Show that \(S_n \cong \left\langle (12), (123\cdots n)\right\rangle\) and also that \(S_n \cong \left\langle (12), (23\cdots n)\right\rangle\)
Solution
Let \[ \sigma=(1\ 2\ \cdots\ n),\qquad \tau=(2\ 3\ \cdots\ n). \] Since \(S_n\) is generated by the adjacent transpositions \((k\ k+1)\), \(1\le k\le n-1\), and also by the transpositions \((1\ j)\), \(2\le j\le n\), it suffices to show that the first pair generates every adjacent transposition and the second pair generates every \((1\ j)\).
1. The permutations \((12)\) and \(\sigma\) generate \(S_n\).
For \(0\le k\le n-2\), conjugation gives \[ \sigma^k(12)\sigma^{-k}=(k+1\ k+2). \] Thus \(\langle(12),\sigma\rangle\) contains every adjacent transposition \[ (12),(23),\ldots,(n-1\ n). \] Every permutation is a product of adjacent transpositions, so \[ \langle(12),\sigma\rangle=S_n. \]
2. The permutations \((12)\) and \(\tau\) generate \(S_n\).
For \(0\le k\le n-2\), \[ \tau^k(12)\tau^{-k}=(1\ \tau^k(2)). \] As \(k\) varies, \(\tau^k(2)\) runs through \(2,3,\ldots,n\). Hence \[ \langle(12),\tau\rangle \] contains every transposition \((1j)\) with \(2\le j\le n\).
These transpositions generate \(S_n\), since for distinct \(i,j>1\), \[ (ij)=(1i)(1j)(1i). \] Therefore \[ \langle(12),\tau\rangle=S_n. \]