 and 132. (b) [5 points] Why does the Euclidean algorithm work? In other words: why do you always get the gcd of a and b by running the Euclidean algorithm on the pair (a, b)? Solution. (a) Compute, get the answer 12. (b) Mainly because the gcd of the two numbers doesn t change at any step. When the two numbers become equal, they will also be equal to their gcd. To be completely rigorous, one should also argue that the process terminates. One way to do that is simply to say that the sum of the two numbers is strictly decreased at each step, so the process started on (a, b) will end after at most a + b steps. (Of course, this is a terrible approximation of the number of steps, but it doesn t matter).

4 4 MATH 150A Question 3. [10 points] Consider the groups G 1 = Z/36Z, G 2 = Z/6Z Z/6Z and G 3 = Z/4Z Z/9Z. Is G 1 isomorphic to G 2? Is G 1 isomorphic to G 3? Is G 2 isomorphic to G 3? Prove all your answers. You are allowed to use any theorems. Solution. G 1 is not isomorphic to G 2. G 1 is cyclic, so it contains elements of order 36. On the other hand, the order of any element in G 2 is at most lcm[6, 6] = 6. Hence the groups are not isomorphic. G 1 is isomorphic to G 3. This simply follows from a theorem proved in class, since 4 and 9 are coprime. G 2 is not isomorphic to G 3. This follows from the previous two.

5 FINAL EXAM 5 Question 4. [10 points] Let G be group with 100 elements, G a group with 99 elements and let ϕ be a group homomorphism from G to G. Prove that ϕ(x) = 1 G, for all x G. Here, 1 G denotes the identity element of G. Solution. This was both on the homework and on the practice problems, with different numbers. We claim that im(ϕ) divides both G and G. The fact that im(ϕ) divides G follows automatically from Lagrange s Theorem, since im(ϕ) is a subgroup of G. For the other half, we have from the index of the kernel Theorem that im(ϕ) = [G : ker(ϕ)], which divides G again by Lagrange s Theorem. Hence im(ϕ) divides gcd( G, G ) = gcd(100, 99) = 1, i.e. im(ϕ) = 1. Therefore, im(ϕ) = {1 G }, completing the proof.

6 6 MATH 150A Question 5. [15 points] (a) [5 points] What is the center of a group? (b) [10 points] If H is a subgroup of a group G, consider the following statement: The center of H is equal to the intersection of H with the center of G. Prove that if G = GL 2 (R) and H = SL 2 (R), then the statement above is true. However, prove that if G = GL 2 (R) and H is the subgroup generated by the matrix [ ] 0 1, 1 0 then the statement is false. Notes. Recall that GL n (R) and SL n (R) are the general linear group, respectively special linear group; they consist of invertible n n matrices and n n matrices of determinant 1 respectively. The composition law is always matrix multiplication. If you find it useful, you may invoke the homework question stating that the center of GL n (R) consists precisely of the nonzero multiples of the identity matrix. However, this is somewhat of a red herring it is quite likely to lead you astray. Solution. (a) It s the set of all elements of G which commute with all elements of G. It turns out it s actually a (normal) subgroup of G. (b) As a preliminary remark, Z(G) H Z(H): all elements in H which commute with all elements in G obviously commute with all elements in H. Therefore, the issue is whether Z(G) H Z(H), i.e. whether any element of H which commutes with all elements of H in fact commutes with all elements of G. Case 1. H = SL 2 (R). This is actually quite fun. The main observation is this: if a matrix A commutes with another matrix B, then it commutes with all scalar multiples of B. In particular, we already know that any A Z(H) commutes with all matrices of positive determinant, so we re almost done. Since K, the group of 2 2 matrices of positive determinant, is a subgroup of index 2 of G, there are just two possibilities: Z(A) is either K or G. However, it can t be K because, for instance, A commutes with [ ] [ ] = [ ] 1 0, 0 1 which has negative determinant. Hence A commutes with all matrices in G, i.e. A Z(G), as required. Case 2. H is generated by the matrix in the problem. In this case, H has order 2, so H = Z(H). Hence, if the statement was true, then H Z(G). However, this is not true; either by the homework problem, or you can simply check that the matrix in the statement doesn t commute, for instance, with [ ] Hence the statement in false in this case. Note. Of course, in reality, the center of SL 2 (R) is simply { I 2, I 2 }. Some of you actually proved this on the exam, which is quite impressive. The point of the question is that it can be avoided.

