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Let $\sigma,\tau$ be two cycles in $S_n$ such that

  • $\sigma$ and $\tau$ have different length;
  • $\sigma$ and $\tau$ are not disjoint.

Then is it always true that $\sigma\circ \tau\neq \tau\circ\sigma$?

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  • $\begingroup$ Note that if we drop one of the conditions, then we can have that $\sigma\circ \tau=\tau\circ\sigma$. $\endgroup$ – Groups Sep 13 '15 at 3:30
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Yes, this is correct. Treating $\sigma$ and $\tau$ as permutations on a set, we can notice that the different lengths condition allows us to assume one of these permutations has a fixed point that the other does not. Assuming there exists some $s$ such that $\tau(s)=s$ but $\sigma(s)\neq s$ we can prove the following:

There exists an $s'$ such that $\tau(s')=s'$ but $\tau(\sigma(s'))\neq \sigma(s')$.

To prove this, we let $n$ be the least $n$ such that $\sigma^n(s)$ isn't a fixed point of $\tau$. Then, $s'=\sigma^{n-1}(s)$ satisfies the condition.

The rest is easy. We start with our second condition on $s'$: $$\tau(\sigma(s'))\neq \sigma(s')$$ and then substitute $s'=\tau(s')$ to get: $$\tau(\sigma(s'))\neq \sigma(\tau(s'))$$ meaning that $\tau\circ \sigma$ and $\sigma\circ \tau$ are different as they evaluate differently on $s'$.

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Yes. Two cycles that generate the same cyclic subgroup must have the same length, so it follows from the following result:

Suppose $\sigma$ and $\tau$ are cycles in $S_n$. Then $\sigma$ and $\tau$ commute iff one of the following holds:

  1. $\sigma$ and $\tau$ are disjoint, or
  2. $\langle\sigma\rangle = \langle\tau\rangle$.

Proof Suppose $\sigma$ is a cycle in $S_n$, say $\sigma = (a_1,\ldots,a_m)$, and $\tau$ commutes with $\sigma$. Then $$\sigma = \tau\sigma\tau^{-1} = (\tau(a_1),\ldots,\tau(a_m)),$$ so there is an integer $k$ such that $\tau(a_i) = a_{i+k} = \sigma^k(a_i)$ for $i = 1,\ldots,m$ where the subscripts are taken mod $m$. It follows that $\tau = \sigma^k\rho$ where $\rho$ fixes $a_i$ for $i=1,\ldots,m$.

Now suppose $\tau$ is also a cycle. Then one of $\sigma^k$ and $\rho$ must be trivial. If $\sigma^k$ is trivial then $\tau$ and $\sigma$ are disjoint cycles. If $\rho$ is trivial then $\sigma^k$ is a cycle, so $\gcd(k,m) = 1$ and $\langle\sigma\rangle = \langle\sigma^k\rangle = \langle\tau\rangle$.

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