Math 360, Fall 2021, Assignment 7

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 $(\mathbb{Z},+)$.

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Textbook Solution:

1. 42 = 9·4+6, q = 4, r = 6

3. −50 = 8(−7)+6, q = −7, r = 6

To find number of generators of a cyclic group of order $n$, we can consider $\mathbb Z_n$ under addition, which is isomorphic to cyclic group order $n$. $\mathbb Z_n$ has elements $0 \cdots n-1$, and as we know, $n-1 + 1 = 0$, $n$ is a generator. We can see that, $\forall a \in \mathbb Z_n, \text{if} gcd(a, n) = 1$, which means, all coprimes of $(n)$, are generators, because if $gcd(a, n)= 1, \exists b, a + b = 1$, $1$ is always in $\left\langle a \right\rangle$, and 1 is a generator, therefore, $a$ is also the generator.

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, according to $g^{n}$.

19. The polar angle for i is π/2, so it generates a subgroup of 4 elements. $(e^{I\frac{\pi}{2}} = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}) = i)$

33. The Klein 4-group, D4

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.

Solution:

2. Let G be a cyclic group with a generator g∈G. Namely, we have G = ⟨g⟩ (every element in G is some power of g.). Let a and b be arbitrary elements in G. Then there exists n,m∈Z such that $a = g^n$ and $b = g^m$. It follows that

$ab = g^n g^m = g^{n+m} = g^{m+n} = g^m g^n = ba$

Hence we obtain ab = ba for arbitrary a,b∈G. Thus G is an abelian group.

3. For a cyclic group with order $\infty$, $\exists g, m, m > 0 , g^m = e = 0$. However, here, $g^m = m \cdot g, m \cdot g = 0$ only when $g = 0$, $0 + 0 = 0$, no chance to generate infinite group. Another case is $m=1$, then $g = \pm 1$. In this case, the cyclic group is isomorphic to $(\mathbb Z, +)$, which is countably infinite.

4. (a) $2: 2+2 = 4, 2+2+2=6 \cdot$ (b)$5: 5+5+5=15, 5+5+ \cdot + 5 = 5 \cdot 7 = 35$ (c) 3: same as above

5. $gcd(k_1,k_2,k_3 \cdot k_m)$