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I know there are already a lot of answers to this question , but I just want to clear a doubt.

For the question above we consider $ \sigma \neq 1 \in S_n$ , now , there exists some $\alpha$ , $\beta$ $\in \{1,2,3,....,n\}$ such that $\sigma(\alpha) = \beta$.

Since $n \geq 3$ , there exists $\gamma \in \{1,2,3,....,n\}$ where $ \gamma \neq \alpha$ and $\gamma \neq \beta$.

So , we take $\tau = (\beta \gamma)$ and prove that $\sigma \tau \neq \tau \sigma$ , where $ \sigma = (\alpha \beta)$.

My question is , $\sigma$ and $\tau$ we've taken , wouldn't commute anyway , since they are not disjoint , so what's the point of taking $\tau = (\beta \gamma)$ ?

Can anyone help ?

User9523
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  • Your notation doesn't make sense. Is $\beta$ an element of ${1,\dots,n}$, or is it an element of $S_n$? Or, are there two different $\beta$s? – Ben Grossmann Nov 23 '15 at 14:08
  • $\beta$ is in ${1,2,3....,n}$. – User9523 Nov 23 '15 at 14:14
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    You cannot just say that $\sigma=(\alpha\beta)$. You have no idea about what $\sigma$ is, except that you've chosen $\alpha, \beta$ such that $\sigma=(\alpha\beta\cdots)(\cdots)\cdots$ in disjoint cycle notation. Also, there are permutations that are not disjoint, but permute anyway, such as $(1234)$ and $(13)(24)$, so you cannot use that argument. At least not alone. – Arthur Nov 23 '15 at 14:22
  • Oh, I see, that's cycle notation later. My mistake. – Ben Grossmann Nov 23 '15 at 14:28

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