I am doin abstract algebra problems, but unfortunately the book we are sing for the class is quite poor and leaves out lots of definitions and explainations, so I am not even sure if I completely understand the “rules of the game.” Only when I thought I started getting it, i realised that trying my generalization in partial cases I get wrong results; also wolfram alpha find another result. I am not sure whats write and whats wrong now.

Here is my problem and attempted solution:

(a) Find $ (1 \ 2 \dots n)(1 \ 2) (1 \ 2 \dots n)^{-1}$ . \end{problem}

\textbf{Solution.}

\begin{eqnarray*} \nonumber (1 \ 2 \dots n)(1 \ 2) (1 \ 2 \dots n)^{-1} &=&(1 \ 2 \ \dots n)(1 \ 2) (n \dots 2 \ 1) \ &=& (1 \ 3 \ 4 \ \dots [n-1] \ n)(1 \ 2)(1 \ 2) (n \dots 2 \ 1) \ &=& (1 \ 3 \ 4 \ \dots [n-1] \ n) (n \ [n-1]\dots 4 \ 3 \ 2 \ 1) \ &=& (1 \ 3){(3 \ 4) \dots ([n-1] \ n) (n \ [n-1])\dots (4 \ 3)}(3 \ 2)(2 \ 1) \ \text{[that nice chain-cancellation]}&=& (1 \ 3)\cancel{(3 \ 4)} \cancel{\dots} \cancel{([n-1] \ n)}\cancel{ (n \ [n-1])}\cancel{\dots}\cancel{ (4 \ 3)}(3 \ 2)(2 \ 1) \ &=& (1 \ 3)(3 \ 2)(2 \ 1) \ &=& (1 \ 2\ 3)(2 \ 1) \ &=& (1 \ 3)(1 \ 2)(2 \ 1) \ &=& (1 \ 3) \end{eqnarray*} \vspace{0.3cm}

(b) Find $ (1 \ 2 \dots n)^2(1 \ 2) (1 \ 2 \dots n)^{-2}$ .

\begin{eqnarray*} \nonumber (1 \ 2 \dots n)^2(1 \ 2) (1 \ 2 \dots n)^{-2} &=&(1 \ 2 \ \dots n)(1 \ 2 \ \dots n)(1 \ 2) (n \dots 2 \ 1)(n \dots 2 \ 1) \ \text{[substituting from part (a)]} &=&(1 \ 2 \ \dots n)\bigg[(1 \ 2 \ \dots n)(1 \ 2) (n \dots 2 \ 1)\bigg](n \dots 2 \ 1) \ &=& (1 \ 2 \ \dots n ) \bigg[(1 \ 3)\bigg] (n \dots 2 \ 1) \ &=& (1 \ 2 \ \dots n ) (1 \ 3) (n \dots 2 \ 1) \ &=& (1 \ 4 \ \dots n ) (1 \ 2 \ 3)(1 \ 3) (n \dots 2 \ 1) \ &=& (1 \ 4 \ \dots n ) ( 2 \ 3 \ 1)(1 \ 3) (n \dots 2 \ 1) \ &=& (1 \ 4 \ \dots n ) ( 2 \ 3)(3 \ 1)(1 \ 3) (n \dots 2 \ 1) \ &=& (1 \ 4 \ \dots n ) ( 2 \ 3) (n \dots 2 \ 1) \ \text{[since the these cycles are disjoint]}&=& ( 2 \ 3)(1 \ 4 \ \dots n ) (n \dots 2 \ 1) \ &=& ( 2 \ 3)(1 \ 4)(4 \ 5) \dots ([n-1] \ n ) (n \ [n-1]) \dots(5 \ 4)(4 \ 3)(3 \ 2)( 2 \ 1) \ \text{[again, chain-cancellation]}&=& ( 2 \ 3)(1 \ 4)\cancel{(4 \ 5)}\cancel{ \dots}\cancel{ ([n-1] \ n )}\cancel{ (n \ [n-1])}\cancel{ \dots}\cancel{(5 \ 4)}(4 \ 3)(3 \ 2)( 2 \ 1) \ &=& ( 2 \ 3)(1 \ 4)(4 \ 3)(3 \ 2)( 2 \ 1) \ &=& ( 2 \ 3)(1 \ 4)(4 \ 3 \ 2 \ 1) \ &=& ( 2 \ 3)(1 \ 4)(4 \ 1)(4 \ 3 \ 2 ) \ &=& ( 2 \ 3)(4 \ 3 \ 2 ) \ &=& ( 2 \ 3)( 3 \ 2 \ 4) \ &=& ( 2 \ 3)( 3 \ 2)(2 \ 4) \ &=& (2 \ 4) \ \end{eqnarray*}

\clearpage

(c) Explain how to obtain $ (k \ k+1)$ from $ (1 \ 2 \dots n)$ and $ (1 \ 2)$ , for $ 1 \leq k < n$ .

\textbf{Solution.}

(d) Find $ (1 \ 2 )(2 \ 3)(1 \ 2)$ .

\textbf{Solution.}

\vspace{0.3cm}

\begin{equation} \nonumber (1 \ 2)(2 \ 3)(1 \ 2) = (1 \ 2 \ 3)(1 \ 2) = (1 \ 3)(1 \ 2) (1 \ 2) = (1 \ 3) \end{equation}

(e) Explain how to obtain any two cycle $ (j \ j+1)$ in $ S_n$ .

\textbf{Solution.}

\clearpage

(f) Use parts (a) to (e) and Lemma 3.7.1 to prove Lemma 3.7.5.

\textbf{Solution.}

Lemma 3.7.1 is given below.

\textbf{Lemma 3.7.1.} For $ n>1$ , $ (1 \ 2)$ and $ (1 \ 2 \dots n)$ generate $ S_n$ .

\begin{proof} By induction.

As in Lemma 3.7.1, we first consider the base case, i.e. $ n = 2$ , then we have $ S_2$ , since the identity $ \epsilon = (1 \ 2) (1 \ 2) = (1 \ 2) (2 \ 1) = (1)$ and $ \epsilon = (1 \ 2 ) ( 1 \ 2 ) = (2 \ 1) (1 \ 2) = (2)$ .