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\begin{document}
	
Page 112: 1, 3, 39, 40

\ssc{Exercise 1}{

Let \afa \eql \mt{\left[
  \begin{array}{cccccc}
  1 & 2 & 3 & 4 & 5 & 6 \\
  2 & 1 & 3 & 5 & 4 & 6
  \end{array}
\right]}
and
\bta \eql \mt{\left[
  \begin{array}{cccccc}
  1 & 2 & 3 & 4 & 5 & 6 \\
  6 & 1 & 2 & 4 & 3 & 5
  \end{array}
\right]}

Compute each of the following:

\balist 
\item \uf{\afa}{-1}

\mt{\left[
  \begin{array}{cccccc}
  2 & 1 & 3 & 5 & 4 & 6 \\
  1 & 2 & 3 & 4 & 5 & 6
  \end{array}
\right]}

\item \bta\afa

\mt{\left[
  \begin{array}{cccccc}
  1 & 2 & 3 & 4 & 5 & 6 \\
  1 & 6 & 2 & 3 & 4 & 5
  \end{array}
\right]}

\item \afa\bta

\mt{\left[
  \begin{array}{cccccc}
  1 & 2 & 3 & 4 & 5 & 6 \\
  6 & 2 & 1 & 5 & 3 & 4
  \end{array}
\right]}

\elist

}

\ssc{Exercise 3}{

Write each of the following permutations as a product of disjoint cycles:

\balist
\item (1235)(413)

1 \lra 3 \lra 5 \lra 1

(15)

2 \lra 2 \lra 3

3 \lra 4 \lra 4

4 \lra 1 \lra 2

(234)

\textbf{(15)(234)}

\item (13256)(23)(46512)

1 \lra 2 \lra 3 \lra 2

2 \lra 4 \lra 4 \lra 4

3 \lra 3 \lra 2 \lra 5

4 \lra 6 \lra 6 \lra 1

5 \lra 1 \lra 1 \lra 3

6 \lra 5 \lra 5 \lra 6

\textbf{(124)(35)(6)}

\item (12)(13)(23)(142)

1 \lra 4 \lra 4 \lra 4 \lra 4

2 \lra 1 \lra 1 \lra 3 \lra 3

3 \lra 3 \lra 2 \lra 2 \lra 1

4 \lra 2 \lra 3 \lra 1 \lra 2

\textbf{(1423)}
\elist

}

\ssc{Exercise 39}{

In \uw{S}{4}, find a cyclic subgroup of order 4 and a noncyclic subgroup of order 4.

\textbf{\abk{(1234)} \eql \bk{e, (1234), \uf{(1234)}{2}, \uf{(1234)}{3}}}

\textbf{The set \abk{(1234)} under composition is a cyclic group of order 4 and a subgroup of \uw{S}{4}.}

For the noncyclic subgroup of order 4:

\textbf{Let S \eql \bk{e, (12), (34), (12)(34)}.}

}

\ssc{Exercise 40}{

In \uw{S}{3}, find elements \afa and \bta such that \av{ \afa} \eql 2, \av{ \bta} \eql 2, and \av{ \afa\bta} \eql 3.

\textbf{\uw{S}{3} \eql \bk{e, (23), (12), (132), (123), (13)}
}

\textbf{Let \afa \eql (12), \bta \eql (23)
}

\textbf{Notice: \afa \eql 213, \uf{\afa}{2} \eql 123, \bta \eql 132, \uf{\bta}{2} \eql 123.
}

\textbf{Notice also: (\afa\bta) \eql 312, \uf{(\afa\bta)}{2} \eql 231, and \uf{(\afa\bta)}{3} \eql 123.
}

\textbf{Hence, \afa \eql (12) and \bta \eql (23) is a solution.
}}

\end{document}