Math 360, Fall 2013, Assignment 4

No doubt many people feel that the inclusion of mathematics among the arts is unwarranted. The strongest objection is that mathematics has no emotional import. Of course this argument discounts the feelings of dislike and revulsion that mathematics induces....

- Morris Kline, Mathematics in Western Culture

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

1. Subgroup.
2. Improper subgroup.
3. Trivial subgroup.
4. Cyclic subgroup (generated by a particular element).
5. Cyclic group.
6. Generator (of a cyclic group).
7. Order (of a group).
8. Order (of an element of a group).
9. GCD (of two integers).
10. Relatively prime.

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

1. Theorem concerning integer division (Theorem 6.3 in the text).
2. Classification of cyclic groups (Theorem 6.10).
3. Classification of subgroups of \(\mathbb{Z}\) (Corollary 6.7).
4. Classification of subgroups of \(\mathbb{Z}_n\) (Theorem 6.14).

Do the following problems:

1. Section 5, problems 7, 9, 22, 23, and 42.
2. Section 6, problems 1, 3, 5, 9, 23, 27, and 30.

--------------------End of assignment--------------------

Questions:

1. Can someone please recap me again what does " aleph_0" mean again? I found it in the example for the order of a group?

aleph_0 (\(\aleph_0\), pronounced aleph-naught), is the cardinality of the positive integers (and the even integers, the odd integers, and the integers, the rational numbers, etc.) It's the largest cardinality where you can still count the elements of the underlying set. You can count the positive integers by going 1,2,3... etc., and you are guaranteed to eventually reach any arbitrary integer. You can do the same for any set of cardinality \(\aleph_0\). But for larger sets (the real numbers) you can't count the elements. There is no sequence of real numbers where you're guaranteed to eventually reach any arbitrary real number.

Book Problems:

1. 5.7

   Which of the sets\[\mathbb{R}, \mathbb{Q}^+, 7\mathbb{Z}, i\mathbb{R}, \pi\mathbb{Q}, \{\pi^n|n\in\mathbb{Z}\}\] are subgroups of the group \(\mathbb{C}^*\) of nonzero complex numbers under multiplication? \(i\mathbb{R}\) is the set of pure complex numbers, and \(\pi\mathbb{Q}\) is the set of rational multiples of pi.

2. 5.9

   Is the set of diagonal \(n\times n\) matrices with no zeros on the diagonal a subgroup of \(GL(n,\mathbb{R})\), which is the set of invertible \(n\times n\) matrices with real number entries?

3. 5.22

   Describe the elements of the set generated by:

   \(\left [\begin{array}{cc}0 & -1\\ -1 & 0\end{array}\right ]\)

   (as a subgroup of \(GL(n,\mathbb{R})\).) (So the group operation is matrix multiplication.)

4. 5.23

   Do the same thing as for 5.22, but with the matrix:

   \(\left [ \begin{array}{cc} 1 & 1\\ 0 & 1\end{array} \right ]\)

5. 5.42

   Let \(\phi : G \rightarrow G^{\prime}\) be a group isomorphism. Prove that if \(G\) is cyclic, then \(G^{\prime}\) is cyclic.

6. 6.1

   Find the quotient and remainder of \(42/9\).

7. 6.3

   Find the quotient and remainder of \(-50/8\).

8. 6.5

   Find the greatest common divisor of 32 and 24.

9. 6.9

   Find the number of generators of a cylic group with order 8.

10. 6.23

    Find all the subgroups of \(\mathbb{Z}_36\), and draw the subgroup diagram.

11. 6.27

    Find all the orders of the subgroups of \(\mathbb{Z}_{12}\).

12. 6.30

    Correct the definition of the italicized term:

    An element \(a\) of a group \(G\) has order \(n\in \mathbb{Z}^{+}\) if and only if \(a^n=e\).

Solutions:

Definitions

1. Subgroup.

   Definition:

   Let \(H\) be a subset of a group \((G,*)\) \(H\) is also a subgroup of \(G\) if it is closed under \(*\), and it is a group under \(*\).

   Example:

   The integers under addition are a subgroup of the real numbers under addition.

   Non-Example:

   The positive integers are not a subgroup of the integers under addition, because the positive integers lack an identity and inverses.

2. Improper Subgroup.

   Definition:

   If we have a group \(G\), \(G\) is the improper subgroup of itself.

   Example:

   The integers with addition are the improper subgroup of the integers with addition.

   Non-Example:

   The even integers with addition are not the improper subgroup of the integers with additon.

3. Trivial Subgroup.

   Definition:

   If we have a group \(G\), then the trivial subgroup of \(G\) is the subgroup consisting of only the identity element.

   Example:

   The group \((0,+)\) is the trivial subgroup of the integers with addition.

   Non-Example:

   The even integers are not the trivial subgroup of the integers with addition.

