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 $\sigma\circ \tau\neq \tau\circ\sigma$?

  • $\begingroup$ 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

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 $s'$.


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 $\langle\sigma\rangle = \langle\sigma^k\rangle = \langle\tau\rangle$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.