Here is another way to picture cyclic subgroups and cosets, one that is accessible

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1 MAT 6200, chapter 3a Decimal Wheels Here is another way to picture cyclic subgroups and cosets, one that is accessible to middle school children. Using a calculator, expand the fractions /7, 2/7, 3/7, 4/7, 5/7, and 6/7 as repeating decimals. 7 = = = = = = We see that each of these decimal numbers consists of the same six digits and in the same order, just starting in a different place in the sequence. We can display this cyclic behavior by placing the six digits on a decimal wheel.

2 7 4 The 7 ths Decimal Wheel From the wheel, the decimal representation of any proper fraction with denominator 7 can be read off by going around the wheel. For instance, to find the decimal expansion of 4/7, start at the digit 5 and go around clockwise to obtain Let s try denominator 3. 2

3 3 = = = = = = = = = = = = This time there are two distinct classes of decimal digits H 2H

4 So, what is it about the fractions 3, 0 3, 9 3, 2 3, 3 3, and* 4 3 that makes them belong together, and similarly for the fractions on the second decimal wheel? To answer this question using elementary arithmetic, we expand 3 as a decimal by long division, keeping track of the remainder at each step, as in homework problem 7 of chapter The remainders 0, 9, 2, 3, 4, are exactly the numerators, in the correct order, of the six fractions on the first decimal wheel. In the language of group theory, these numerators comprise the cyclic subgroup H generated by the congruence class of 0 in the group Z * 3. The six elements on the second wheel represent the coset 2H. If you were presenting this topic to school children, would one of them notice that any pair of opposite decimal digits adds up to 9, while any pair of opposite fractions 4

5 adds up to? (This result is called Midy s Theorem it works for all prime denominators p provided the period of /p is even.) Homework. How many irreducible proper fractions are there with denominator 39? List them. (This is another way to understand what ϕ(39) means.) 2. For each irreducible fraction x, use a calculator to expand the fraction as a 39 repeating decimal. What is the common period of these decimal representations? 3. Group the fractions together that have the same six decimal digits. Display the answers as four decimal wheels. 4. In this problem the order of [0] in Z 39 is 6 and the index of the cyclic subgroup H!=! [0] is 4. Label the four decimal wheels with the names of the appropriate cosets, as above for the example of denominator (a) Can you spot the inverse of 0 mod (39) on the decimal wheel for H? (b) Compute the cyclic subgroup generated by [4] in Z * 39, and compare the answer with the elements of H in the opposite order. 6. (a) Use the method of al- jabr and al- muqabala to solve the equation [0][x ]=[5] in Z * 39. 5

6 (b) Intuitively, the solution to the equation 0x = 5 ought to be x = ½. The fraction ½, however, is not an element of Z * 39. But [2] is! Is your answer to part (a) equal to [2]? 6. Computer problem. Is 980 the world s coolest rational number? Use a computer to expand this fraction as a repeating decimal. The period is 98. (To see why it turns out this way, write 980 = 99 = ) Internet problem: Cyclic numbers. We know that = The whole 7 number is an example of a cyclic number. This means that if we multiply by each of, 2, 3, 4, 5, and 6, the answer will always comprise the same six digits, in the same order, just starting in a different place. For instance = 28574, corresponding to the decimal expansion of 2/7. All cyclic numbers arise in this way. Suppose that P is a prime number for which [0] has maximal order P mod(p). In other words, the cyclic subgroup generated * by [0] in the group Z 39 is the whole group, and there are no other cosets. This is equivalent to the condition that the period of the repeating decimal representation of /P is the maximal possible length, P. For example, /7 = (period 6), and

7 is a cyclic number. (Although this is a whole number, leading 0s are OK in this theory.) Find another cyclic number. Check that it really is cyclic by multiplying it by 2, 3, 4,, P. Practice exam - - chapters 3 and 3a. Let (G, ) be a group with identity e and let a,b,c G. Prove the following cancellation property. If a b!=!a c then b = c. Indicate where you use the associative property, the defining property of inverses, and the defining property of the identity. 2. The following set of four matrices forms a group under the operation of matrix multiplication. (You do not have to prove this.) (*! ) +*,!,!!,* -.* (a) What is the identity element? (b) What is the inverse of? (c) Use the method of al- jabr and al- muqabala to solve the equation AX!=!B 7

