Math 360, Fall 2021, Assignment 7

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Do not imagine that mathematics is hard and crabbed and repulsive to commmon sense. It is merely the etherealization of common sense.

- Lord Kelvin

Read:

  1. Section 6.

Carefully define the following terms, then give one example and one non-example of each:

  1. Multiplicative notation (for a general group).
  2. Additive notation (for a general abelian group).
  3. Cyclic group.
  4. Generator (of a cyclic group).

Carefully state the following theorems (you need not prove them):

  1. Laws of exponents.
  2. Laws of multiples (i.e. the restatement of the laws of exponents in additive notation).
  3. Theorem concerning integer division.
  4. Classification of cyclic groups.

Solve the following problems:

  1. Section 6, problems 1, 3, 9, 10, 17, 19, 33, 34, 35, 36, and 37.
  2. Prove that every cyclic group is abelian. (Hint: every element has the form $g^i$ for some fixed generator $g$; now use the laws of exponents.)
  3. Prove that every cyclic group is countable (i.e. either finite or countably infinite; you may utilize the classification of cyclic groups even though we have not yet completed its proof in class).
  4. Show that each of the following subgroups of $(\mathbb{Z},+)$ can be generated by a single non-negative integer: (a) $\left\langle 4, 6\right\rangle$, (b) $\left\langle 15, 35\right\rangle$, and (c) $\left\langle 12, 18, 27\right\rangle$.
  5. (Challenge) Following the pattern of the three parts of the last problem, try to guess a general formula for a single non-negative generator for the subgroup $\left\langle k_1,k_2,\dots,k_m\right\rangle$ of $(\mathbb{Z},+)$.
--------------------End of assignment--------------------

Questions:

Solutions:

    1. Textbook Solution

1. 42 = 9·4+6, q = 4, r = 6

3. −50 = 8(−7)+6, q = −7, r = 6

9. 1, 3, 5, and 7 are relatively prime to 8 so the answer is 4.

10. 1, 5, 7, and 11 are relatively prime to 12 so the answer is 4.

17. gcd(25, 30) = 5 and 30/5 = 6 so <25> has 6 elements.

19. The polar angle for i is π/2, so it generates a subgroup of 4 elements.

33. The Klein 4-group

34. <\mathbb{R}, +>

35. $\mathbb{Z}_2$

36. No such example exists. Every infinite cyclic group is isomorphic to <\mathbb{Z}, +> which has just two generators, 1 and -1.

37. $\mathbb{Z}_8$ has generators 1, 3, 5, and 7.