Let $$σ = \begin{pmatrix}1 &2 &3 &4 &5 &6 &7 &8 &9 &10\\ 3 &4 &2 &8 &5 &7 &6 &10 &1& 9 \end{pmatrix} $$
Find $σ^{2345}$ and express it as $σ^k$, with $k$ minimal.
2026-04-02 14:20:29.1775139629
Basic Symmetric Group Question
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By straightforward calculation, we see that \begin{align*} 1&\rightarrow 3\rightarrow 2\rightarrow 4\rightarrow 8\rightarrow 10\rightarrow 9\rightarrow1&&(7\mbox{-cycle)}\\ 5&\rightarrow 5&&(1\mbox{-cycle)}\\ 6&\rightarrow 7\rightarrow 6&&(2\mbox{-cycle)} \end{align*} and thus $\sigma^{14}=\begin{pmatrix}1&2&3&4&5&6&7&8&9&10\\ 1&2&3&4&5&6&7&8&9&10\end{pmatrix}=\iota$ (identity symbol). Hence we conclude that $$\sigma^{2345}=\sigma^7\circ(\sigma^{14})^{167} =\sigma^7\circ\iota^{167} =\sigma^7 =\begin{pmatrix}1&2&3&4&5&6&7&8&9&10\\ 1&2&3&4&5&7&6&8&9&10\end{pmatrix}$$ and $k=7$.