8 where A!=! and B!=!. 2. (a) Find the order of the multiplicative group G = Z * 37. (b) Compute the elements of the cyclic subgroup H = [0] 37 generated by the congruence class [0] mod(37). What is the order of the element [0] in Z * 37? Is this result compatible with Lagrange s Theorem? (c) We have a theorem that asserts that if the order of [0] mod(p) is k, then P is a factor of (k ones). Check to see if this theorem is true for P = 37. (d) How many cosets will the subgroup H = [0] 37 have in the group G = Z * 37? (e) Compute the cosets 2H and 3H. (f) Add together the three elements of H. Add together the three elements of 2H. Add together the three elements of 3H. (g) Fill in the blank below to form a conjecture about the sum of the elements of a coset of this type. Conjecture. Let P be a prime number and let [x ] be an element of Z * P of order k. Denote by H the cyclic subgroup of Z * P generated by [x ], and let yh be a coset of H. Then the sum of (the representatives of) the k elements of yh is congruent to mod(p). 8

9 (h) Check your conjecture in part (g) against the example P = 4, H = [0] 4, for the coset H itself. 3. (a) Expand the fraction /89 as a repeating decimal by long division, keeping track of the partial remainder at each step. (b) Put the partial remainders obtained in part (a) on a wheel as indicated (c) Write the last digit of each remainder in order, staring with the remainder 0. Compare with the decimal digits of the quotient of the division. (Caveat: this neat trick works only for numbers ending in 9, like 89.) (d) If x is an element of order k in a group G, then for all exponents j we have x j x k j!=!x k!=!e, the identity in the group. Use this observation to match up the six elements on the remainder wheel with their inverses in the group Z * 89. 9

10 Solutions to practice exam!. (a) The identity element is the matrix I!=!.! Check:! a c! b = d a c b d (b) The inverse of is. Check: =. (c) To solve the equation AX =B, multiply both sides by the inverse of A. X = ( * ) + X - =, ( * ) + - X =, (associativeproperty) X = (propertyofinverses) 0

11 X!=! (property!of!the!identity) X!=!!=!A (definition!of!matrix!multiplication) But that was the hard way, the method of plain old matrix algebra. The method of abstract algebra is to work with A s, B s and C s as long as possible without caring what the letters stand for. Problem: Solve the equation AX!=!B in any group. Solution: Multiply both sides by the inverse of A and apply the three group properties to achieve the solution X!=!A B We have solved the equation without knowing anything about matrices at all. For a numerical solution we have one more step: X!=!A B!=!!=! where the matrix A was determined in part (a). 2. (a) 37 is prime, so Z 37 =ϕ(37)=36.

12 (b) H!=! [0] 37!=!{[0],[0] 2,[0] 3,...}! =! {[0] 37,[00] 37,[000] 37,...}!=!{[0],[26],[] } The order of [0] is 3. This is compatible with Lagrange s Theorem because 3 is a factor of 36. (c) Yes, this checks out because 3 is a factor of. ( = 37 3.) (d) There will be 36/3 = 2 cosets, counting H itself. (e) H = {,0,26} (omitting brackets for less cumbersome notation) 2H = {2, 20,5}, where 5=2 26 reduced mod(37) 3H = {3, 30, 4} (f) = 37 0mod(37) = 37 0mod(37) = 37 0mod(37). (g) Conjecture. Let P be a prime number and let [x ] be an element of Z * P of order k. Denote by H the cyclic subgroup of Z * P generated by [x ], and let yh be a coset of H. Then the sum of (the representatives of) the k elements of yh is congruent to 0 mod(p). (At least, that s the way it turned out in this example.) (h) H = {[0] 4,[0] 2 4,...} )=){[0],[8],[6],[37],[]} = 82 = mod(4) (check!) 2

13 3. (a) (b) (c) Remainders: 0, , 72, 9, Last digits: Quotient:

14 (d) [0] 6 = [] [0] [0] [0] and [9] are inverses [0] [0] 2 [00] and [72] are inverses [0] 3 [55] is its own inverse Lawrence Brenton Mathematics Department Wayne State University 4