Permutation Groups

Find \((a_1, a_2, \cdots , a_n)^{-1}\).

\(Solution.\) \((a_n, a_{n - 1}, \cdots , a_1)\). \(\blacksquare\)

Prove that in \(A_n\) with \(n \geq 3\), any permutation is a product of cycles of length \(3\).

\(Proof.\) \(A_n\) consists of even Permutation, so if any product of 2 transpositions equal to product of cycles of length \(3\), then any permutation is a product of cycles of length \(3\).

For \(i \not = j \not = s \not = t\), \((ij)(ij) = 1\), \((ij)(js) = (ijs)\), \((ij)(st) = (ijs)(jst)\), thus any type of product of 2 transpositions can be represented by cycles of length \(3\). \(\blacksquare\)

Let \(\alpha \in S_n\) for \(n \geq 3\). If \(\alpha\beta = \beta\alpha\) for all \(\beta \in S_n\), prove that \(\alpha\) must be the identity permutation; hence, the center of \(S_n\) is the trivial subgroup.

\(Proof.\) Suppose \(\sigma = (1\ 2\ 3 \cdots k)\) is a cycle of \(\alpha\), it's obvious that \(C_{S_k}(\sigma) = \langle \sigma \rangle\), but \(|\langle \sigma \rangle| = k < |S_{k}| = k!\), thus \(\exists \beta \in S_k\) and \(\alpha\beta \not = \beta\alpha\). Hence the center of \(S_n\) must be the trivial subgroup. \(\blacksquare\)