4. Cyclic Subgroup.

   Definition:

   The cyclic subgroup of a group \(G\) generated by \(a\in G\) is the subgroup \(< a>\) of \(G\) such that \(< a > = \{a^n | n\ in \mathbb{Z}\}\)

   Example:

   The even integers are a cyclic subgroup of the integers, generated by 2 (or -2).

   Non-Example:

5. Generator of a Cyclic Group.

   Definition:

   A generator of a group \(G\) is an element \(a\in G\) such that every element of \(G\) is a power of \(a\), i.e. \(\forall b\in G: \exists n\in \mathbb{Z}: b = a^n\)

   Example:

   1 is a generator of the integers under addition.

   Non-Example:

   2 is not a generator of the integers under addition.

6. Order of a Group.

   Definition:

   The order of a group is just the cardinality of the underlying set.

   Example:

   The order of the integers is \(\aleph_0\).

   Non-Example:

   The order of \(\mathbb{Z}/0\) is not 0.

7. GCD of Two Integers.

   Definition:

   The GCD of two integers is the greatest common divisor of the integers. For integers \(a,b\) their GCD is the greatest integer \(g\) such that \(a=ng\) and \(b=mg\) (i.e. \(g\) is the biggest integer that goes into both of the other integers evenly.

   Example:

   The GCD of 10 and 15 is 5.

   Non-Example:

   The GCD of 10 and 16 is not 5.

8. Relatively Prime.

   Definition:

   Two integers are relatively prime if their greatest common divisor is 1.

   Example:

   6 and 55 are relatively prime.

   Non-Example:

   6 and 9 are not relatively prime.

Theorems

1. Theorem Concerning Integer Division

   For any integers \(a\) and \(b\), there are integers \(q\) and \(r\), such that \(0 \leq r < b\), and \(a=bq+r\).

2. Classification of Cyclic Groups

   Every cyclic group is isomorphic either to \(\mathbb{Z}\), or two \(\mathbb{Z}_n\). Specifically, infinite cyclic groups are isomorphic to \(\mathbb{Z}\), and groups of order \(n\) are isomorphic to \(\mathbb{Z}_n\).

3. Classification of Subgroups of \(\mathbb{Z}\)

   The subgroups of \(\mathbb{Z}\) under addition are the subgroups \(n\mathbb{Z}\), i.e. the subgroups of \(\mathbb{Z}\) are the cyclic subgroups generated by the elements of \(\mathbb{Z}\). (So the even integers, the multiples of 3, and 4, and 5, etc.)

4. Classification of Subgroups of \(\mathbb{Z}_n\)

   Take a cyclic group \(G\) with generator <matha</math>. This means that every element of \(G\) is a "power" of \(a\), i.e. \(\forall g\in G: \exists s \in \mathbb{Z}: g = a^s\). Furthermore, \(g\) generates a cyclic subgroup of \(G\) (possibly a trivial or improper subgroup, but a subgroup). This subgroup, \(<g>\), contains \(n/gcd(n,s))\) elements, and is therefore isomorphic to \(\mathbb{Z}_{n/gcd(n,s)}\). So for the cyclic group \(\mathbb{Z}_6 = <1>\), the subgroup generated by \(2=2*1 = 1^2\) is of order \(6/gcd(6,2) = 6/2 = 3\) elements, and is isomorphic to \(\mathbb{Z}_3\). We checked this in class\[<2> = \{1,2,4\}\].

Book Problem Solutions

1. 5.7

   \(\mathbb{R}\) is not a subgroup - it is not a group at all, because 0 has no multiplicative inverse. \(\mathbb{Q}^{+}\) is a subgroup, because all the positive rational numbers have inverses, 1 is an identity, and multiplication is associative. \(7\mathbb{Z}\) is not a subgroup - this is all the integer multiples of 7, which don't have inverses or an identity under multiplication. \(i\mathbb{R}\) is a not a subgroup - it is not closed under multiplication. \(\pi\mathbb{Q}\) is not a subgroup - it contains no identity or inverses. \(\pi^n\) is a subgroup - \(\pi^0\) is the identity, \(\pi^-a\) is the inverse of \(\pi^a\), multiplication is already associative, and \(\pi^n * \pi^m = \pi^{m+n}\), so it's closed.

2. 5.9

   This is not a subgroup, as it is not closed. Consider:

   \(\left [ \begin{array}{cc} 1 & -1 \\ -1 & 1\end{array}\right ] \times \left [ \begin{array}{cc} 1 & 1\\ 1 & 1\end{array}\right ] = \left[\begin{array}{cc} 0 & 0\\ 0 & 0\end{array} \right ]\)

3. 5.22

   There are only two elements in this group:

   \(\left [ begin{array}{cc} 0 & -1\\ -1 & 0 \end{array} \right ],\left [ \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right ]\)

4. 5.23

5. 5.42

6. 6.1

7. 6.3

8. 6.5

9. 6.9

10. 6.23

11. 6.27

12. 6.30