7 FINAL EXAM 7 Question 6. [15 points] (a) [5 points] If G is a group and H a subgroup, under what circumstances can you define the quotient group G/H? Just name the condition no justification required. (b) [5 points] Is Q isomorphic to Z (Q/Z)? (c) [5 points] Is C isomorphic to R (C/R)? Note. All groups Z, Q, R, C and their quotients are taken with usual addition. Solution. (a) If H is normal. (b) No, they re not isomorphic. This was also on the second midterm. The element (0, 0.5) of Z (Q/Z) has order 2, whereas Q contains no element of order 2. (c) Yes, they are isomorphic. By the first isomorphism theorem applied to the imaginary part homomorphism Im : C R, we have C/R = R. On the other hand, the map (x, y) x + iy is an isomorphism R R = C. Hence, as desired. C = R R = R (C/R), Note. In case you re wondering whether R is isomorphic to Q (R/Q), the answer is yes, but the proof is counterintuitive and hardly related to what we did in this course. Strangely enough, R, R/Q and R Q are all isomorphic, so R = Q R = Q (R/Q).

8 8 MATH 150A Question 7. [10 points] (a) [5 points] What is the centralizer of an element of a group? (b) [5 points] Does there exist a group of order 100 whose center has order 20? Solution. (a) It s the set of all elements of the group which commute with the given element. It turns out to be a subgroup. (b) No. This is almost identical to the proof of the fact that p-groups of order p 2 are commutative and to a related problem on the practice list. Assume by way of contradiction that such a group G existed. Choose an arbitrary element x / Z(G). Then Z(G) Z(x) G. The inclusion on the left is strict because x Z(x), but x / Z(G) by assumption. By Lagrange s Theorem, Z(x) is a multiple of 20 strictly bigger than 20 and a divisor of 100. So the only possibility is Z(x) = 100, i.e. Z(x) = G. Read differently, this just says x Z(G), which is a contradiction.

9 FINAL EXAM 9 Question 8. [10 points] (a) [4 points] What is the group of (all) symmetries of a cube? Your answer should be of the form It is a group of order [...], isomorphic to the group [...]. No proof required. (b) [3 points] Let A, B, C, D be the vertices of a regular tetrahedron. Can you find four (small and disjoint) cubes of equal volumes, centered at the vertices A, B, C, D, such that all symmetries of the tetrahedron ABCD are also symmetries of the (4-piece) solid obtained by taking the union of the four cubes? (c) [3 points] Same question with a regular icosahedron instead of a regular tetrahedron: can you find 12 (small and disjoint) cubes of equal volumes, centered at the 12 vertices of the regular icosahedron, such that all symmetries of the icosahedron are also symmetries of the (12-piece) solid obtained by taking the union of the 12 cubes? Solution. (a) It s a group of order 48, isomorphic to S 4 C 2. (b) Yes, it is possible. The key here is to remember the geometric trick discussed in class, which underlies the relation between the symmetry group of a tetrahedron and the symmetry group of a cube. The key point is that any regular tetrahedron can be inscribed in a cube, i.e. the four vertices of the tetrahedron become four of the cube s vertices. With this picture in mind, the problem becomes obvious: just take the faces of the 4 small cubes to be parallel to the faces of the auxiliary cube and the condition is obviously satisfied. (c) No, it s not possible. In plain language, the icosahedron has some 2kπ/5-rotational symmetries (which fix two vertices), which a cube does not. More scientifically, consider the action of the group of orientation-preserving symmetries of the icosahedron on the set of vertices. The stabilizer H of any vertex is isomorphic to C 5. However, H will also be a subgroup of the group of symmetries of the small cube at that vertex. Hence, by Lagrange s theorem and part (a), H = 5 divides 48, contradiction.

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