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 $\sigma\tau=\tau\sigma$.
Best Answer
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.