I'm trying to prove that in a symmetric group two disjoint cycles commute. But I suspect that something is not right about my proof (a sense of vagueness). Some hints would be appreciated.

Here's my proof:

Let $\sigma=(s_1 s_2 ... s_n)$, $\tau=(t_1 t_2 ... t_m)$ for some integers $m, n$. That is, $s_i \neq t_j$ for any $i \in [1, n]$, $j \in [1, m]$. Now, $\sigma\tau=(s_1 ... s_n)(t_1 ... t_m)$ is a cycle of disjoint permutations, which cannot be represented by any other disjoint permutations. Thus $\sigma\tau=\tau\sigma$.


Your proof would be something like

Let $s_i$ be in $\{s_1,s_2,\ldots,s_n\}$ and $t_j$ be in $\{t_1,t_2,\ldots,t_m\}$. Then $\sigma\circ\tau(s_i)=\sigma(\tau(s_i))=\sigma(s_i)=s_{i+1}$ and $\tau(\sigma(s_i))=\tau(s_{i+1})=s_{i+1}$. Similarly for $t_j$. Also note both functions have the same domain. Thus the two functions are the same.

  • 2
    $\begingroup$ I know it's been almost two years, but I would also add that for $s\in\operatorname{dom}(\sigma)\setminus(S\cup T)$ the equality holds too, where $S=\{s_1,\dots,s_n\}$ and $T=\{t_1,\dots,t_m\}$. $\endgroup$ – Sha Vuklia Jun 26 '17 at 14:56

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