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4. Consider the subgroup H = {1, 10, 26} of U37. (a) Find and compare the orders of 6 E U37 and 6H E U37/H. (b) Repeat part (a) with 34 E U37 and 34H E U37/H. It might help to write 34 as -3 modulo 37. (c) Repeat part (a) with 4 E U37 and 4H E U37/H. (d) What conjecture do you have?

4. Consider the subgroup H = {1, 10, 26} of U37. (a) Find and compare the orders of 6 E U37 and 6H E U37/H. (b) Repeat part (a) with 34 E U37 and 34H E U37/H. It might help to write 34 as -3 modulo 37. (c) Repeat part (a) with 4 E U37 and 4H E U37/H. (d) What conjecture do you have?

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Chapter 22. Quotient Group Examples 3. Consider the subgroup H = {1, 10, 26} of U37. Verify that H is equal to (10) {10k | ke Z}, i.e., the cyclic subgroup generated by 10. (This computation is re- ferred to in Section 22.2.) 228 4. Consider the subgroup H = {1, 10, 26} of U37. (a) Find and compare the orders of 6 € U37 and 6H E U37/H. + (b) Repeat part (a) with 34 € U37 and 34H € U37/H. It might help to write 34 as -3 modulo 37. (c) Repeat part (a) with 4 € U37 and 4H € U37/H. (d) What conjecture do you have? 5. Consider the subgroup H = {1, 10, 26} of U37. Let a € U37 with ord(a) = 18. Show that (aH)¹8 = 1H. What does this say about the order of aH in U37/H? Explain. 6. Our friends are working on Exercise #4: Elizabeth: Phew! I just found that ord (4) = 18 in U37. Anita: Great! So ord (4H) must be 18 as well, since (4H)18 = 418H = 1H. Elizabeth: But do we know if 18 is the smallest positive exponent for 4H? Anita: Sure. If n is less than 18, then (4H)" = 4" H can't equal 1H, because 4" #1. How would you respond to Anita? F. Prove: Consider the subgroup H = {1, 10, 26) of U37. Let aH E U37/H where a € U37. Then ord (aH) in U37/H is a divisor of ord(a) in U37. Consider the subgroup H = {1, 10, 26} of U37. (a) Find (15H)-¹, i.e., the multiplicative inverse of 15H in U37/H. () Find (28)-1
Transcribed Image Text:Chapter 22. Quotient Group Examples 3. Consider the subgroup H = {1, 10, 26} of U37. Verify that H is equal to (10) {10k | ke Z}, i.e., the cyclic subgroup generated by 10. (This computation is re- ferred to in Section 22.2.) 228 4. Consider the subgroup H = {1, 10, 26} of U37. (a) Find and compare the orders of 6 € U37 and 6H E U37/H. + (b) Repeat part (a) with 34 € U37 and 34H € U37/H. It might help to write 34 as -3 modulo 37. (c) Repeat part (a) with 4 € U37 and 4H € U37/H. (d) What conjecture do you have? 5. Consider the subgroup H = {1, 10, 26} of U37. Let a € U37 with ord(a) = 18. Show that (aH)¹8 = 1H. What does this say about the order of aH in U37/H? Explain. 6. Our friends are working on Exercise #4: Elizabeth: Phew! I just found that ord (4) = 18 in U37. Anita: Great! So ord (4H) must be 18 as well, since (4H)18 = 418H = 1H. Elizabeth: But do we know if 18 is the smallest positive exponent for 4H? Anita: Sure. If n is less than 18, then (4H)" = 4" H can't equal 1H, because 4" #1. How would you respond to Anita? F. Prove: Consider the subgroup H = {1, 10, 26) of U37. Let aH E U37/H where a € U37. Then ord (aH) in U37/H is a divisor of ord(a) in U37. Consider the subgroup H = {1, 10, 26} of U37. (a) Find (15H)-¹, i.e., the multiplicative inverse of 15H in U37/H. () Find (28)-1
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