I am doing abstract algebra problems, but unfortunately, the book we are using for the class is quite poor and leaves out lots of definitions and explanations, 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 finds another result. I am not sure what's right and what's 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 $\epsilon = (1 \ 2 ) ( 1 \ 2 ) = (2 \ 1) (1 \ 2) = (2)$.
Inthese problems, I'd rather simply calculate the effect of the product one by one on the individual elements on$ \{1,\ldots,n\}$
a) By definition, $(1\,2)$ maps $1\mapsto 2$, $2\mapsto 1$, and $k\mapsto k$ otherwise. Similarly, $(1\,2\,\ldots\,n)$ maps $k\mapsto k+1$ for $1\le k<n$ and $n\mapsto 1$, and therefore $(1\,2\,\ldots\,n)^{-1}$ maps $1\mapsto n$ and $k\mapsto k-1$ for $2\le k\le n$.
Hence (assuming $n\ge 3$) the composition $(1\,2\,\ldots\,n)(1\,2)(1\,2\,\ldots\,n)^{-1}$ maps $1\mapsto n\mapsto n\mapsto 1$, $2\mapsto 1\mapsto 2\mapsto 3$, $3\mapsto 2\mapsto 1\mapsto 2$, and for $4\le k\le n$, $k\mapsto k-1\mapsto k-1\mapsto k$. We conclude that the result is $(2\,3)$.