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I'm trying to prove that two $\tau$-cycles commute provided that these cycles are disjoints. Hungerford in his book says the following remark about this fact:

Intuitively clearly this is true, but how can we prove this formally?

Thanks in advance

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    $\begingroup$ Probably induction on the length of the two cycles. Painful and pointless. $\endgroup$ – Frank Apr 2 '14 at 21:19
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    $\begingroup$ @Frank maybe this can be interesting for a beginner like me. $\endgroup$ – user42912 Apr 2 '14 at 21:27
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$\sigma\tau=\tau\sigma$ because they are the same function -- namely $\sigma\tau(i)=\tau\sigma(i)$ for all $i\in\{1,\ldots,n\}$.

Proof. First assume $\sigma(i)\ne i$. Then $\tau(i)=i$ by definition because $\sigma$ and $\tau$ are disjoint, and therefore $\sigma(\tau(i))=\sigma(i)$. On the other hand, because permutations are injective, $\sigma(i)\ne i$ means that $\sigma(\sigma(i))\ne\sigma(i)$, so $\tau(\sigma(i))=\sigma(i)$, again because $\sigma$ and $\tau$ are disjoint. Since $\sigma\tau(i)$ and $\tau\sigma(i)$ both equal $\sigma(i)$, they are equal.

Next assume $\sigma(i)=i$. Then it may be that $\tau(i)\ne i$, in which case proceed as in the previous case with $\tau$ and $\sigma$ interchanged.

Finally if $\sigma(i)=i$ and $\tau(i)=i$, then obviously $\sigma\tau(i)=i=\tau\sigma(i)$.

Since there's always one of these cases that holds, and $\sigma\tau(i)=\tau\sigma(i)$ in each of them, it holds always.

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    $\begingroup$ Excellent answer, thank you very much. $\endgroup$ – user42912 Apr 3 '14 at 1:09
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You can probably do this more formally by considering the permutations as elements which act on a set, namely the set of letters $\{1, \ldots, n\}$. In such a case, a cycle cyclically permutes a sub-collection of letters $\{\ell_1, \ldots, \ell_r\}$, and the fact that two cycles commute tell us that these sets which are permuted are disjoint. It should be then clear that the order in which we perform this permutation is irrelevant.

I should note that it is important that the representation of the group elements as permutations of letters is a faithful representation, otherwise this doesn't work. But this is certainly the case here.

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    $\begingroup$ "consider permutations as elements which act on a set"? This action is the very definition of permutation (and hence faithfulness is indeed clear). $\endgroup$ – Hagen von Eitzen Apr 2 '14 at 21:42
  • $\begingroup$ I know that, but I just thought that I would spell that out, since otherwise what I was saying isn't true. $\endgroup$ – Simon Rose Apr 3 '14 at 0:35

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