Month: April 2023

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202304131118 Problem 2.1.1

For each of the following parts, tell whether it specifies \theta as a permutation on \mathbf{X_5}=\{ 1,2,3,4,5\}. If not, explain how Definition 1 is not satisfied.

(a) \theta:1\mapsto 1,2\mapsto 3,3\mapsto 4,4\mapsto 5,5\mapsto 1.
(b) \theta:1\mapsto 5,2\mapsto 1,3\mapsto 2,4\mapsto 3,5\mapsto 4.
(c) \theta:1\mapsto 1,2\mapsto 2,3\mapsto 3,4\mapsto 4,5\mapsto 5.

Extracted from A. P. Hillman. (1999). Abstract Algebra A First Undergraduate Course.

Roughwork.

(a)

The arrow form

\theta:1\mapsto 1,2\mapsto 3,3\mapsto 4,4\mapsto 5,5\mapsto 1

may also be expressed as in function form

\begin{aligned} \theta (1) & = 1\\ \theta (2) & = 3\\ \theta (3) & = 4\\ \theta (4) & = 5\\ \theta (5) & = 1\\ \end{aligned}

For any finite non-empty set S, A(S) the set of all \textrm{1--1} transformations (mappings) of S onto S forms a group called permutation group and any element of A(S), i.e., a mapping from S onto itself, is called permutation.

From Wikibooks on Permutation groups

As is seen, function \theta here is \textrm{\scriptsize{NOT}} injective (i.e., \textrm{\scriptsize{NOT}} \textrm{1--1}), for \theta sends both 1 and 5 to 1. Hence it cannot specify a permutation.

(b)

The arrow form

\theta:1\mapsto 5,2\mapsto 1,3\mapsto 2,4\mapsto 3,5\mapsto 4

or two-line form

\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 5 & 1 & 2 & 3 & 4 \end{pmatrix}

or cyclic form

(15432)

also in diagram

is obviously a one-to-one function (/an injection). Thus function \theta here specifies a permutation.

(c)

In arrow form

\theta:1\mapsto 1,2\mapsto 2,3\mapsto 3,4\mapsto 4,5\mapsto 5

or matrix form

\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \end{pmatrix}

or cyclic form

()

is the identity permutation.

This problem is not to be attempted.

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202304131021 Problem 1.2.1

Which of the following integers are in the set 6\mathbb{Z} of integral multiples of 6?

(a) 10;
(b) -10;
(c) 12;
(d) -12;
(e) 4002;
(f) -4002;
(g) 4003;
(h) -4003.

Extracted from A. P. Hillman. (1999). Abstract Algebra A First Undergraduate Course.

Roughwork.

(a), (b)

\begin{aligned} \pm 10\div 6 & = \frac{\pm 10}{6} \\ & = \pm 1\frac{2}{3}\enspace\notin \mathbb{Z} \\ \end{aligned}

\begin{aligned} \because \quad & 6\nmid \pm 10 \\ \therefore \quad & \mathrm{\pm 10}\notin 6\mathbb{Z} \\ \end{aligned}

Because 6 is not a divisor (/factor) of \pm 10 in \mathbb{Z}, integers 10 and -10 are \textrm{\scriptsize{NOT}} in the set 6\mathbb{Z} of integral multiples of 6.

(c), (d)

\begin{aligned} \pm 12\div 6 & = \frac{\pm 12}{6} \\ & = \pm 2\enspace\in \mathbb{Z} \end{aligned}

\begin{aligned} \because\quad & 6\mid \pm12 \\ \therefore\quad & \mathrm{\pm 12}\in 6\mathbb{Z} \\ \end{aligned}

Since 6 is an integral divisor of \pm 12, integers 12 and -12 are in the set 6\mathbb{Z} of integral multiples of 6.

(e), (f)

\begin{aligned} \pm 4002\div 6 & = \frac{\pm 4002}{6} \\ & = \pm 667\enspace \in\mathbb{Z} \\ \end{aligned}

\begin{aligned} \because\quad & 6\mid \pm 4002 \\ \therefore\quad & \mathrm{\pm 4002}\in 6\mathbb{Z} \\ \end{aligned}

As \pm 4002 is a multiple of 6 in \mathbb{Z}, integers 4002 and -4002 are in the set 6\mathbb{Z} of integral multiples of 6.

(g), (h)

\begin{aligned} \pm 4003\div 6 & = \frac{\pm 4003}{6} \\ & = \pm 667\frac{1}{6}\enspace \notin\mathbb{Z} \\ \end{aligned}

\begin{aligned} \because\quad & 6\nmid \pm 4003 \\ \therefore\quad & \mathrm{\pm 4003} \notin 6\mathbb{Z} \\ \end{aligned}

Now that \pm 4003 is not an integral multiple of 6, integers 4003 and -4003 are \textrm{\scriptsize{NOT}} in the set 6\mathbb{Z} of integral multiples of 6.

This problem is not to be attempted.